# Bubbles in discrete time models

Martin Herdegen  University of Warwick, United Kindom
<EMAIL_ADDRESS>and  Dörte Kreher  Humboldt-Universität zu Berlin,
Germany<EMAIL_ADDRESS>
###### Abstract.

We introduce a new definition of bubbles in discrete-time models based on the
discounted stock price losing mass at some finite drawdown under an equivalent
martingale measure. We provide equivalent probabilistic characterisations of
this definition and give examples of discrete-time martingales that are
bubbles and those that are not. In the Markovian case, we provide sufficient
analytic conditions for the presence of bubbles. We also show that the
existence of bubbles is directly linked to the existence of a non-trivial
solution to a linear Volterra integral equation of the second kind involving
the Markov kernel. Finally, we show that our definition of bubbles in discrete
time is consistent with the strict local martingale definition of bubbles in
continuous time in the sense that a properly discretised strict local
martingale in continuous time is a bubble in discrete time.

###### Key words and phrases:

Bubble, strict local martingale, Markov martingale, Volterra integral equation

###### 2010 Mathematics Subject Classification:

60G42, 60J10, 91G99, 45D05

We thank Johannes Ruf as well as Paolo Guasoni, the co-editor, and two
anonymous referees for helpful comments on previous versions.

Over the last 20 years, a mathematical theory of bubbles for continuous time
models has been developed based on the concept of strict local martingales,
cf. the seminal papers by Loewenstein and Willard [12], Cox and Hobson [4],
Jarrrow, Protter and Shimbo [9, 10] as well as the survey article by Protter
[17] and the references therein. In economic terms, an asset price bubble
exists if the fundamental value of the asset deviates from its current price.
If the fundamental value is understood to be the expectation of the
(discounted) price process $S=(S_{t})_{t\geq 0}$ under the equivalent local
martingale measure $\mathbb{P}$, then the asset $S$ has a bubble if
$\mathbb{E}^{\mathbb{P}}[S_{T}]<\mathbb{E}^{\mathbb{P}}[S_{0}]$ for some fixed
time $T>0$, i.e., if $S$ is a strict local $\mathbb{P}$-martingale, that is a
local $\mathbb{P}$-martingale that fails to be a $\mathbb{P}$-martingale.

While many implications and extensions of this definition have been discussed
in the literature (see e.g. Ekström and Tysk [6], Bayraktar, Kardaras and Xing
[1], Biagini, Föllmer and Nedelcu [2], Herdegen and Schweizer [7]), the strict
local martingale definition of bubbles has no direct analogue in discrete time
models. The reason for this is that a nonnegative local martingale
$S=(S_{k})_{k\in\mathbb{N}_{0}}$ in discrete time with $S_{0}\in L^{1}$ is
automatically a (true) martingale. Hence, a definition of bubbles based on
strict local martingales is void. Also trying to define a bubble in discrete
time as a martingale $S=(S_{k})_{k\in\mathbb{N}_{0}}$ that is not uniformly
integrable does not lead to a meaningful concept as this would imply that
virtually all relevant models such as the standard binomial model (considered
on an unbounded time horizon) are bubbles, which seems absurd.

Despite the above negative results, the goal of this paper is to introduce a
new definition of bubbles for discrete-time models on an unbounded time
horizon – keeping the standard assumption that the discounted stock price is a
martingale. This definition has to satisfy at least two conditions.

  * (I)

It should _split_ martingales that are not uniformly integrable into two
sufficiently rich classes: those that are bubbles and those that are not. In
particular, standard discrete-time models with i.i.d. returns like the
binomial model should not be bubbles.

  * (II)

It should be _consistent_ with the strict local martingale definition in
continuous times in the sense that a properly discretised strict local
martingale in continuous time is a bubble in discrete time, and conversely,
that a local martingale in continuous time is a strict local martingale if all
properly discretised martingales are bubbles in discrete time.

To the best of our knowledge, there has been no attempt in the extant
literature to extend the martingale theory of bubbles to discrete time. The
only slight exception is Roch [18] who introduced the notion of asymptotic
asset price bubbles using the concept of weakly convergent discrete time
models (“large financial market”). More precisely, he showed that even if the
price process is a martingale in a sequence of weakly convergent discrete time
models, it can have properties similar to a bubble in that the fundamental
value in the asymptotic market can be lower than the current price in the
asymptotic market. In contrast to [18], our approach is non-asymptotic.

To motivate our definition of a bubble in a discrete time model, consider a
non-sophisticated investor who follows a simple buy-and-hold strategy to
invest into an asset with (discounted) price process
$S=(S_{k})_{k\in\mathbb{N}_{0}}$. The investor buys the asset at time $k=0$
and hopes that it will rise and rise. When the (discounted) asset price drops
for the first time, the investor fears to lose money and sells the asset.
Denoting by $\tau_{1}:=\inf\\{j>0:\ S_{j}<S_{j-1}\\}$ the time of the first
drawdown of the asset, the fundamental value of $S$ at time $0$ (viewed with
regard to the first drawdown) is $\mathbb{E}^{\mathbb{P}}[S_{\tau_{1}}]$,
where $\mathbb{P}$ denotes an equivalent martingale measure. As the process
$S$ is a nonnegative supermartingale, we always have
$\mathbb{E}^{\mathbb{P}}[S_{\tau_{1}}]\leq S_{0}$. If
$\mathbb{E}^{\mathbb{P}}[S_{\tau_{1}}]<S_{0}$, the fundamental value of the
asset (viewed with regard to the first drawdown) is lower than its initial
price and hence $S$ might be considered a bubble. Indeed, if the market is
complete, the predictable representation theorem implies that a sophisticated
investor might choose a dynamic trading strategy $\vartheta$ whose
(discounted) value process
$V(\vartheta)=(V_{k}(\vartheta))_{k\in\mathbb{N}_{0}}$ satisfies
$V_{0}(\vartheta)=\mathbb{E}^{\mathbb{P}}[S_{\tau_{1}}]<S_{0}$ and
$V_{\tau}(\vartheta)=S_{\tau}$.

Of course, the requirement that $S$ loses mass at the _first_ drawdown is
somewhat arbitrary and unrealistic. For this reason, our precise definition of
a bubble in discrete time is more general and only requires $S$ to lose mass
at the $k$-th drawdown for some $k\in\mathbb{N}$. While this definition is
very simple, it leads to a rich theory.

In Section 1, we provide several equivalent probabilistic characterisations
for a nonnegative discrete-time martingale to be a bubble. We also provide
necessary and sufficient characterisations for a discrete-time martingale with
independent increments to have a bubble. In particular, we show that the
standard binomial model does not have a bubble, which implies that Condition
(I) above is satisfied.

In Sections 2, we look at the special case that $S$ is a Markov martingale. We
provide characterisations for the presence or absence of bubbles, depending on
the _probability_ of going down, $a(x)=\mathbb{P}_{x}[S_{1}<x]$, and the
_relative recovery_ when going down,
$b(x)=\mathbb{E}_{x}[\tfrac{S_{1}}{x}\mathbf{1}_{\\{S_{1}<x\\}}]$. Loosely
speaking, it turns out that $S$ is a bubble if and only if $b(x)$ converges to
$0$ fast enough as $x\to\infty$. To make this precise, however, is quite
involved. While we are able to give sufficient conditions in the general case,
we provide necessary and sufficient conditions in the case of complete
markets.

In Section 3, we continue our study of Markov martingales by looking more
closely at the underlying Markov kernel. We show that the existence of bubbles
for $S$ is directly linked to the existence of a non-trivial nonnegative
solution to a linear Volterra integral equation of the second kind involving
the Markov kernel. Among others, this allows us to give some additional
sufficient conditions for the existence of bubbles that cannot be covered with
the results from Section 2.

Finally, in Section 4, we discuss how our definition of a bubble in discrete
time relates to the strict local martingale definition in continuous time. We
show that when discretising a positive continuous strict local martingale
along sequences of stopping times in a certain somewhat canonical class, one
obtains a bubble in discrete time. Conversely, we show that a positive
continuous local martingale is a strict local martingale if for all localising
sequences in the same class, the corresponding discretised martingales are
bubbles. This shows that Condition (II) above is also satisfied.111As pointed
out by one referee, our concept of a discrete-time bubble has a clear economic
foundation which the strict local martingale definition of bubbles in
continuous time maybe lacks – at least at a first glance. Hence, by showing
that appropriately discretised strict local martingales yield discrete-time
bubbles, we provide additional and new support to modelling continuous-time
bubbles by strict local martingales. To prove these discretisation results we
rely on the deep change of measure techniques first employed by Delbaen and
Schachermayer [5] and further developed by Pal and Protter [15], Kardaras,
Kreher, and Nikeghbali [11], and Perkowski and Ruf [16] that allow to turn the
inverse of a nonnegative strict local martingale into a true martingale under
a locally dominating probability measure. Some technical proofs of this
section are shifted to the appendix.

##  1\. Definition and characterisation of bubbles

In this section, we introduce our definition of a bubble in discrete time and
provide equivalent probabilistic characterisations of this concept.

######  Definition 1.1.

Let
$(\Omega,\mathcal{F},\mathbb{F}=(\mathcal{F}_{k})_{k\in\mathbb{N}_{0}},\mathbb{P})$
be a filtered probability space. A nonnegative
$(\mathbb{P},\mathbb{F})$-martingale $S=(S_{k})_{k\in\mathbb{N}_{0}}$ is
called a _bubble_ if

(1.1) $\mathbb{E}\left[S_{\tau_{k}}\right]<\mathbb{E}\left[S_{0}\right]$

for some $k\in\mathbb{N}$, where $\tau_{0}:=0$ and
$\tau_{k}:=\inf\\{j>\tau_{k-1}:\ S_{j}<S_{j-1}\\},\ k\in\mathbb{N}$, denotes
the $k$-th _drawdown_ of $S$. We also call $\mathbb{P}$ a _bubble measure_ for
$S$.

Some comments on the above definition are in order.

######  Remark 1.2.

(a) We include the possibility that $\mathbb{P}[\tau_{k}=\infty]>0$ with
positive probability. Since $S$ is a nonnegative martingale, it converges
$\mathbb{P}\text{-a.s.}$ to some integrable random variable $S_{\infty}$ by
Doob’s supermartingale convergence theorem. Hence, $S_{\tau_{k}}$ is well
defined in any case.

(b) If $S$ is a Markov process, $S$ will be a bubble if and only if
$\mathbb{E}[S_{\tau_{1}}]<\mathbb{E}[S_{0}]$, cf. Section 2 below. In general,
however, the above definition does not contain any redundancy, thinking for
example of dynamics with a change point.

(c) By the stopping theorem for uniformly integrable martingales, $S$ can only
be a bubble if it is not uniformly integrable. Example 1.9, however, shows
that not every martingale that fails to be uniformly integrable is a bubble.
This can be compared to the strict local martingale definition in continuous
time. Also, it emphasizes that the precise definition of the stopping time
$\tau_{k}$ in (1.1) is important and cannot naively be replaced by an
arbitrary stopping time.

(d) Our bubbles are strictly speaking _$\mathbb{P}$ -bubbles_ since the
definition depends on the choice of the (equivalent) martingale measure
$\mathbb{P}$. In incomplete markets, it is possible to have a _$\mathbb{P}$
-bubble_ under some equivalent martingale measure (EMM) $\mathbb{P}$ but not a
_$\tilde{\mathbb{P}}$ -bubble_ under a different EMM $\tilde{\mathbb{P}}$. In
this sense, our definition is not robust with respect to the choice of EMM.
Note that exactly the same issue arises in the strict local martingale
definition of bubbles in continuous time. Using similar ideas as in [7], one
could introduce the notion of a _strong bubble_ in discrete time. We leave the
details of this to future work.

(e) Definition 1.1 can be reformulated in the following way: For some
$k\in\mathbb{N}$, the stopped processes $S^{\tau_{k}}$ fails to be uniformly
integrable. This reformulation allows to apply results from the literature
that are proven for general càdlàg (local) martingales. For example, Hulley
and Ruf [8] provide necessary and sufficient characterizations for a càdlàg
(local) martingale to be uniformly integrable. However, the conditions given
in [8] require to calculate the distribution of the supremum of $S^{\tau_{k}}$
and the distribution of the jumps of $S^{\tau_{k}}$ at certain hitting times,
both of which are difficult to get hold of. For this reason, they are not very
useful in our discrete time setup.

We proceed to give a first simple example of a bubble in a complete market
model.

######  Example 1.3.

Define the process $S=(S_{k})_{k\in\mathbb{N}_{0}}$ and the probability
measure $\mathbb{P}$ recursively by $S_{0}=s_{0}>\tfrac{1}{2}$, and for
$k\in\mathbb{N}$,

$\displaystyle{\mathbb{P}}\left[S_{k}=2S_{k-1}-\frac{1}{2}\;\middle|\;S_{k-1}>\frac{1}{2}\right]={\mathbb{P}}\left[S_{k}=\frac{1}{2}\;\middle|\;S_{k-1}>\frac{1}{2}\right]=\frac{1}{2}$
$\displaystyle{\mathbb{P}}\left[S_{k}=\frac{1}{2}\;\middle|\;S_{k-1}=\frac{1}{2}\right]=1.$

Then $S$ is a $\mathbb{P}$-martingale for its natural filtration,
$\tau_{1}<\infty$ $\mathbb{P}\text{-a.s.}$, and
$\mathbb{E}[S_{\tau_{1}}]=\frac{1}{2}<s_{0}=\mathbb{E}[S_{0}]$, i.e., $S$ is a
bubble.

The following result provides two equivalent characterisations of bubbles. The
first one shows that a nonnegative martingale $S$ is a bubble if and only if
there exists a deterministic time $k\in\mathbb{N}_{0}$ such that $S$ loses
mass at the first drawdown after $k$. The second one provides a limit
characterisation. This second characterisation is particularly useful for
checking whether or not a martingale $S$ is a bubble.

######  Theorem 1.4.

Let $S=(S_{k})_{k\in\mathbb{N}_{0}}$ be a nonnegative martingale. For
$k\in\mathbb{N}_{0}$, define the stopping time $\tilde{\tau}_{k}$ by

$\tilde{\tau}_{k}:=\inf\\{j>k:S_{j}<S_{j-1}\\}.$

Then the following are equivalent:

  1. (a)

$S$ is a bubble.

  2. (b)

There exists $k\geq 0$ such that
$\mathbb{E}[S_{\tilde{\tau}_{k}}]<\mathbb{E}[S_{k}]$.

  3. (c)

There exists $k\geq 0$ such that
$\lim_{n\to\infty}\mathbb{E}[(S_{n}-S_{\infty})\mathbf{1}_{\\{S_{k}\leq
S_{k+1}\leq\dots\leq S_{n}\\}}]>0$.

###### Proof.

(a) $\Rightarrow$ (b). Suppose that (b) is not true, i.e.,
$\mathbb{E}[S_{\tilde{\tau}_{k}}]=\mathbb{E}[S_{k}]$ for all $k\geq 0$. Then
for all $k\geq 0$, the stopped process $S^{\tilde{\tau}_{k}}$ is a right-
closed supermartingale which does not lose mass at $\infty$ and hence is
uniformly integrable. We proceed by induction to show that
$\mathbb{E}[S_{\tau_{\ell}}]=\mathbb{E}[S_{0}]$ for all $\ell\geq 0$, whence
(a) is not true. The induction basis $\ell=0$ is trivial. For the induction
step, suppose that $\mathbb{E}[S_{\tau_{\ell-1}}]=\mathbb{E}[S_{0}]$ for some
$\ell\geq 1$. Then using the stopping theorem for the uniformly integrable
martingale $S^{\tilde{\tau}_{k}}$ in the third equality, we obtain

$\displaystyle\mathbb{E}[S_{\tau_{\ell}}]$
$\displaystyle=\sum_{k=\ell-1}^{\infty}\mathbb{E}\left[S_{\tau_{\ell}}\mathbf{1}_{\\{\tau_{\ell-1}=k\\}}\right]+\mathbb{E}[S_{\infty}\mathbf{1}_{\\{\tau_{\ell-1}=\infty\\}}]$
$\displaystyle=\sum_{k=\ell-1}^{\infty}\mathbb{E}\left[S_{\tilde{\tau}_{k}}\mathbf{1}_{\\{\tau_{\ell-1}=k\\}}\right]+\mathbb{E}[S_{\tau_{\ell-1}}\mathbf{1}_{\\{\tau_{\ell-1}=\infty\\}}]$
$\displaystyle=\sum_{k=\ell-1}^{\infty}\mathbb{E}\left[S_{\tau_{\ell-1}}\mathbf{1}_{\\{\tau_{\ell-1}=k\\}}\right]+\mathbb{E}[S_{\tau_{\ell-1}}\mathbf{1}_{\\{\tau_{\ell-1}=\infty\\}}]=\mathbb{E}[S_{\tau_{\ell-1}}]=\mathbb{E}[S_{0}].$

(b) $\Rightarrow$ (a). This follows from the fact that
$\mathbb{E}[S_{k}]=\mathbb{E}[S_{0}]$ by the martingale property of $S$
together with the fact that $\tilde{\tau}_{k}\leq\tau_{k+1}$.

(b) $\Leftrightarrow$ (c). The equivalence follows from the following
calculation, which uses the martingale property of $S$ in the third equality:

$\displaystyle\mathbb{E}\left[S_{\tilde{\tau}_{k}}\right]$
$\displaystyle=\lim_{n\to\infty}\sum_{\ell=k+1}^{n}\mathbb{E}\left[S_{\ell}\mathbf{1}_{\\{S_{k}\leq
S_{k+1}\leq\dots\leq
S_{\ell-1}>S_{\ell}\\}}\right]+\mathbb{E}\left[S_{\infty}\mathbf{1}_{\\{S_{k}\leq
S_{k+1}\leq\dots\\}}\right]$
$\displaystyle=\lim_{n\to\infty}\sum_{\ell=k+1}^{n}\left(\mathbb{E}\left[S_{\ell}\mathbf{1}_{\\{S_{k}\leq\dots\leq
S_{\ell-1}\\}}\right]-\mathbb{E}\left[S_{\ell}\mathbf{1}_{\\{S_{k}\leq\dots\leq
S_{\ell}\\}}\right]\right)$
$\displaystyle\qquad+\mathbb{E}\left[S_{\infty}\mathbf{1}_{\\{S_{k}\leq
S_{k+1}\leq\dots\\}}\right]$
$\displaystyle=\lim_{n\to\infty}\sum_{\ell=k+1}^{n}\left(\mathbb{E}\left[S_{\ell-1}\mathbf{1}_{\\{S_{k}\leq\dots\leq
S_{\ell-1}\\}}\right]-\mathbb{E}\left[S_{\ell}\mathbf{1}_{\\{S_{k}\leq\dots\leq
S_{\ell}\\}}\right]\right)$
$\displaystyle\qquad+\mathbb{E}\left[S_{\infty}\mathbf{1}_{\\{S_{k}\leq
S_{k+1}\leq\dots\\}}\right]$ (1.2)
$\displaystyle=\mathbb{E}\left[S_{k}\right]+\lim_{n\to\infty}\mathbb{E}[(S_{\infty}-S_{n})\mathbf{1}_{\\{S_{k}\leq
S_{k+1}\leq\dots\leq S_{n}\\}}].$

This finishes the proof. ∎

The characterisation (c) in Theorem 1.4 is generally the most useful one to
decide whether or not $S$ is a bubble. The following corollary strengthens
this characterisation. It shows directly that bounded martingales fail to be
bubbles. (Of course, this follows directly from the fact that a bounded
martingale is uniformly integrable.)

######  Corollary 1.5.

Let $S=(S_{k})_{k\in\mathbb{N}_{0}}$ be a nonnegative martingale. Then the
following are equivalent:

  1. (a)

$S$ is a bubble.

  2. (b)

There exist $x\geq 0$ and $k\geq 0$ such that

$\lim_{n\to\infty}\mathbb{E}[(S_{n}-S_{\infty})\mathbf{1}_{\\{x\leq S_{k}\leq
S_{k+1}\leq\dots\leq S_{n}\\}}]>0.$

  3. (c)

For all $x\geq 0$, there exists $k\geq 0$ such that

$\lim_{n\to\infty}\mathbb{E}[(S_{n}-S_{\infty})\mathbf{1}_{\\{x\leq S_{k}\leq
S_{k+1}\leq\dots\leq S_{n}\\}}]>0.$

###### Proof.

It is clear that (c) implies (b), and (b) a fortiori implies condition (c) of
Theorem 1.4, which by Theorem 1.4 gives (a). So it remains to prove (a)
$\Rightarrow$ (c). So fix $x\geq 0$. Since $S$ is a bubble, by Theorem 1.4(b)
and the martingale property of $S$, there exists $\ell\geq 0$ such that
$\mathbb{E}[S_{\tau^{\ell}_{1}}]<\mathbb{E}[S_{\ell}]=\mathbb{E}[S_{0}]$.

Define the stopping times $\sigma_{\ell,x}$ and $\tau^{\sigma_{\ell,x}}_{1}$
by

$\displaystyle\sigma_{\ell,x}:=\inf\\{k\geq\ell:S_{k}\geq
x\\}\quad\text{and}\quad\tau^{\sigma_{\ell,x}}_{1}:=\inf\\{j>\sigma_{\ell,x}:S_{j}<S_{j-1}\\}.$

Then $\sigma_{\ell,x}$ is the first hitting time of $[x,\infty)$ after $\ell$
and $\tau^{\sigma_{\ell,x}}_{1}$ the first drawdown of $S$ after
$\sigma_{\ell,x}$. Since $\sigma_{\ell,x}\geq\ell$, it follows that
$\tau^{\sigma_{\ell,x}}_{1}\geq\tau^{\ell}_{1}$.

By the definition of $\sigma_{\ell,x}$, the stopped process
$S^{\sigma_{\ell,x}}$ is uniformly integrable because

$S^{\sigma_{\ell,x}}_{k}\leq\max(S_{0},\ldots,S_{\ell},S_{\sigma_{\ell,x}},x)\in
L^{1}.$

This implies that $\mathbb{E}[S_{\sigma_{\ell,x}}]=\mathbb{E}[S_{0}]$. Since
$\tau^{\sigma_{\ell,x}}_{1}\geq\tau_{1}^{\ell}$, this in turn implies that

$\mathbb{E}\left[S_{\tau^{\sigma_{\ell,x}}_{1}}\right]\leq\mathbb{E}\left[S_{\tau^{\ell}_{1}}\right]<\mathbb{E}\left[S_{0}\right]=\mathbb{E}\left[S_{\sigma_{\ell,x}}\right].$

A similar calculation as in (1.2) shows that

$\mathbb{E}\big{[}S_{\tau^{\sigma_{\ell,x}}_{1}}\big{]}=\mathbb{E}\left[S_{\sigma_{\ell,x}}\right]+\lim_{n\to\infty}\mathbb{E}\Big{[}(S_{\infty}-S_{\sigma_{\ell,x}+n})\mathbf{1}_{\\{S_{\sigma_{\ell,x}}\leq\dots\leq
S_{\sigma_{\ell,x}+n}\\}}\mathbf{1}_{\\{\sigma_{\ell,x}<\infty\\}}\Big{]}.$

This together with the tower property of conditional expectations and
dominated convergence yields

$\mathbb{E}\left[\lim_{n\to\infty}\mathbb{E}\left[(S_{\sigma_{\ell,x}+n}-S_{\infty})\mathbf{1}_{\\{S_{\sigma_{\ell,x}}\leq
S_{\sigma_{\ell,x}+1}\leq\dots\leq
S_{\sigma_{\ell,x}+n}\\}}\,\middle|\,\mathcal{F}_{\sigma_{\ell,x}}\right]\mathbf{1}_{\\{\sigma_{\ell,x}<\infty\\}}\right]>0.$

We may deduce that there is $k\geq\ell$ such that

$\mathbb{E}\left[\lim_{n\to\infty}\mathbb{E}\left[(S_{k+n}-S_{\infty})\mathbf{1}_{\\{S_{k}\leq
S_{k+1}\leq\dots\leq
S_{k+n}\\}}\mathbf{1}_{\\{\sigma_{\ell,x}=k\\}}\,\middle|\,\mathcal{F}_{k}\right]\right]>0.$

Using that $S_{k}\geq x$ on $\\{\sigma_{\ell,x}=k\\}$, this implies that

$\mathbb{E}\left[\lim_{n\to\infty}\mathbb{E}\left[(S_{k+n}-S_{\infty})\mathbf{1}_{\\{x\leq
S_{k}\leq S_{k+1}\leq\dots\leq
S_{k+n}\\}}\,\middle|\,\mathcal{F}_{k}\right]\right]>0.$

Now dominated convergence and the tower property of conditional expectations
give (c). ∎

While characterisations (b) or (c) in Corollary 1.5 are an improvement of
Theorem 1.4(c), they still depend on $S_{\infty}$, of which we generally do
not have a good knowledge. The following corollary provides a mild condition
on $S$ under which the bubble behaviour of $S$ can be characterised without
involving $S_{\infty}$.

######  Proposition 1.6.

Let $S=(S_{k})_{k\in\mathbb{N}_{0}}$ be a nonnegative martingale and $x>0$.
Suppose that

(1.3) $\sum_{k=0}^{\infty}{\mathbb{P}}\left[S_{k}<x\text{ or
}S_{k}>S_{k+1}\;\middle|\;\mathcal{F}_{k}\right]=\infty\;\;\mathbb{P}\text{-a.s.}$

Then for each $k\in\mathbb{N}_{0}$,

(1.4) $\lim_{n\to\infty}\mathbb{E}\left[S_{\infty}\mathbf{1}_{\\{x\leq
S_{k}\leq S_{k+1}\leq\dots\leq S_{n}\\}}\right]=0.$

In particular, $S$ is a bubble if and only if there exists
$k\in\mathbb{N}_{0}$ such that

$\lim_{n\to\infty}\mathbb{E}[S_{n}\mathbf{1}_{\\{x\leq S_{k}\leq
S_{k+1}\leq\dots\leq S_{n}\\}}]>0.$

###### Proof.

The conditional Lemma of Borel–Cantelli implies that

${\mathbb{P}}\left[\liminf_{k\to\infty}\\{x\leq S_{k}\leq
S_{k+1}\\}\right]=1-{\mathbb{P}}\left[\limsup_{k\to\infty}\left(\\{S_{k}<x\\}\cup\\{S_{k}>S_{k+1}\\}\right)\right]=0.$

This implies a fortiori that $\\{x\leq S_{k}\leq S_{k+1}\leq\cdots\\}$ is a
$\mathbb{P}$-null set for each $k\in\mathbb{N}_{0}$. This gives (1.4). The
final claim now follows from Corollary 1.5. ∎

The following result gives simple necessary and sufficient conditions for a
nonnegative martingales with independent increments to be a bubble.

######  Theorem 1.7.

Let $(X_{k})_{k\in\mathbb{N}}$ be an independent sequence of nonnegative
random variables with $\mathbb{E}[X_{k}]=1$ for $k\in\mathbb{N}$. Define the
process $S=(S_{k})_{k\in\mathbb{N}_{0}}$ by $S_{k}=\prod_{\ell=1}^{k}X_{\ell}$
and the filtration $\mathbb{F}=(\mathcal{F}_{k})_{k\in\mathbb{N}_{0}}$ by
$\mathcal{F}_{k}:=\sigma(S_{0},\ldots,S_{k})$. Moreover, for $k\in\mathbb{N}$
set

$a_{k}:=\mathbb{P}[X_{k}<1]\in[0,1),\quad
b_{k}:=\mathbb{E}\left[X_{k}\mathbf{1}_{\\{X_{k}<1\\}}\right]\in[0,a_{k}].$

Then $S$ is a positive $(\mathbb{P},\mathbb{F})$-martingale. It is a bubble if
and only if

$\sum_{k=1}^{\infty}a_{k}=\infty\quad\text{and}\quad\sum_{k=1}^{\infty}b_{k}<\infty.$

###### Proof.

It is clear by construction that $S$ is a nonnegative
$(\mathbb{P},\mathbb{F})$-martingale.

First, we argue that if $\sum_{k=1}^{\infty}a_{k}<\infty$, then $S$ is
uniformly integrable and hence cannot be a bubble. To this end, note that

$\infty>\sum_{k=0}^{\infty}a_{k}=\sum_{k=0}^{\infty}\mathbb{P}[X_{k}<1]\geq\sum_{k=0}^{\infty}\mathbb{E}\big{[}(1-\sqrt{X_{k}})\mathbf{1}_{\\{X_{k}<1\\}}\big{]}\geq\sum_{k=0}^{\infty}\mathbb{E}\big{[}(1-\sqrt{X_{k}})\big{]}.$

By Kakutani’s theorem (see, e.g. [19, Theorem 14.12(v)]) this implies that $S$
is uniformly integrable.

Next, if $\sum_{k=1}^{\infty}a_{k}=\infty$, by independence of the $X_{k}$,

$\sum_{k=0}^{\infty}{\mathbb{P}}\left[S_{k}<1\text{ or
}S_{k}>S_{k+1}\;\middle|\;\mathcal{F}_{k}\right]\geq\sum_{k=0}^{\infty}{\mathbb{P}}\left[X_{k+1}<1\;\middle|\;\mathcal{F}_{k}\right]=\sum_{k=1}^{\infty}a_{k}=\infty.$

This together with Corollary 1.5 implies that $S$ is a bubble if and only if
there exists $k\geq 0$ such that

$\displaystyle\lim_{n\to\infty}\mathbb{E}[S_{n}\mathbf{1}_{\\{1\leq S_{k}\leq
S_{k+1}\leq\dots\leq S_{n}\\}}]$ $\displaystyle=\mathbb{P}[S_{k}\geq
1]\prod_{\ell=k+1}^{\infty}\mathbb{E}\left[X_{k}\mathbf{1}_{\\{X_{k}\geq
1\\}}\right]$
$\displaystyle=P[S_{k}\geq\varepsilon]\prod_{\ell=k+1}^{\infty}(1-b_{\ell})>0.$

Since $\mathbb{P}[S_{k}\geq 1]>0$ and $(1-b_{\ell})>0$ for all
$\ell\in\mathbb{N}$, this is equivalent to
$\prod_{\ell=1}^{\infty}(1-b_{\ell})>0$, which in turn is equivalent to
$\sum_{k=1}^{\infty}b_{k}<\infty$. ∎

We illustrate the above theorem by two examples. The first one gives bubble in
a time-dependent binomial-type model, where the downward jumps get more and
more severe.

######  Example 1.8.

Let $(X_{k})_{k\in\mathbb{N}}$ be a sequence of independent random variables
with $P[X_{k}=\frac{1}{k}]=\frac{1}{k}$ and
$P[X_{k}=1+\frac{1}{k}]=1-\frac{1}{k}$. Define the process
$S=(S_{k})_{k\in\mathbb{N}_{0}}$ by $S_{k}:=\prod_{\ell=1}^{k}X_{\ell}$ and
the filtration $\mathbb{F}=(\mathcal{F}_{k})_{k\in\mathbb{N}_{0}}$ by
${\mathcal{F}_{k}:=\sigma(S_{0},\ldots,S_{k})}$. Then $a_{k}=\frac{1}{k}$ and
$b_{k}=\frac{1}{k^{2}}$. Hence, $\sum_{k=1}^{\infty}a_{k}=\infty$ and
$\sum_{k=1}^{\infty}b_{k}<\infty$, whence $S$ is a bubble.

The second example shows that a martingale with i.i.d. returns is never a
bubble. In particular an standard binomial model is never a bubble, which
coincides with our intuition.

######  Example 1.9.

Let $(X_{k})_{k\in\mathbb{N}}$ be a sequence of i.i.d. random variables that
are nonnegative and satisfy $\mathbb{E}[X_{k}]=1$ and $\mathbb{P}[X_{k}\neq
1]>0$. Define the process $S=(S_{k})_{k\in\mathbb{N}_{0}}$ by
$S_{k}:=\prod_{\ell=1}^{k}X_{\ell}$ and the filtration
$\mathbb{F}=(\mathcal{F}_{k})_{k\in\mathbb{N}_{0}}$ by
${\mathcal{F}_{k}:=\sigma(S_{0},\ldots,S_{k})}$. Then $S$ is a martingale but
fails to be uniformly integrable because
$S_{n}=\exp\left(\sum_{k=1}^{n}\log(X_{k})\right)\stackrel{{\scriptstyle
a.s.}}{{\longrightarrow}}0$ by the strong law of large numbers since
$\mathbb{E}[\log(X_{k})]<1$. However, setting
$b_{k}:=\mathbb{E}[X_{k}\mathbf{1}_{\\{X_{k}<1\\}}]=b>0$, we obtain

$\sum_{k=1}^{\infty}b_{k}=\infty.$

Thus, $S$ fails to be a bubble.

##  2\. Characterization of bubble measures for Markov chains

Throughout this section, we suppose that $S=(S_{k})_{k\in\mathbb{N}_{0}}$ is a
positive Markov martingale with transition kernel
$K:(0,\infty)\times\mathcal{B}_{(0,\infty)}\to[0,\infty)$, starting from
$S_{0}=x>0$. Our goal is to determine under which conditions on the kernel
$K(x,\,\mathrm{d}y)$ the measure $\mathbb{P}_{x}$ is a bubble measure. To this
end, we define the functions $a,b:(0,\infty)\to[0,1)$ by

(2.1) $\displaystyle a(x)$
$\displaystyle:=\mathbb{P}_{x}\left[S_{1}<x\right]=\int_{[0,x)}K(x,\mathrm{d}y),$
(2.2) $\displaystyle b(x)$
$\displaystyle:=\mathbb{E}_{x}\left[\frac{S_{1}}{x}\mathbf{1}_{\\{S_{1}<x\\}}\right]=\int_{[0,x)}\frac{y}{x}K(x,\mathrm{d}y).$

Here, $a(x)$ denotes the _probability_ of a downward jump and $b(x)$ the
_relative recovery_ in case of a downward jump.

First, we show we show that $S$ cannot be a bubble unless the relative
recovery function $b$ converges to zero at infinity.

######  Proposition 2.1.

Assume that $\liminf_{x\to\infty}b(x)>0$. Then $S$ fails to be a bubble under
$\mathbb{P}_{x}$ for any $x\in(0,\infty)$.

###### Proof.

There exists $x_{0}>0$ and $\varepsilon\in(0,1)$ such that
$b(x)\geq\varepsilon$ for all $x\geq x_{0}$. Pick $x\in(0,\infty)$. By
Corollary 1.5 and using that $S_{\infty}\geq 0$, it suffices to show that for
each $k\in\mathbb{N}_{0}$,

$\lim_{n\to\infty}\mathbb{E}_{x}\left[S_{n}\mathbf{1}_{\\{x_{0}\leq S_{k}\leq
S_{k+1}\leq\cdots\leq S_{n}\\}}\right]=0.$

So let $k<n$. Then by the Markov property and the definition of $b$,

$\displaystyle\mathbb{E}_{x}\left[S_{n}\mathbf{1}_{\\{x_{0}\leq S_{k}\leq
S_{k+1}\leq\cdots\leq S_{n}\\}}\right]$
$\displaystyle\quad=\mathbb{E}_{x}\left[\mathbb{E}_{x}\left[\frac{S_{n}}{S_{n-1}}\mathbf{1}_{\\{S_{n}\geq
S_{n-1}\\}}\,\middle|\,\mathcal{F}_{n-1}\right]S_{n-1}\mathbf{1}_{\\{x_{0}\leq
S_{k}\leq S_{k+1}\leq\cdots\leq S_{n-1}\\}}\right]$
$\displaystyle\quad=\mathbb{E}_{x}\left[(1-b(S_{n-1}))S_{n-1}\mathbf{1}_{\\{x_{0}\leq
S_{k}\leq S_{k+1}\leq\cdots\leq S_{n-1}\\}}\right]$
$\displaystyle\quad\leq(1-\varepsilon)\mathbb{E}_{x}\left[S_{n-1}\mathbf{1}_{\\{x_{0}\leq
S_{k}\leq S_{k+1}\leq\cdots\leq S_{n-1}\\}}\right]$ (2.3)
$\displaystyle\quad\leq(1-\varepsilon)^{n-k}\mathbb{E}_{x}\big{[}S_{k}\mathbf{1}_{\\{x_{0}\leq
S_{k}\\}}\big{]}\leq(1-\varepsilon)^{n-k}x.$

Now the claim follows by letting $n\to\infty$. ∎

We proceed to formulate a mild condition on the function $a$, which allows us
to characterize the bubble behaviour of $S$ without involving $S_{\infty}$.

######  Assumption 2.2.

There exists $x_{a}>0$ such that $\inf_{x\in[x_{a},y]}a(x)>0$ for all
$y>x_{a}$.

######  Remark 2.3.

Assumption 2.2 is in particular fulfilled if $a$ is positive and lower semi-
continuous.

######  Proposition 2.4.

Suppose Assumption 2.2 is satisfied for some $x_{a}>0$. Let $x^{\prime}\geq
x_{a}$. Then for each $k\in\mathbb{N}_{0}$,

(2.4)
$\lim_{n\to\infty}\mathbb{E}_{x}\left[S_{\infty}\mathbf{1}_{\\{x^{\prime}\leq
S_{k}\leq S_{k+1}\leq\dots\leq S_{n}\\}}\right]=0.$

Moreover, $S$ is a bubble under $\mathbb{P}_{x}$ if and only if there exists
$k\in\mathbb{N}_{0}$ such that

$\lim_{n\to\infty}\mathbb{E}_{x}\left[S_{n}\mathbf{1}_{\\{x^{\prime}\leq
S_{k}\leq S_{k+1}\leq\cdots\leq S_{n}\\}}\right]>0.$

###### Proof.

By Proposition 1.6, it suffices to check that $\mathbb{P}_{x}\text{-a.s.}$

$\sum_{k=0}^{\infty}\mathbb{P}_{x}\left[S_{k}<x^{\prime}\text{ or
}S_{k}>S_{k+1}\,\middle|\,\mathcal{F}_{k}\right]=\sum_{k=0}^{\infty}(\mathbf{1}_{\\{S_{k}<x^{\prime}\\}}+a(S_{k})\mathbf{1}_{\\{S_{k}\geq
x^{\prime}\\}})=\infty.$

We now distinguish two cases: If
$\omega\in\limsup_{k\to\infty}\\{S_{k}<x^{\prime}\\}$, it follows that
$\sum_{k=0}^{\infty}\mathbf{1}_{\\{S_{k}(\omega)<x^{\prime}\\}}=\infty$. If
$\omega\in\liminf_{k\to\infty}\\{S_{k}\geq x^{\prime}\\}\cap\\{\sup_{k\geq
0}S_{k}<\infty\\}$, then
$\sum_{k=0}^{\infty}a(S_{k}(\omega))\mathbf{1}_{\\{S_{k}(\omega)\geq
x\\}})=\infty$ by the assumption on $a$. Since $\sup_{k\geq 0}S_{k}<\infty$
$\mathbb{P}_{x}\text{-a.s.}$ by Doob’s martingale convergence theorem, the
claim follows. ∎

We next aim to give a sufficient characterisation for $S$ being a bubble. To
this end, we need to slightly relax the definition of the function $b$ from
(2.2): For $\varepsilon>0$, define the function
$b_{\varepsilon}:(0,\infty)\to(0,1]$ by

$b_{\varepsilon}(x):=\mathbb{E}_{x}\left[\frac{S_{1}}{x}\mathbf{1}_{\\{S_{1}<x(1+\varepsilon)\\}}\right]=\int_{[0,x(1+\varepsilon))}\frac{y}{x}K(x,\mathrm{d}y).$

######  Theorem 2.5.

Suppose Assumption 2.2 is satisfied and there exists $\varepsilon>0$ and
$x_{b}>0$ such that the function $b_{\varepsilon}$ is nonincreasing for $x\geq
x_{b}$ and satisfies
$\int_{\log(x_{b})}^{\infty}b_{\varepsilon}(\exp(x))\,\mathrm{d}x<\infty$.
Then $S$ is a bubble under each $\mathbb{P}_{x}$, for which $S$ is not
$\mathbb{P}_{x}$-a.s. bounded.

###### Proof.

Suppose that $S$ is not $\mathbb{P}_{x}$-a.s. bounded. Let $x_{a}$ be the
constant in Assumption 2.2. We may assume without loss of generality that
$x_{b}\geq x_{a}$. By Proposition 2.4, it suffices to check that there is
$k\in\mathbb{N}_{0}$ such that

$\displaystyle\lim_{n\to\infty}\mathbb{E}_{x}\left[S_{n}\mathbf{1}_{\\{x_{b}\leq
S_{k}\leq S_{k+1}\leq\cdots\leq S_{k+n}\\}}\right]$
$\displaystyle\quad\geq\lim_{n\to\infty}\mathbb{E}_{x0}\bigg{[}S_{k}\mathbf{1}_{\\{S_{k}\geq
x_{b}\\}}\prod_{j=1}^{n}\frac{S_{k+j}}{S_{k+j-1}}\mathbf{1}_{\\{S_{k+j}\geq
S_{k+j-1}(1+\varepsilon)\\}}\bigg{]}>0.$

Since $S$ is not $\mathbb{P}_{x}$-a.s. bounded, there exists
$k\in\mathbb{N}_{0}$ with $\mathbb{E}_{x}[S_{k}\mathbf{1}_{\\{S_{k}\geq
x_{b}\\}}]>0$. Fix $n\in\mathbb{N}$. By the definition of $b_{\varepsilon}$ ,
we obtain for $j\in\\{1,\ldots n\\}$

$\displaystyle\mathbb{E}_{x}\bigg{[}\frac{S_{k+j}}{S_{k+j-1}}\mathbf{1}_{\\{S_{j+k-1}(1+\varepsilon)\leq
S_{j+k}\\}}\,\bigg{|}\,\mathcal{F}_{k+j-1}\bigg{]}=(1-b_{\varepsilon}(S_{k+j-1}))\;\;\mathbb{P}_{x}\text{-a.s.}$

This together with the tower property of conditional expectations and the fact
that $b_{\varepsilon}$ is nonincreasing for $x\geq x_{b}$ gives

$\displaystyle\mathbb{E}_{x}\bigg{[}S_{k}\mathbf{1}_{\\{S_{k}\geq
x_{b}\\}}\prod_{j=1}^{n}\frac{S_{k+j}}{S_{k+j-1}}\mathbf{1}_{\\{S_{k+j}\geq
S_{k+j-1}(1+\varepsilon)\\}}\bigg{]}$
$\displaystyle=\mathbb{E}_{x}\bigg{[}S_{k}\mathbf{1}_{\\{S_{k}\geq
x_{b}\\}}\bigg{(}\prod_{j=1}^{n-1}\frac{S_{k+j}}{S_{k+j-1}}\mathbf{1}_{\\{S_{k+j}\geq
S_{k+j-1}(1+\varepsilon)\\}}\bigg{)}(1-b_{\varepsilon}(S_{k+n-1}))\bigg{]}$
$\displaystyle\geq\mathbb{E}_{x}\bigg{[}S_{k}\mathbf{1}_{\\{S_{k}\geq
x_{b}\\}}\bigg{(}\prod_{j=1}^{n-1}\frac{S_{k+j}}{S_{k+j-1}}\mathbf{1}_{\\{S_{k+j}\geq
S_{k+j-1}(1+\varepsilon)\\}}\bigg{)}(1-b_{\varepsilon}(x_{b}(1+\varepsilon)^{n-1})\bigg{]}$
$\displaystyle\geq\mathbb{E}_{x}\bigg{[}S_{k}\mathbf{1}_{\\{S_{k}\geq
x_{b}\\}}\prod_{j=0}^{n-1}(1-b_{\varepsilon}(x_{b}(1+\varepsilon)^{j}))\bigg{]}$
$\displaystyle=\mathbb{E}_{x}\left[S_{k}\mathbf{1}_{\\{S_{k}\geq
x_{b}\\}}\right]\prod_{j=0}^{n-1}(1-b_{\varepsilon}(x_{b}(1+\varepsilon)^{j})).$

Thus, it remains to show that
$\prod_{j=0}^{\infty}(1-b_{\varepsilon}(x_{b}(1+\varepsilon)^{j}))>0$. The
latter condition is equivalent to
$\sum_{j=0}^{\infty}b_{\varepsilon}(\exp(\log(x_{b})+\log(1+\varepsilon)j))<\infty$,
which is equivalent to
$\int_{\log(x_{b})}^{\infty}b_{\varepsilon}(\exp(x))\,\mathrm{d}x<\infty$. ∎

We illustrate the above result first by two examples. The first one is a
“smooth” version of Example 1.3; the second one is an example of a “discrete
diffusion” for the log price.

######  Example 2.6.

Assume that the Markov kernel is given by

$K(x,\mathrm{d}y)=\begin{cases}\frac{1}{2}\mathbf{1}_{(0,1)}(y)\,\mathrm{d}y+\frac{1}{2}\mathbf{1}_{(2x-1,2x)}(y)\,\mathrm{d}y&\text{if
}x>1,\\\ \frac{1}{2x}\mathbf{1}_{(0,2x)}(y)\,\mathrm{d}y&\text{if }x\leq
1.\end{cases}$

Then $a(x)=\frac{1}{2}$ and for $\varepsilon\in(0,1)$,

$b_{\varepsilon}(x)=\tfrac{(1+\varepsilon)^{2}}{4}\mathbf{1}_{\\{x\leq
1\\}}+(1-x(1-\tfrac{(1+\varepsilon)^{2}}{4}))\mathbf{1}_{\\{1\leq
x\leq\tfrac{1}{1-\varepsilon}\\}}+\tfrac{1}{4x}\mathbf{1}_{\\{x>\tfrac{1}{1-\varepsilon}\\}}.$

By Theorem 2.5, $S$ is a bubble under $\mathbb{P}_{x}$ for all $x>0$.

######  Example 2.7.

Let $(Z_{k})_{k\in\mathbb{N}}$ be a sequence of i.i.d. standard normal random
variables and $\sigma:\mathbb{R}\to(0,\infty)$ a measurable function such that
$\sigma(x)$ is nondecreasing for large values of $x$. Define the process
$(X_{k})_{k\in\mathbb{N}_{0}}$ recursively by $X_{0}:=0$ and

$X_{k+1}:=X_{k}+\sigma(X_{k})Z_{k}-\frac{\sigma^{2}(X_{k})}{2},\quad
k\in\mathbb{N}_{0}.$

Then the process $S=(S_{k})_{k\in\mathbb{N}_{0}}$ defined by
$S_{k}:=\exp(X_{k})$ for $k\in\mathbb{N}_{0}$ is a Markov martingale.

Denoting by $\Phi$ the cdf of a standard normal random variable, it is not
difficult to check that for $x>0$ and $\varepsilon\geq 0$,

$a(x)=\Phi\left(\frac{\sigma(\log(x))}{2}\right),\qquad
b_{\varepsilon}(x)=\Phi\left(\frac{\log(1+\varepsilon)}{\sigma(\log(x))}-\frac{\sigma(\log(x))}{2}\right)$

where $b_{0}=b$. Thus, by Proposition 2.1, for $S$ to be a bubble it is
necessary that $\sigma(x)\to\infty$ as $x\to\infty$. Moreover, by Theorem 2.5,
a sufficient condition for $S$ to be a bubble is given by

(2.5)
$\int^{\infty}_{x_{0}}\Phi\left(\frac{\log(2)}{\sigma(x)}-\frac{\sigma(x)}{2}\right)\,\mathrm{d}x<\infty\quad\text{for
some }x_{0}\in\mathbb{R}.$

Denoting the pdf of a standard normal random variable by $\varphi$, using
Mill’s ratio and the fact that
$|(\frac{\log(2)}{\sigma(x)}-\frac{\sigma(x)}{2})^{2}-(\frac{\sigma(x)}{2})^{2}|\leq(\frac{\log(2)}{\sigma(x)})^{2}+\log(2)$
is uniformly bounded for all $x$ sufficiently large as $\sigma$ is
nondecreasing, it is not difficult to check that (2.5) is equivalent to

(2.6)
$\int^{\infty}_{x_{0}}\frac{1}{\sigma(x)}\varphi\left(-\frac{\sigma(x)}{2}\right)\,\mathrm{d}x<\infty\quad\text{for
some }x_{0}\in\mathbb{R}.$

######  Remark 2.8.

It is insightful to compare Example 2.7 to the continuous-time theory of
bubbles. To this end, recall that in continuous time, the process
$S=(S_{t})_{t\geq 0}$ given by $S_{t}:=\exp(X_{t})$, where $X=(X_{t})_{t\geq
0}$ solves the SDE

$dX_{t}=\sigma(X_{t})dW_{t}-\frac{1}{2}\sigma^{2}(X_{t})dt,$

is a strict local martingale and hence a bubble if and only if for some
$x_{0}\in\mathbb{R}$,

(2.7) $\int^{\infty}_{x_{0}}\frac{dx}{\sigma^{2}(x)}<\infty,$

cf. [14, Corollary 4.3]. While (2.6) and (2.7) both say that $S$ is a bubble
if and only if $\sigma(x)\to\infty$ fast enough as $x\to\infty$, the exact
rate of increase of $\sigma$ required for a bubble is quite different as (2.7)
is a much stronger requirement than (2.6) on the growth of $\sigma$. The
reason for this is that the discretisation of the diffusion model in
continuous time should not be done along a deterministic time grid, but along
certain sequences of stopping times; see Section 4 below.

While Proposition 2.1 and Theorem 2.5 give useful general sufficient
conditions for the absence or presence of a bubble, respectively, these
conditions are not necessary. In the complete Markov case, we can give a
necessary and sufficient characterisation of bubbles under mild assumptions on
the functions $a$ and $b$.

######  Theorem 2.9.

Suppose the Markov kernel is given by

$K(x,\mathrm{d}y)=a(x)\delta_{\frac{b(x)x}{a(x)}}(\mathrm{d}y)+(1-a(x))\delta_{\frac{(1-b(x))x}{1-a(x)}}(\mathrm{d}y),$

where $0\leq b(x)<a(x)<1$ and
$0<\liminf_{x\to\infty}a(x)\leq\limsup_{x\to\infty}a(x)<1$. Moreover, suppose
that there exists $x_{b}>0$ such that the function $b$ is nonincreasing for
$x\geq x_{b}$. Then $S$ is a bubble under $\mathbb{P}_{x}$ if and only if $S$
is not $\mathbb{P}_{x}$-a.s. bounded and
$\int_{\log(x_{b})}^{\infty}b(\exp(x))\,\mathrm{d}x<\infty$.

###### Proof.

Suppose that $S$ is not $\mathbb{P}_{x}$-a.s. bounded. We may focus on the
case that $\lim_{x\to\infty}b(x)=0$. Indeed, otherwise it follows that
$\lim_{x\to\infty}b(x)>0$ and hence
$\int_{\log(x_{b})}^{\infty}b(\exp(x))\,\mathrm{d}x=\infty$ and $S$ fails to
be a bubble by Proposition 2.1.

Since $\lim_{x\to\infty}b(x)=0$ and
$0<\liminf_{x\to\infty}a(x)\leq\limsup_{x\to\infty}a(x)<1$, after potentially
enlarging $x_{b}$, we may assume that there exists $0<c<C$ such that
$1+c\leq\frac{1-b(x)}{1-a(x)}\leq 1+C$ for all $x\geq x_{b}$. By Proposition
2.4, $S$ is a bubble under $\mathbb{P}_{x}$ if and only if there exists $k\geq
0$ such that

$\displaystyle\lim_{n\to\infty}\mathbb{E}_{x}\left[S_{n}\mathbf{1}_{\\{x_{b}\leq
S_{k}\leq S_{k+1}\leq\cdots\leq S_{k+n}\\}}\right]>0.$

Using that $\\{S_{j}\leq
S_{j+1}\\}=\\{S_{j+1}=\frac{1-b(S_{j})}{1-a(S_{j})}S_{j}\\}$ for
$j\in\mathbb{N}_{0}$ and arguing as in the proof of Theorem 2.5 gives for each
$0\leq k\leq n$,

$\displaystyle\mathbb{E}_{x}\left[S_{k}\mathbf{1}_{\\{S_{k}\geq
x_{b}\\}}\right]\prod_{j=0}^{n-1}(1-b(x_{b}(1+C)^{j}))\geq\mathbb{E}_{x}\left[S_{n}\mathbf{1}_{\\{x_{b}\leq
S_{k}\leq S_{k+1}\leq\cdots\leq S_{k+n}\\}}\right]$
$\displaystyle\qquad\qquad\qquad\geq\mathbb{E}_{x}\left[S_{k}\mathbf{1}_{\\{S_{k}\geq
x_{b}\\}}\right]\prod_{j=0}^{n-1}(1-b(x_{b}(1+c)^{j})).$

Now using that for $\gamma\in\\{c,C\\}$,
$\prod_{j=0}^{\infty}(1-b(x_{b}(1+\gamma)^{j})>0$ if and only if
$\sum_{j=0}^{\infty}b(\exp(\log(x_{b})+\log(1+\gamma)j))<\infty$, which in
turn is equivalent to
$\int_{\log(x_{b})}^{\infty}b(\exp(x))\,\mathrm{d}x<\infty$, the claim
follows. ∎

##  3\. A fixed point equation associated to a Markovian bubble

In this section, we continue our study of Markov martingales, taking a more
analytic perspective. We assume throughout that
$S=(S_{k})_{k\in\mathbb{N}_{0}}$ is a positive Markov martingale with kernel
$K:(0,\infty)\times\mathcal{B}_{(0,\infty)}\to[0,\infty)$, starting from
$S_{0}=x>0$. The key object of this section is the _default function_ of $S$.

######  Definition 3.1.

The Borel-measurable function $M_{S}:(0,\infty)\to[0,\infty)$, defined by

(3.1)
$M_{S}(x):=\lim_{n\to\infty}\mathbb{E}_{x}\left[(S_{n}-S_{\infty})\mathbf{1}_{\\{S_{n}\geq
S_{n-1}\geq\dots\geq S_{1}\geq x\\}}\right],$

is called the _default function_ of $S$.

It follows from (1.2) (with $k=0$) that
$M_{S}(x)=\mathbb{E}_{x}[S_{\tau_{1}}]$, so that $M_{S}$ measures the loss of
mass at the first drawdown of $S$. It is clear that $\mathbb{P}_{x}$ is a
bubble measure for $S$ if $M_{S}(x)>0$. The following two results show that
$M$ essentially fully characterises the bubble behaviour of $S$ under
$\mathbb{P}_{x}$ for all $x>0$.

######  Proposition 3.2.

Suppose $M_{S}(x)=0$ for all $x\geq x^{\prime}>0$. Then $M_{S}(x)=0$ for all
$x>0$, and $\mathbb{P}_{x}$ fails to be a bubble measure for $S$ for any
$x>0$.

###### Proof.

Fix $x>0$. Then for all $k\in\mathbb{N}_{0}$, the Markov property of $S$, the
choice of $x^{\prime}$ and dominated convergence give

$\displaystyle\lim_{n\to\infty}\mathbb{E}_{x}[(S_{n}-S_{\infty})\mathbf{1}_{\\{x^{\prime}\leq
S_{k}\leq S_{k+1}\leq\dots\leq S_{n}\\}}]$
$\displaystyle\qquad=\mathbb{E}_{x}\left[\lim_{n\to\infty}\mathbb{E}_{S_{k}}[(S_{n}-S_{\infty})\mathbf{1}_{\\{S_{k}\leq
S_{k+1}\leq\dots\leq S_{n}\\}}]\mathbf{1}_{\\{S_{k}\geq x^{\prime}\\}}\right]$
$\displaystyle\qquad=\mathbb{E}_{x}\left[M_{S}(S_{k})\mathbf{1}_{\\{S_{k}\geq
x^{\prime}\\}}\right]=0.$

Thus, Corollary 1.5 implies that $S$ is not a bubble under $\mathbb{P}_{x}$
and so $M_{S}(x)=0$. ∎

######  Proposition 3.3.

Suppose that $\liminf_{x\to\infty}M_{S}(x)>0$. Then $S$ is a bubble under
$\mathbb{P}_{x}$ for all $x>0$ for which $S$ is not $\mathbb{P}_{x}$-a.s.
bounded.

###### Proof.

Fix $x>0$ and suppose that $S$ is not $\mathbb{P}_{x}$-a.s. bounded. By
hypothesis, there exists $x^{\prime}\geq x$ such that $M_{S}(y)>0$ for all
$y\geq x^{\prime}$. Since $S$ is not $\mathbb{P}_{x}$-a.s. bounded, there is
$k\geq 0$ such that $\mathbb{P}_{x}[S_{k}\geq x^{\prime}]>0$. Then by the
Markov property of $S$, the choice of $x^{\prime}$ and dominated convergence,

$\displaystyle\lim_{n\to\infty}\mathbb{E}_{x}[(S_{n}-S_{\infty})\mathbf{1}_{\\{x^{\prime}\leq
S_{k}\leq S_{k+1}\leq\dots\leq S_{n}\\}}]$
$\displaystyle\qquad=\mathbb{E}_{x}\left[\lim_{n\to\infty}\mathbb{E}_{S_{k}}[(S_{n}-S_{\infty})\mathbf{1}_{\\{S_{k}\leq
S_{k+1}\leq\dots\leq S_{n}\\}}]\mathbf{1}_{\\{S_{k}\geq x^{\prime}\\}}\right]$
$\displaystyle\qquad=\mathbb{E}_{x}\left[M_{S}(S_{k})\mathbf{1}_{\\{S_{k}\geq
x^{\prime}\\}}\right]>0.$

Thus, $S$ is a bubble under $\mathbb{P}_{x}$ by Corollary 1.5. ∎

In the remainder of this section, we seek to characterise the function $M_{S}$
in an analytic way and provide conditions for it to be non-zero.

First, we show that $M_{S}$ solves a fixed point equation, more precisely a
homogeneous _Volterra integral equation of the second kind_ , cf. Brunner [3]
for a textbook treatment.

######  Lemma 3.4.

The default function $M_{S}$ is a solution to the Volterra integral equation

(3.2) $M_{S}(x)=\int_{[x,\infty)}M_{S}(y)K(x,\mathrm{d}y),\quad x>0.$

###### Proof.

Fix $x>0$. Using the definition of $M_{S}$, dominated convergence and the
Markov property of $S$, we obtain

$\displaystyle\int_{[x,\infty)}M_{S}(y)K(x,\mathrm{d}y)$
$\displaystyle=\mathbb{E}_{x}\left[M_{S}(S_{1})\mathbf{1}_{\\{S_{1}\geq
x\\}}\right]$
$\displaystyle=\mathbb{E}_{x}\left[\lim_{n\to\infty}\mathbb{E}_{S_{1}}\left[(S_{n}-S_{\infty})\mathbf{1}_{\\{S_{n}\geq
S_{n-1}\geq\dots\geq S_{1}\geq x\\}}\right]\right]$
$\displaystyle=\lim_{n\to\infty}\mathbb{E}_{x}\left[(S_{n}-S_{\infty})\mathbf{1}_{\\{S_{n}\geq
S_{n-1}\geq\dots\geq S_{1}\geq x\\}}\right]=M_{S}(x).$

∎

Note that (3.2) is non-standard in that the domain is non-compact. Therefore,
we cannot apply standard existence and uniqueness results for Volterra
integral equations, cf. [3, Chapter 8]. In fact, existence is anyway not an
issue since the zero function always solves (3.2). Since the bubble case
corresponds to (3.2) having a non-zero (nonnegative) solution, we are actually
interested in _non-uniqueness_ , i.e., the case that (3.2) has multiple
nonnegative solutions. By homogeneity of (3.2), we then always have infinitely
many solutions and so it is clear that we need an additional condition to pin
down the default function $M_{S}$.

It follows from the definition of $M_{S}$, that $M_{S}(x)\leq x$ for all
$x>0$. So we consider nonnegative solutions to (3.2) that are dominated by the
identity. To this end, denote by $\mathcal{I}$ all Borel-measurable functions
$M:(0,\infty)\to[0,\infty)$ satisfying $M(x)\leq x$ for all $x>0$. Using that

$\displaystyle
0\leq\int_{[x,\infty)}M(y)K(x,\mathrm{d}y)\leq\int_{(0,\infty)}M(y)K(x,\mathrm{d}y)\leq\int_{(0,\infty)}yK(x,\mathrm{d}y)=x$

for all $M\in\mathcal{I}$ and $x>0$, we can define the map
$\mathcal{K}:\mathcal{I}\to\mathcal{I}$ by

$\mathcal{K}(M)(x)=\int_{[x,\infty)}M(y)K(x,\mathrm{d}y),\quad x>0.$

Then the nonnegative solutions to (3.2) dominated by the identity are
precisely given by fixed points of $\mathcal{K}$.

While the map $\mathcal{K}$ is in general not a contraction (and therefore
(3.2) may have multiple solutions on $\mathcal{I}$), it is _monotone_ , and
this property will prove crucial for our subsequent analysis.

######  Proposition 3.5.

The map $\mathcal{K}$ is monotone on $\mathcal{I}$.

###### Proof.

Let $M_{1},M_{2}\in\mathcal{I}$ with $M_{1}\leq M_{2}$. Then monotonicity of
the integral gives for $x>0$,

$\mathcal{K}(M_{1})(x)=\int_{[x,\infty)}M_{1}(y)K(x,\mathrm{d}y)\leq\int_{[x,\infty)}M_{2}(y)K(x,\mathrm{d}y)=\mathcal{K}(M_{2})(x).$

∎

Due to monotonicity of $\mathcal{K}$, it is very useful to consider
_subsolution_ and _supersolutions_ to (3.2) on $\mathcal{I}$.

######  Definition 3.6.

A function $M\in\mathcal{I}$ is called a _subsolution_ to (3.2) if

$M(x)\leq\int_{[x,\infty)}M(y)K(x,dy),\quad x>0.$

It is called a _supersolution_ to (3.2) if

$M(x)\geq\int_{[x,\infty)}M(y)K(x,dy),\quad x>0.$

The following result shows that we can construct from each sub- or
supersolution a solution to (3.2) by Picard iteration. To this end, for
$n\in\mathbb{N}_{0}$, define $\mathcal{K}^{n}(M)$ recursively by
$\mathcal{K}^{0}(M):=M$ and
$\mathcal{K}^{n}(M):=\mathcal{K}(\mathcal{K}^{n-1}(M))$ for $n\geq 1$.

######  Proposition 3.7.

Let $M\in\mathcal{I}$ be a sub- or supersolution to (3.2). Then the limit
$\mathcal{K}^{\infty}(M)=\lim_{n\to\infty}K^{n}(M)$ exists and is a solution
to (3.2). Moreover,

  * •

if $M$ is a subsolution, then the sequence
$(\mathcal{K}^{n}(M))_{n\in\mathbb{N}}$ is nondecreasing and
$\mathcal{K}^{\infty}(M)$ is the smallest solution dominating $M$;

  * •

if $M$ is a supersolution, then the sequence
$(\mathcal{K}^{n}(M))_{n\in\mathbb{N}}$ is nonincreasing and
$\mathcal{K}^{\infty}(M)$ is the largest solution dominated by $M$.

###### Proof.

We only consider the case that $M$ is a subsolution; the proof for the case
that $M$ is a supersolution is analogous. If $M$ is a subsolution, the
sequence $(\mathcal{K}^{n}(M))_{n\in\mathbb{N}_{0}}$ is nondecreasing in $M$
by monotonicity of $\mathcal{K}$. Hence the limit $\lim_{n\to\infty}K^{n}(M)$
exists and is in $\mathcal{I}$ since each $\mathcal{K}^{n}(M)$ is in
$\mathcal{I}$. Moreover, it follows from monotone convergence that

$\displaystyle\mathcal{K}^{\infty}(M)(x)$
$\displaystyle=\lim_{n\to\infty}\mathcal{K}^{n}(M)(x)=\lim_{n\to\infty}\int_{[x,\infty)}\mathcal{K}^{n-1}(M)(y)K(x,\mathrm{d}y)$
$\displaystyle=\int_{[x,\infty)}\mathcal{K}^{\infty}(M)(y)K(x,\mathrm{d}y),\quad
x>0,$

whence $\mathcal{K}^{\infty}(M)$ is a solution to (3.2).

Moreover, let $\tilde{M}\in\mathcal{I}$ be any solution to (3.2) dominating
$M$. It suffices to show that $\tilde{M}\geq\mathcal{K}^{n}(M)$ for all
$n\in\mathbb{N}_{0}$. We argue by induction. The induction basis is trivial.
For the induction step, suppose that $n\geq 1$ and
$\tilde{M}\geq\mathcal{K}^{n-1}(M)$. Then by the induction hypothesis and the
definition of $\mathcal{K}^{n}(M)$, for $x>0$,

$\tilde{M}(x)=\int_{[x,\infty)}\tilde{M}(y)K(x,\mathrm{d}y)\geq\int_{[x,\infty)}\mathcal{K}^{n-1}(M)(y)K(x,\mathrm{d}y)=\mathcal{K}^{n}(M)(x).$

∎

We note the following important corollary.

######  Corollary 3.8.

The largest solution to (3.2) on $\mathcal{I}$ is given by
$\mathcal{K}^{\infty}(\operatorname{id})$, where $\operatorname{id}$ denotes
the identity function.

It follows from Lemma 3.4 and Corollary 3.8 that the default function $M_{S}$
is dominated by $\mathcal{K}^{\infty}(\operatorname{id})$. Under a mild
assumption on the kernel $K$, we can assert that $M_{S}$ coincides with
$\mathcal{K}^{\infty}(\operatorname{id})$. Thus, in this case, we can
characterise the default function $M_{s}$ as the maximal solution to (3.2)
dominated by the identity.

######  Theorem 3.9.

Suppose that Assumption 2.2 is satisfied for any $x_{a}>0$. Then $M_{S}$ is
the maximal solution to (3.2) dominated by the identity. It is given by
$M_{S}=\mathcal{K}^{\infty}(\operatorname{id})$. Moreover, $M_{S}(x)<x$ for
all $x>0$ and if $M_{S}\neq 0$, then

(3.3) $\limsup_{x\to\infty}\frac{M_{S}(x)}{x}=1.$

###### Proof.

We first show by induction that for each $n\in\mathbb{N}_{0}$,

$\mathcal{K}^{n}(\operatorname{id})(x)=\mathbb{E}_{x}\left[S_{n}\mathbf{1}_{\\{S_{n}\geq
S_{n-1}\geq\dots\geq S_{1}\geq S_{0}\\}}\right],\quad x>0.$

The induction basis $n=0$ follows from the martingale property of $S$. For the
induction step, let $n\geq 1$. By the definition of
$\mathcal{K}^{n}(\operatorname{id})$ and the Markov property of $S$, we obtain

$\displaystyle\mathcal{K}^{n}(\operatorname{id})(x)$
$\displaystyle=\int_{[x,\infty)}\mathcal{K}^{n-1}(\operatorname{id})(y)K(x,\mathrm{d}y)=\mathbb{E}_{x}\left[\mathcal{K}^{n-1}(\operatorname{id})(S_{1})\mathbf{1}_{\\{S_{1}\geq
x\\}}\right]$
$\displaystyle=\mathbb{E}_{x}\left[\mathbb{E}_{S_{1}}\left[S_{n}\mathbf{1}_{\\{S_{n}\geq
S_{n-1}\geq\dots\geq S_{1}\\}}\right]\mathbf{1}_{\\{S_{1}\geq x\\}}\right]$
$\displaystyle=\mathbb{E}_{x}\left[S_{n}\mathbf{1}_{\\{S_{n}\geq
S_{n-1}\geq\dots\geq S_{1}\geq S_{0}\\}}\right].$

Hence, the definitions of $\mathcal{K}^{\infty}(\operatorname{id})$ and
$M_{S}$ together with (2.4) give

$\displaystyle\mathcal{K}^{\infty}(\operatorname{id})(x)$
$\displaystyle=\lim_{n\to\infty}\mathbb{E}_{x}\left[S_{n}\mathbf{1}_{\\{S_{n}\geq
S_{n-1}\geq\dots\geq S_{1}\geq S_{0}\\}}\right]$
$\displaystyle=\lim_{n\to\infty}\mathbb{E}_{x}\left[(S_{n}-S_{\infty})\mathbf{1}_{\\{S_{n}\geq
S_{n-1}\geq\dots\geq S_{1}\geq S_{0}\\}}\right]=M_{S}(x),\quad x>0.$

It follows from Corollary 3.8 that $M_{S}$ is the maximal solution to (3.2)
dominated by the identity.

Moreover, $M_{S}\leq\operatorname{id}$, the fact that
$\mathbb{P}_{x}[S_{1}<x]=a(x)>0$ for all $x>0$ and the martingale property of
$S$ give

$\displaystyle
M_{S}(x)=\mathbb{E}_{x}\big{[}M_{S}(S_{1})\mathbf{1}_{\\{S_{1}\geq
x\\}}\big{]}\leq\mathbb{E}_{x}\big{[}S_{1}\mathbf{1}_{\\{S_{1}\geq
x\\}}\big{]}<\mathbb{E}_{x}\big{[}S_{1}\big{]}=x.$

Finally, to establish (3.3), denote the function $H_{S}:(0,\infty)\to[0,1]$ by
$H_{S}(x)=\sup_{y\geq x}\tfrac{M_{S}(x)}{x}$. If $M_{S}\neq 0$, it follows
from Proposition 3.2 that $H_{S}(x)>0$ for all $x>0$. It suffices to show that
$H_{S}(x)=1$ for all $x>0$. Seeking a contradiction, suppose there exists
$x^{\prime}>0$ such that $H_{S}(x^{\prime})<1$. Define the function
$M:(0,\infty)\to[0,\infty)$ by

$M(y)=\begin{cases}0&\text{if }x<x^{\prime},\\\
\frac{M_{S}(x)}{H_{S}(x^{\prime})}&\text{if }x\geq x^{\prime}.\end{cases}$

Then $M\in\mathcal{I}$ by the fact that
$H_{S}(x^{\prime})\geq\tfrac{M_{S}(x)}{x}$ for $x\geq x^{\prime}$. Moreover,
$M$ is a subsolution to (3.2) and $M(x)>M_{S}(x)$ for $x\geq x^{\prime}$. This
together with Proposition 3.7 implies that $\mathcal{K}^{\infty}(M)$ is a
solution to (3.2) dominated by the identity. Since
$\mathcal{K}^{\infty}(M)(x)\geq M(x)>M_{S}(x)$ for $x>x^{\prime}$, this is in
contradiction to $M_{S}$ being the maximal solution to (3.2) dominated by the
identity. ∎

The following corollary shows that if we can find a non-trivial subsolution to
(3.2), then $S$ is a bubble.

######  Corollary 3.10.

Suppose Assumption 2.2 is satisfied for any $x_{a}>0$. If $M\in\mathcal{I}$ is
a subsolution to (3.2), then $M\leq M_{S}$. If in addition
$\liminf_{x\to\infty}M(x)>0$, then $S$ is a bubble under $\mathbb{P}_{x}$ for
all $x>0$ for which $S$ is not $\mathbb{P}_{x}$-a.s. bounded.

###### Proof.

Proposition 3.7 and Theorem 3.9 give $M\leq\mathcal{K}^{\infty}(M)\leq M_{S}$.
The additional claim then follows from Proposition 3.3. ∎

A typical candidate for a subsolution in Corollary 3.10 is the call function
$M(x)=(x-L)^{+}$ for some $L>0$. This is illustrated by the following example.
Note that this example cannot be addressed with the results from Section 2.

######  Example 3.11.

Suppose that for all $x>0$, $K(x,\mathrm{d}y)=k(x,y)\,\mathrm{d}y$, where the
density $k$ satisfies

$k(x,y)=\frac{2}{3(x+1)}\mathbf{1}_{[x,2x]}(y)\quad\text{for }y\geq x>0,$

and $k(x,y)$ on $\\{(x,y):0<y<x\\}$ is chosen such that $K$ is a martingale
kernel. Then the function $a$ from (2.1) satisfies
$a(x)=1-\int_{[x,\infty)}k(x,y)\,\mathrm{d}y=\tfrac{3+x}{3+3x}\geq\frac{1}{3}$
so that Assumption 2.2 is satisfied for any $x_{a}>0$. Consider
$M(x):=(x-3)^{+}$. Then trivially
$\int_{x}^{\infty}M(y)k(x,y)\,\mathrm{d}y\geq 0=M(x)$ for $x\leq 3$ and

$\int_{x}^{\infty}M(y)k(x,y)\,\mathrm{d}y=x-3\frac{x}{x+1}\geq x-3=M(x)\text{
for }x>3.$

It follows that $M$ is subsolution to (3.2) and we may deduce that $S$ has a
bubble under $P_{x}$ for all $x>0$ by Corollary 3.10 since $S$ is not
$\mathbb{P}_{x}$-a.s. bounded for any $x>0$.

While Theorem 3.9 provides a characterisation of the default function $M$, it
does not provide a criterion to decide whether (3.2) has a non-trivial, i.e.,
a non-zero nonnegative solution dominated by the identity. Moreover, it does
not provide a criterion to decide whether a given candidate solution $M$ to
(3.2) is indeed maximal. Under a stronger assumption on the kernel $K$, we can
provide a sufficient criterion for the existence of non-trivial solutions to
(3.2) dominated by the identity. Moreover, we obtain a local uniqueness result
in this case. To this end, recall the definitions of the functions $a$ and $b$
from (2.1) and (2.2), respectively. Moreover, denote by $\|\cdot\|_{\sup}$ the
supremum norm.

######  Theorem 3.12.

Suppose $\inf_{x>0}a(x)>0$. Then the following are equivalent:

  1. (a)

$\sup_{x>0}xb(x)<\infty$.

  2. (b)

For all $L>0$ sufficiently large, the call function $M(x)=(x-L)^{+}$ is a
subsolution to (3.2).

  3. (c)

The default function $M_{S}$ is non-trivial and satisfies
$\|\operatorname{id}-M_{S}\|_{\sup}<\infty$.

Moreover, if one of the above conditions is satisfied, $M_{S}$ is the unique
solution to (3.2) among all solutions $M\in\mathcal{I}$ satisfying
$\|\operatorname{id}-M\|_{\sup}<\infty$.

###### Proof.

(a) $\Rightarrow$ (b). Set $\alpha:=\inf_{x>0}a(x)>0$,
$\beta:=\sup_{x>0}xb(x)$ and $L\geq\frac{\beta}{\alpha}$. Then the call
function $M(x):=(x-L)^{+}$ satisfies
$\int_{[x,\infty)}M(y)K(x,\mathrm{d}y)\geq 0=M(x)$ for $x\leq L$ and

$\displaystyle\int_{[x,\infty)}M(y)K(x,\,\mathrm{d}y)$
$\displaystyle=x(1-b(x))-L(1-a(x))=M(x)-\beta(x)x+\alpha(x)L$
$\displaystyle\geq M(x)-\beta+\alpha L\geq M(x),\quad x>L.$

Hence, $M$ is a subsolution to (3.2).

(b) $\Rightarrow$ (c). Let $L>0$ such that $M(x)=(x-L)^{+}$ is a subsolution
to (3.2). Then $M_{S}\geq M$ by Corollary 3.10, whence $M_{S}$ is non-trivial
and satisfies
$\|\operatorname{id}-M_{S}\|_{\sup}<\|\operatorname{id}-M\|_{\sup}=L$.

(c) $\Rightarrow$ (a). Since
$\mathcal{K}(\operatorname{id})\geq\mathcal{K}^{\infty}(\operatorname{id})=M_{S}$
by Proposition 3.7 and Theorem 3.9, it follows that
$\|\operatorname{id}-\mathcal{K}(\operatorname{id})\|_{\sup}\leq\|\operatorname{id}-M_{S}\|_{\sup}<\infty$.
Now the claim follows from the fact that
$xb(x)=\int_{(0,x)}yK(x,\mathrm{d}y)=x-\mathcal{K}(\operatorname{id})(x)=\operatorname{id}(x)-\mathcal{K}(\operatorname{id})(x)$
for $x>0$.

For the additional claim, set
$\mathcal{I}_{\sup}:=\\{M\in\mathcal{I}:\|\operatorname{id}-M\|_{\sup}<\infty\\}$.
Then $\mathcal{I}_{\sup}$ is a complete metric space for the metric generated
by the supremum norm. Moreover, $\mathcal{K}$ maps $\mathcal{I}_{\sup}$ to
itself. Indeed, for each $M\in\mathcal{I}_{\sup}$, by (b), there exists
$L\geq\|\operatorname{id}-M\|_{\sup}$ such that $\tilde{M}(x):=(x-L)^{+}$ is a
subsolution to (3.2). Hence by monotonicity of $\mathcal{K}$ and the fact that
$\tilde{M}\leq M$ is a subsolution,

$\|\operatorname{id}-\mathcal{K}(M)\|_{\sup}\leq\|\operatorname{id}-\mathcal{K}(\tilde{M})\|_{\sup}\leq\|\operatorname{id}-\tilde{M}\|_{\sup}=L<\infty.$

Finally, we show that $\mathcal{K}$ is a contraction on $\mathcal{I}_{\sup}$.
Let $M_{1},M_{2}\in\mathcal{I}_{\sup}$. Then

$\displaystyle|\mathcal{K}(M_{1})(x)-\mathcal{K}(M_{2})(x)|$
$\displaystyle=\bigg{|}\int_{[x,\infty)}(M_{1}(y)-M_{2}(y))K(x,\mathrm{d}y)\bigg{|}$
$\displaystyle\leq\|M_{1}-M_{2}\|_{\sup}\int_{[x,\infty]}K(x,\mathrm{d}y)$
$\displaystyle\leq(1-\alpha)\|M_{1}-M_{2}\|_{\sup}.$

Taking the supremum over $x$ shows that $\mathcal{K}$ is indeed a contraction
since $\alpha>0$. Now Banach’s fixed point theorem implies that $\mathcal{K}$
contains a unique fixed point, and by (c), this fixed point is $M_{S}$. ∎

We proceed to illustrate Theorem 3.12 by an example.

######  Example 3.13.

Suppose that for all $x>0$, $K(x,\mathrm{d}y)=k(x,y)\,\mathrm{d}y$, where the
density satisfies

$k(x,y)=\frac{e}{2}\cdot\frac{1-e^{-x}}{1-e^{-y}}\frac{1}{x}e^{-y/x}\quad\text{for
}y\geq x>0,$

and $k(x,y)$ on $\\{(x,y):0<y<x\\}$ is chosen such that $K$ is a martingale
kernel. Note that

$\int_{x}^{\infty}k(x,y)\,\mathrm{d}y<\frac{e}{2}\int_{x}^{\infty}\frac{1}{x}e^{-y/x}\,\mathrm{d}y=\frac{1}{2}.$

This implies in particular that $a(x)\geq 1/2$ for all $x>0$. In this case,
the fixed point equation (3.2) is given by

$\displaystyle M(x)$
$\displaystyle=\int_{x}^{\infty}M(y)\cdot\frac{e}{2}\cdot\frac{1-e^{-x}}{1-e^{-y}}\cdot\frac{1}{x}\cdot
e^{-y/x}\,\mathrm{d}y$
$\displaystyle\Leftrightarrow\quad\frac{M(x)x}{1-e^{-x}}$
$\displaystyle=\frac{e}{2}\int_{x}^{\infty}\frac{M(y)}{1-e^{-y}}\cdot
e^{-y/x}\,\mathrm{d}y.$

As can easily be checked, the function $M_{\lambda}(x):=\lambda x(1-e^{-x})$,
$x>0$, is a solution in $\mathcal{I}$ to (3.2) for any $\lambda\in[0,1]$. As
$M_{S}$ is the largest solution to (3.2) dominated by the identity, this
yields the candidate $M_{1}(x)=x(1-e^{-x})$. Since
$\|\operatorname{id}-M_{1}\|_{\sup}=e^{-1}<\infty$, it follows from Theorem
3.12 that $M_{S}=M_{1}$.

Combing Theorem 3.12 with Proposition 3.3, we get the following existence
results for bubbles. Note that this result covers cases that cannot be treated
with the theory of Section 2.

######  Corollary 3.14.

Suppose that $\inf_{x>0}a(x)>0$ and $\sup_{x>0}xb(x)<\infty$. Then $S$ is a
bubble under $\mathbb{P}_{x}$ for all $x>0$ for which $S$ is not
$\mathbb{P}_{x}$-a.s. bounded.

##  4\. Relation to the strict local martingale definition of asset price
bubbles in continuous time models

In this final section, we discuss how our definition of bubbles in discrete
time relates to the strict local martingale definition of bubbles in
continuous time. To approach this question, one first has to discretise a
positive continuous local martingale $X=(X_{t})_{t\geq 0}$ in continuous time
in such a way that it becomes a discrete time martingale. Of course, there are
many ways to do this and we choose a somewhat canonical construction. To wit,
we consider localising sequences $(\tau_{n})_{n\in\mathbb{N}}$ of stopping
times with $\tau_{n}\to\infty$ $\mathbb{P}$-a.s. such that for each $n$,
$\tau_{n}$ and the stopped process $X^{\tau_{n}}$ are uniformly bounded. We
then define the discrete time process $S=(S_{n})_{n\in\mathbb{N}}$ by
$S_{n}:=X_{\tau_{n}}$. Then $S$ is a martingale by the stopping theorem and
satisfies $S_{\infty}=X_{\infty}$ $\mathbb{P}\text{-a.s.}$, which implies that
$S$ is uniformly integrable if and only if $X$ is uniformly integrable.

The simplest way to get localising sequences as above is to choose two
increasing sequences of positive real numbers $a=(a_{n})_{n\in\mathbb{N}}$ and
$b=(b_{n})_{n\in\mathbb{N}}$ converging to infinity and to define the sequence
$(\tau^{a,b}_{n})_{n\in\mathbb{N}}$ of stopping times by $\tau^{a,b}_{0}:=0$
and

(4.1) $\tau^{a,b}_{n}:=\inf\\{t\geq 0:\ X_{t}\geq b_{n}\\}\wedge a_{n},\quad
n\in\mathbb{N}.$

Then $(\tau^{a,b}_{n})_{n\in\mathbb{N}}$ is a localizing sequence of stopping
times for $X$ with $\tau^{a,b}_{n}\leq a_{n}$ and $\sup_{t\geq
0}X^{\tau^{a,b}_{n}}_{t}\leq b_{n}$.

In the special case that $X$ is a Markov process, we would like to stop in
such a way that the discrete time process $S$ is again a Markov process. In
this case, the simplest way to get localising sequences as above is to choose
two constants $\alpha,\beta>0$ and to define the sequence of stopping times
$(\tau^{\alpha,\beta}_{n})_{n\in\mathbb{N}}$ by $\tau^{\alpha,\beta}_{0}:=0$
and

(4.2) $\tau^{\alpha,\beta}_{n}:=\inf\left\\{t\geq\tau^{\alpha,\beta}_{n-1}:\
X_{t}\geq(1+\beta)X_{\tau^{\alpha,\beta}_{n-1}}\right\\}\wedge\left(\tau^{\alpha,\beta}_{n-1}+\alpha\right),\quad
n\in\mathbb{N}.$

In this case it is still true that
$(\tau^{\alpha,\beta}_{n})_{n\in\mathbb{N}}$ is a localizing sequence of
stopping times for $X$ and $\tau^{\alpha,\beta}_{n}$ and
$X^{\tau^{\alpha,\beta}_{n}}$ are uniformly bounded.

Our first goal in this section is to show that if $X$ is a continuous positive
strict local martingale, then the discrete time process $S$ is a bubble for
either choice of stopping times.

The proof of this result relies on the following deep characterisation of
strict local martingales in continuous time; cf. [13, 5, 11]. Let
$X=(X_{t})_{t\geq 0}$ be a positive càdlàg local $\mathbb{P}$-martingale with
$X_{0}=x$. Then under some technical assumptions on the probability space and
the underlying filtration, there exists a probability measure $\mathbb{Q}$
with $\mathbb{Q}|_{\mathcal{F}_{t}}\gg\mathbb{P}|_{\mathcal{F}_{t}}$ for all
$t\geq 0$ such that $Y:=1/X$ is a nonnegative true $\mathbb{Q}$-martingale,
and for all bounded stopping times $\tau$ and $A\in\mathcal{F}_{\tau}$,

$\mathbb{P}[A]=x\cdot\mathbb{E}^{\mathbb{Q}}\left[Y_{\tau}\mathbf{1}_{A}\right].$

Especially, we have the identity

$\mathbb{E}^{\mathbb{P}}[X_{0}-X_{t}]=x\cdot\mathbb{Q}[Y_{t}=0],\quad t\geq
0,$

i.e., $X$ is a strict local martingale on $[0,t]$ if and only if
$\mathbb{Q}[Y_{t}=0]>0$.

With this, we have the following two results.

######  Proposition 4.1.

Let $X=(X_{t})_{t\geq 0}$ be a continuous positive strict local
$\mathbb{P}$-martingale. Let $a=(a_{n})_{n\in\mathbb{N}}$ and
$b=(b_{n})_{n\in\mathbb{N}}$ be increasing sequences of positive real numbers
converging to $\infty$. Define the sequence of stopping times
$(\tau^{a,b}_{n})_{n\in\mathbb{N}}$ by (4.1) and set
$S^{a,b}_{n}=X_{\tau^{a,b}_{n}}$ for $n\in\mathbb{N}_{0}$. Then the measure
$\mathbb{P}$ is a bubble measure for the discrete time martingale
$S^{a,b}=(S^{a,b}_{n})_{n\in\mathbb{N}_{0}}$.

######  Proposition 4.2.

Let $X=(X_{t})_{t\geq 0}$ be a continuous-time positive strict local Markov
martingale under the measure $\mathbb{P}_{x}$. Let $\alpha,\beta>0$ and define
the sequence of stopping times $(\tau^{\alpha,\beta}_{n})_{n\in\mathbb{N}}$ by
(4.2) and set $S^{\alpha,\beta}_{n}=X_{\tau^{\alpha,\beta}_{n}}$ for
$n\in\mathbb{N}_{0}$. Then the measure $\mathbb{P}_{x}$ is a bubble measure
for the discrete time Markov martingale
$S^{\alpha,\beta}=(S^{\alpha,\beta}_{n})_{n\in\mathbb{N}_{0}}$.

We only establish the proof of Proposition 4.1. The proof of Proposition 4.2
is similar and left to the reader.

###### Proof of Proposition 4.1.

Since $X$ is a local martingale with respect to its natural filtration and
$(\tau^{a,b}_{n})_{n\in\mathbb{N}}$ is adapted to this filtration, we may
assume without loss of generality that $X$ is the canonical process on
$C([0,\infty),(0,\infty])$ with $X_{0}=x>0$. Set $\tau_{\infty}:=\inf\\{t\geq
0:X_{t}=\infty\\}$. Then $\tau^{a,b}_{n}<\tau_{\infty}$ for all
$n\in\mathbb{N}$.

Then there exists a measure $\mathbb{Q}$ on $C([0,\infty),(0,\infty])$, with
$\mathbb{Q}|_{\mathcal{F}_{t}}\gg\mathbb{P}|_{\mathcal{F}_{t}}$ for all $t\geq
0$, such that $Y:=1/X$ is a nonnegative $\mathbb{Q}$-martingale and
$\mathbb{E}^{\mathbb{P}}\left[X_{\tau}\mathbf{1}_{A}\right]=x_{0}\cdot\mathbb{Q}[A]$
for each bounded stopping time $\tau<\tau_{\infty}$ and
$A\in\mathcal{F}_{\tau}$. Let
$k:=\min\\{n\in\mathbb{N}:\mathbb{E}[X_{0}-X_{a_{n}}]>0\\}$. Then for $n\geq
k$, using that $\tau^{a,b}_{n}$ is bounded by $a_{n}$ and
$\tau^{a,b}_{n}<\tau_{\infty}$, we obtain Taking the limit as $n\to\infty$ on
the left hand side, it follows that $\mathbb{P}$ is a bubble measure for $S$
by Theorem 1.4. ∎

We proceed to illustrate Proposition 4.2 by an Example.

######  Example 4.3.

Let $(X_{t})_{t\geq 0}$ be the three-dimensional inverse Bessel process, i.e.,
$X$ is the unique strong solution to the SDE
$\,\mathrm{d}X_{t}=-X_{t}^{2}\mathrm{d}B_{t}$, where $B=(B_{t})_{t\geq 0}$ is
a $\mathbb{P}_{x}$-Brownian motion. The process $Y:=1/X$ is
$\mathbb{Q}_{1/x}$-Brownian motion, stopped when it reaches zero.

Fix $\alpha,\beta>0$. We proceed to calculate the Markov kernel
$K(x,\,\mathrm{d}y)$ for $S^{\alpha,\beta}$ under $\mathbb{P}_{x}$. Denote by
$W$ a standard Brownian motion starting at zero and by $\Phi$ the cdf of
standard normal random variable. Using the reflection principle of Brownian
motion and denoting the running supremum and the running infimum of a process
$Z$ by $\overline{Z}$ and $\underline{Z}$, respectively, we obtain

$\displaystyle\mathbb{P}_{x}\left[S^{\alpha,\beta}_{1}=(1+\beta)x\right]=\mathbb{P}_{x}\left[X_{\tau^{\alpha,\beta}_{1}}=(1+\beta)x\right]$
$\displaystyle\qquad\qquad=x\mathbb{E}^{\mathbb{Q}}_{1/x}\left[Y_{\tau^{\alpha,\beta}_{1}}\mathbf{1}_{\\{Y_{\tau^{\alpha,\beta}_{1}}=\frac{1}{(1+\beta)x}\\}}\right]=\frac{1}{1+\beta}\mathbb{Q}_{1/x}\left[\underline{Y}_{\alpha}\leq\frac{1}{(1+\beta)x}\right]$
$\displaystyle\qquad\qquad=\frac{1}{1+\beta}\mathbb{Q}\left[\overline{W}_{\alpha}\geq\frac{1}{x}-\frac{1}{(1+\beta)x}\right]=\frac{2}{1+\beta}\Phi\left(-\frac{\beta}{(1+\beta)x\sqrt{\alpha}}\right).$

Moreover, for $z\in(0,(1+\beta)x)$, using the joint density of Brownian motion
and its running supremum, we obtain

$\displaystyle\mathbb{P}_{x}\left[S^{\alpha,\beta}_{1}\leq
z\right]=\mathbb{P}_{x}\left[X_{\tau^{\alpha,\beta}_{1}}\leq
z\right]=\mathbb{P}_{x}\left[X_{\alpha}\leq
z,\overline{X}_{\alpha}<(1+\beta)x\right]$
$\displaystyle=x\mathbb{E}^{\mathbb{Q}}_{1/x}\left[Y_{\alpha}\mathbf{1}_{\\{Y_{\alpha}\geq\frac{1}{z}\\}}\mathbf{1}_{\\{\underline{Y}_{\alpha}>\frac{1}{(1+\beta)x}\\}}\right]$
$\displaystyle=x\mathbb{E}^{\mathbb{Q}}\left[\left(\frac{1}{x}-W_{\alpha}\right)\mathbf{1}_{\\{W_{\alpha}\leq\frac{1}{x}-\frac{1}{z}\\}}\mathbf{1}_{\\{\overline{W}_{\alpha}<\frac{1}{x}-\frac{1}{(1+\beta)x}\\}}\right]$
$\displaystyle=x\int_{-\infty}^{\frac{1}{x}-\frac{1}{z}}\left(\frac{1}{x}-y\right)\int_{0}^{\frac{\beta}{(1+\beta)x}}\frac{2(2u-y)}{\sqrt{2\pi\alpha^{3}}}\exp\left(-\frac{(2u-y)^{2}}{2\alpha}\right)\,\mathrm{d}u\,\mathrm{d}y$
$\displaystyle=x\int_{-\infty}^{\frac{1}{x}-\frac{1}{z}}\left(\frac{1}{x}-y\right)\int_{\frac{-y}{\sqrt{\alpha}}}^{\frac{\frac{2\beta}{(1+\beta)x}-y}{\sqrt{\alpha}}}\frac{v}{\sqrt{2\pi\alpha}}\exp\left(-\frac{v^{2}}{2}\right)\,\mathrm{d}v\,\mathrm{d}y$
$\displaystyle=x\int_{-\infty}^{\frac{1}{x}-\frac{1}{z}}\frac{1}{\sqrt{2\pi\alpha}}\left(\frac{1}{x}-y\right)\bigg{(}\exp\left(-\frac{y^{2}}{2\alpha}\right)-\exp\bigg{(}-\frac{(\frac{2\beta}{(1+\beta)x}-y)^{2}}{2\alpha}\bigg{)}\bigg{)}\,\mathrm{d}y$
$\displaystyle=x\int_{0}^{z}\frac{1}{\sqrt{2\pi\alpha}w^{3}}\bigg{(}\exp\bigg{(}-\frac{(\frac{1}{x}-\frac{1}{w})^{2}}{2\alpha}\bigg{)}-\exp\bigg{(}-\frac{(\frac{2\beta}{(1+\beta)x}-\frac{1}{x}+\frac{1}{w})^{2}}{2\alpha}\bigg{)}\bigg{)}\,\mathrm{d}w$

Thus, the Markov kernel $K(x,\mathrm{d}y)$ for $S^{\alpha,\beta}$ under
$\mathbb{P}_{x}$ is given by

$\displaystyle K(x,\mathrm{d}y)$
$\displaystyle=\frac{2}{1+\beta}\Phi\left(-\frac{\beta}{(1+\beta)x\sqrt{\alpha}}\right)\delta_{(1+\beta)x}(\mathrm{d}y)$
$\displaystyle+\frac{x}{\sqrt{2\pi\alpha}y^{3}}\bigg{(}e^{-\frac{(\frac{1}{y}-\frac{1}{x})^{2}}{2\alpha}}-e^{-\frac{(\frac{1}{y}+\frac{\beta-1}{(1+\beta)x})^{2}}{2\alpha}}\bigg{)}\mathbf{1}_{(0,(1+\beta)x)}(y)\,\mathrm{d}y.$

This ends the example.

We finish this section by providing a converse to Proposition 4.1.

######  Theorem 4.4.

Let $X=(X_{t})_{t\geq 0}$ be a continuous positive local
$\mathbb{P}$-martingale that converges to zero $\mathbb{P}\text{-a.s.}$ Then
$X$ is a strict local $\mathbb{P}$-martingale if and only if for all sequences
$a=(a_{n})_{n\in\mathbb{N}}$ and $b=(b_{n})_{n\mathbb{N}}$ converging to
infinity, the measure $\mathbb{P}$ is a bubble measure for the discrete time
martingale $S^{a,b}=(S^{a,b}_{n})_{n\in\mathbb{N}_{0}}$.

###### Proof.

First, assume that $X$ is a strict local martingale. The the result follows
from Proposition 4.1

Conversely, suppose that $X$ is a true $\mathbb{P}$-martingale. As in the
proof of Proposition 4.1, we may assume without loss of generality that $X$ is
the canonical process on $C([0,\infty),(0,\infty])$ with $X_{0}=1$. Then there
exists a measure $\mathbb{Q}$ on $C([0,\infty),(0,\infty])$, with
$\mathbb{Q}|_{\mathcal{F}_{t}}\gg\mathbb{P}|_{\mathcal{F}_{t}}$ for all $t\geq
0$, such that $Y:=1/X$ is a positive $\mathbb{Q}$-martingale that converges to
$0$ $\mathbb{Q}$-almost surely and
$\mathbb{E}^{\mathbb{P}}\left[X_{\tau}\mathbf{1}_{A}\right]=\mathbb{Q}[A]$ for
each bounded stopping time $\tau<\tau_{\infty}$ and $A\in\mathcal{F}_{\tau}$.
By Proposition A.1, there exists an increasing sequence
$a=(a_{n})_{n\in\mathbb{N}}$ converging to infinity such that for each
$k\in\mathbb{N}$,

(4.3) $\mathbb{Q}[Y_{a_{k}}\geq Y_{a_{k+1}}\geq\cdots]=0.$

Since $Y$ is positive $\mathbb{Q}\text{-a.s.}$, we can find an increasing
sequence $b=(b_{n})_{n\in\mathbb{N}}$ converging to infinity such that
$\mathbb{Q}[\underline{Y}_{a_{n}}\leq 1/b_{n}]<2^{-n}$ for all
$n\in\mathbb{N}$. By the Borel–Cantelli lemma, this implies that

(4.4) $\mathbb{Q}\left[\underline{Y}_{a_{k}}\leq\frac{1}{b_{k}}\text{ for
infinitely many }k\right]=0.$

Then for any $k\in\mathbb{N}$, using (4.3) and (4.4),and recalling that each
$\tau^{a,b}_{n}$ is bounded by $a_{n}$, we obtain

$\displaystyle\lim_{n\to\infty}\mathbb{E}^{\mathbb{P}}\left[S^{a,b}_{n}\mathbf{1}_{\\{S^{a,b}_{k}\leq
S^{a,b}_{k+1}\leq\cdots\leq
S^{a,b}_{n}\\}}\right]=\lim_{n\to\infty}\mathbb{Q}\left[Y_{\tau^{a,b}_{n}}\leq
Y_{\tau^{a,b}_{n-1}}\leq\cdots\leq Y_{\tau^{a,b}_{k}}\right]$
$\displaystyle\qquad=\mathbb{Q}\left[Y_{\tau^{a,b}_{k}}\geq
Y_{\tau^{a,b}_{k+1}}\geq\cdots\right]$
$\displaystyle\qquad\leq\mathbb{Q}\left[\underline{Y}_{a_{\ell}}\leq\frac{1}{b_{\ell}}\text{
for infinitely many }\ell\right]+\mathbb{Q}\left[Y_{a_{\ell}}\leq
Y_{a_{\ell-1}}\text{ eventually}\right]=0.$

By Theorem 1.4 , this shows that $\mathbb{P}$ is not a bubble measure for
$S^{a,b}$. ∎

##  Appendix A Auxiliary results

######  Proposition A.1.

Let $(M_{t})_{t\geq 0}$ be a positive continuous local martingale that
$\mathbb{P}\text{-a.s.}$ never becomes constant, i.e.
$\mathbb{P}(M_{t}=M_{\infty}\text{ for all }t\geq s)=0$ for all $s\geq 0$.
Then there exists an increasing sequence of nonnegative real numbers
$(a_{k})_{k\in\mathbb{N}}$ with $\lim_{k\to\infty}a_{k}=\infty$ such that for
each $k\in\mathbb{N}$,

$P[M_{a_{k}}\geq M_{a_{k+1}}\geq\cdots]=0.$

###### Proof.

For $n,m\in\mathbb{N}_{0}$, set
$D^{n}_{m}:=\\{n+j2^{-m}:j\in\mathbb{N}_{0}\\}$. Then for each fixed
$n\in\mathbb{N}_{0}$, $D^{n}_{m}$ is increasing in $m$. Set
$D^{n}_{\infty}:=\bigcup_{m\in\mathbb{N}_{0}}D^{n}_{m}$, which is dense in
$[n,\infty)$. Since $M$ is a continuous local martingale that
$\mathbb{P}\text{-a.s.}$ never becomes constant, its paths
$\mathbb{P}\text{-a.s.}$ never become monotone. By continuity of the paths of
$M$, this implies that for each $n\in\mathbb{N}_{0}$,

$\mathbb{P}[M_{s}\geq M_{t}\text{ for all }s,t\in D^{n}_{\infty}\text{ with
}s<t]=0.$

By $\sigma$-continuity of $\mathbb{P}$, for each $n\in\mathbb{N}_{0}$, there
exists $m_{n}\in\mathbb{N}_{0}$ such that

$\mathbb{P}[M_{s}\geq M_{t}\text{ for all }s,t\in D^{n}_{m_{n}}\text{ with
}s<t]\leq 2^{-n}.$

We may assume without loss of generality that the sequence
$(m_{n})_{n\in\mathbb{N}}$ is nondecreasing. Define the set $D$ by

$D:=\bigcup_{n\in\mathbb{N}_{0}}\left(D^{n}_{m_{m}}\cap[n,n+1)\right).$

Then $D\cap[n,\infty)\supset D^{n^{\prime}}_{m_{n^{\prime}}}$ for all
$n^{\prime}\geq n$. This implies that for each $k\in\mathbb{N}_{0}$,

$\mathbb{P}\left[M_{s}\geq M_{t}\text{ for all }t,s\in D\cap[n,\infty)\text{
with }s<t\right]\leq\lim_{n^{\prime}\to\infty}2^{-n^{\prime}}=0.$

If $(a_{k})_{k\in\mathbb{N}}$ is an enumeration of $D$ in increasing order,
the result follows. ∎

######  Remark A.2.

Note that in the situation of Proposition A.1, in general there also exist
sequences $(a_{k})_{k\in\mathbb{N}}$ with $\lim_{k\to\infty}a_{k}=\infty$ such
that

$P[M_{a_{1}}\geq M_{a_{2}}\geq\cdots]>0.$

For example let $M_{t}=\exp(W_{t}-t/2)$, where $W$ is a Brownian motion.
Define the sequence $(a_{k})_{k\in\mathbb{N}}$ by $a_{1}=1$ and
$a_{k}=a_{k-1}+k$. Then using that Brownian motion has independent and
normally distributed increments, denoting the cdf of a standard normal random
variable by $\Phi$, we obtain

$\mathbb{P}[M_{a_{1}}\geq
M_{a_{2}}\geq\cdots]=\prod_{k=1}^{\infty}\Phi\left(\frac{\sqrt{k}}{2}\right)\geq\prod_{k=1}^{\infty}\left(1-\frac{1}{\sqrt{2\pi}\frac{\sqrt{k}}{2}}\exp\left(-\frac{1}{8}k\right)\right)>0.$

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# Latent Painter

Shih-Chieh Su
Amazon
San Diego, CA 92122
<EMAIL_ADDRESS>
Demo and code available at https://latentpainter.github.io/.

###### Abstract

Latent diffusers revolutionized the generative AI and inspired creative art.
When denoising the latent, the predicted original image at each step
collectively animates the formation. However, the animation is limited by the
denoising nature of the diffuser, and only renders a sharpening process. This
work presents Latent Painter, which uses the latent as the canvas, and the
diffuser predictions as the plan, to generate painting animation. Latent
Painter also transits one generated image to another, which can happen between
images from two different sets of checkpoints.

##  1 Introduction

Can you imagine the generated picture having its own painting action? Or
transit between multiple images generated by different pre-trained diffusers?
Enabling such function extends the capability of the denoising diffusion, and
the proposed method works on any existing diffuser.

Recently, denoising diffusers gain a lot of traction in generative AI, for its
high quality outcome without adversarial training [1], its efficiency [2, 3],
content diversity with easy text conditioning [4], and reasonable footprint
[5]. Although the convenience of text-to-image largely spurs the creativity,
little has be studied about the composition of its generated art. This work
presents Latent Painter, which uses the existing diffuser to generate painting
animation along with the output image.

During the diffusion denoising, the latent is denoised step-by-step into the
state representing an image matching the text input. The predicted original
image, which is the progressive estimate $\hat{x_{0}}$ of the reverse
diffusion process in [1], becomes sharper and sharper when the latent being
denoised. Collecting the predicted original images forms an animation about
the sharpening process, where information is updated omnipresently in the same
frame. However, the update is uneven across frames, with higher total pixel
value change toward earlier frames. Latent painter prioritizes the update
locally to just like the brush strokes. Once the released information is close
enough to match the current predicted original, the residue is accumulated for
later updates. This mechanism provides update more evenly over frames.

##  2 Method

Let $Z(x,y,c)$ denote the state of the latent, having shape of width by height
by channels, and can be initiated as zeros. During each inference step $t$,
the scheduler provides an updated denoised latent sample $D_{t}$. The policy
$\mathcal{P}$ decides whether the information difference between $Z$ and
$D_{t}$ qualifies an update and if so, which subset ($\bar{C}$) of all latent
channels $C$ needs to be updated.

$\bar{C}=\mathcal{P}(Z,D_{t})\text{, where }\bar{C}\in C$ (1)

During a channel update, a channel $c$ is chosen out of $\bar{C}$. Then, a
qualifying threshold $\theta$ picks the region $R$ to be stroked on.

$R=\\{(x,y):|Z(x,y,c)-D_{t}(x,y,c)|>\theta\\}$ (2)

Let $G_{c}(x,y)$ denote the information gain at the current channel $c$.

$G_{c}(x,y)=|Z_{c}(x,y)-D_{t,c}(x,y)|$ (3)

The first stroke is placed at the location whose neighborhood
$\mathcal{N}(x,y)$ has the largest information gain.

$p(\hat{x},\hat{y})=\arg\\!\max_{(x,y)}\sum_{(x^{\prime},y^{\prime})\in\mathcal{N}(x,y)}G_{c}(x^{\prime},y^{\prime})$
(4)

The whole neighborhood $\mathcal{N}(x,y)$, presented as the stroke of the
Latent Painter, is then updated with the current scheduler output.

$Z(x,y,c)=D_{t}(x,y,c)\text{, for all }(x,y)\in\mathcal{N}(\hat{x},\hat{y})$
(5)

The stroke-able region in Eq. 2 is then updated to exclude the newly stroke
region. Following the same procedure, the strokes are placed one by one until
$R$ being empty, or an early stopping criteria $\mathcal{E}$ has been reached.
Continue the stroke action in other channels in $\bar{C}$ likewise.

Upon finishing all $\bar{C}$ channels, the painter steps through another
iteration $t$ of the scheduler to get a new $D_{t}$, then starts from Eq. 1 to
get the channels to be painted this iteration. However, there could be very
little difference between the new $D_{t}$ and current $Z$. This is
particularly true toward the end of the denoising process.

To ensure the animation frames having meaningful update, one idea is to
release $D_{t}$ more evenly across strokes, each being presented in one frame.
Therefore, $\mathcal{P}$ requires the total difference between $Z$ and $D_{t}$
being larger than a portion of the largest total difference over all previous
iterations. With $D_{t}$ sufficiently diverse from $Z$, $\mathcal{P}$ then
provides the channels needing update and to be painted, as in Eq. 1.

Note the current scheduler output $D_{t}$ may not fully pass onto the latent
$Z$ at time $t$, when the early stopping condition $\mathcal{E}$ exists. The
un-updated residue is carried over to next scheduler output $D_{t+1}$. The
accumulated residue becomes the momentum for the next strokes.

###  2.1 Cost

Algorithm 1 Latent Painter - Strokes
Given policy $\mathcal{P}$, stroke qualifier $\theta$, stroke neighborhood
$\mathcal{N}$

Initialize painter state $Z$ as zeros of the latent shape $w\times h\times|C|$

for $t$ in diffuser schedule do

Compute $D_{t}$, the diffuser latent space prediction of $x_{0}$ at current
time $t$

Based on policy $\mathcal{P}$, decides the channels to be updated
$\bar{C}=\mathcal{P}(Z,D_{t})$, where $\bar{C}\in C$

for $c\in\bar{C}$ do

Compute the stroke region $R=\\{(x,y):|Z(x,y,c)-D_{t}(x,y,c)|>\theta\\}$

Initialize move cost $M(x,y)$ as $w\times h$ of ones

while $R$ not empty, as step $s$, do

Compute gain $G_{c}(x,y)=|Z_{c}(x,y)-D_{t,c}(x,y)|$ at channel $c$

Compute motivation $V(x,y)=G_{c}(x,y)\cdot M_{c}(x,y)$

Pick the stroke point
$p(\hat{x},\hat{y})=\arg\\!\max_{(x,y)}\sum_{(x^{\prime},y^{\prime})\in\mathcal{N}(x,y)}V(x^{\prime},y^{\prime})$

Stroke to make $Z(x,y,c)=D_{t}(x,y,c)$ for all
$(x,y)\in\mathcal{N}(\hat{x},\hat{y})$

Update $R\leftarrow R$ \$\\{(x,y):(x,y)\in\mathcal{N}(\hat{x},\hat{y})\\}$

Compute move cost $M(x,y)$ as an inverse Gaussian filter centered
$(\hat{x},\hat{y})$

end while

end for

end for

The human painter typically considers optimizing the effort in painting, such
as to stroke the nearby area first, and keep using the same brush and the same
color as much as possible, before changing or cleaning the brush. On the
machine side, while the convolutional layers are trained to capture congruent
patterns, placing the strokes within the same channel allows more congruent
patterns being stroked.

The moving cost of the brush, denoted by $M_{c}(x,y)$, is modeled as the
inverse of a Gaussian kernel centered at the current stroke location. Rather
than the location with largest information gain, the stroke is placed where
having the largest motivation $V$, which is defined as the information gain
modulated by the moving cost,

$V(x,y)=G_{c}(x,y)\cdot M_{c}(x,y)$ (6)

The motivation-based stroke method is presented in Alg. 1.

##  3 Result

Figure 1: Stroke location of Latent Painter. Within each block, the bottom row
shows the predicted original images $\hat{x_{0}}$ after each of the first 12
denoising iterations, each with only one frame. The top row shows the Latent
Painter stroke snapshots, each containing 20-40 strokes (frames), during the
same period. The middle row are the corresponding accumulated stroke map,
stacked over channels. The animations are available at
https://latentpainter.github.io/

Sample outputs from the Latent Painter Strokes (Alg. 1) are shown in Fig. 1.
The samples were first generated with stable diffusion [5] using a text
sentence. The incurred latent series of the predicted original images is fed
into the painter to produce the strokes. From the bottom rows of each block in
Fig. 1, the denoising outputs quickly converge close to the final state in the
first couple iterations, each of which provides only one frame in the
animation.

In contrast, the Latent Painter slows down the rapid update during early
denoising iterations. This prevents frames being updated too quickly, and
helps rendering the new information more evenly over frames. Here, the new
information is released in the form of strokes. Each denoising iteration can
be released in tens to hundreds of strokes (frames). The stroke maps in the
middle rows indicate the regions with large information gap between $D_{T}$
and $Z$, thus being stroked.

The brighter regions in the stroke map are from the accumulated strokes in
those regions, aggregated across channels. Since the Latent Painter Strokes is
driven by the differential content between denoising updates, the stroke map
provides a way to observe the informative regions that require intensive
updates to yield the final detail.

###  3.1 Stroke Content

Figure 2: Stroke content of Latent Painter. Within each block, the rows show
the cumulative stroke output of channel 0, when it was chosen out of $\bar{C}$
for the first three times as the target channel to stroke.

What have been stroked onto the canvas? Do the spatially and temporally nearby
strokes share similar pattern, color or style? Some stroke samples from the
Latent Painter are shown in Fig. 2. The snapshots each row only accumulates
the strokes from the same channel update, which starts from Eq. 2. Within the
same channel, the strokes tend to be congruent in color or style, or both. The
channels of the latent space are different from those at the visible layer,
where the RGB channels each only account for a color.

Ultimately, the cardinality in color and style is bounded by the number of
trainable neurons, i.e. network parameters. Latent Painter paints at the
latent space, which has only four channels per the diffusion system in [5].
Designed to be a compact representation space, the latent space has only four
channels. Thus even within the same channel, the strokes can contain different
styles and colors. However, the denoising process provides the guidance to
differentiate the further detail within the channel. Therefore, one can attain
better congruence via either lifting the early stop condition $\mathcal{E}$ or
ensuring $Z(x,y,c)=D_{t}(x,y,c)$ for all $(x,y)$ at the end of the update of
current $c$.

###  3.2 Beyond Strokes

(a) glow

(b) dissolve vertical

(c) dissolve

(d) predicted original

Figure 3: Extensions from the strokes. Animations and more examples are
available at https://latentpainter.github.io/.

Besides the painting action in Alg. 1, there are other ways to leverage the
differential response between updates. The glow effect collects updates of the
differential latent into the mass center, where the update radiates
concentrically.

In addition to the content-driven fashion, Latent Painter can also paint
regardless of the differential latent response. Some use cases including the
flip effect that mimics page flipping and the fade effect that release the
update uniformly. The dissolve effect, however, can be either content-driven
or random. Some sample trails of the mentioned effects are visualized in Fig.
3 in the form of update heatmap.

###  3.3 Image Transition

Figure 4: Image to image transiting animation guided by the interpolated
latents from [6]. Within each block, the top row presents the stroke heatmap
after the first 12 channel updates, while the second row presents the update
heatmap of glow effect, both cumulative. The bottom row shows the output
status at the corresponding time points.

Existing image transiting animation is based on interpolating the seeded
embeddings between the source and the destination, or interpolating the
latents, or both. However, it takes either more memory or more time to denoise
the interpolated embeddings. Latent Painter animates based on the predicted
original images from only two denoising trails, the source and the
destination. With the chosen painting effect, the update runs the source image
denoising schedule backward, and then the destination schedule forward.
Through the constraints of the update release, the transition time can be
traded with the detail.

When the source and destination images share a certain part of background,
such as in image editing, the interpolated latents between the source latent
and the destination latent can be used as the prediction guidance. This avoids
using the denoising schedules to guide, while it still requires the denoised
latents. Fig 4 illustrates the painting progress to transit the semantically
edited images from [6].

As an extra credit, Latent Painter can transit the generated images from two
different sets of denoising checkpoints, given the same VAE is used to decode
the latent.

##  4 Conclusion

Latent Painter is presented in this work. It turns existing latent diffuser
output into painting actions via evenly releasing the information update
during the denoising iterations. Several extension beyond the stroke algorithm
has been covered. Since the latent space of the stable diffusion [5] has only
four channels, it is possible to investigate the VAE for further painting
behaviors.

## References

  * [1] Jonathan Ho, Ajay Jain, and Pieter Abbeel.  Denoising diffusion probabilistic models.  Advances in neural information processing systems, 33:6840–6851, 2020.
  * [2] Jiaming Song, Chenlin Meng, and Stefano Ermon.  Denoising diffusion implicit models.  arXiv preprint arXiv:2010.02502, 2020.
  * [3] Tero Karras, Miika Aittala, Timo Aila, and Samuli Laine.  Elucidating the design space of diffusion-based generative models.  Advances in Neural Information Processing Systems, 35:26565–26577, 2022.
  * [4] Aditya Ramesh, Prafulla Dhariwal, Alex Nichol, Casey Chu, and Mark Chen.  Hierarchical text-conditional image generation with clip latents.  arXiv preprint arXiv:2204.06125, 2022.
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  * [6] Manuel Brack, Felix Friedrich, Dominik Hintersdorf, Lukas Struppek, Patrick Schramowski, and Kristian Kersting.  Sega: Instructing diffusion using semantic dimensions, 2023.