<a href="https://colab.research.google.com/github/Jun-629/20MA573/blob/master/src/Hw4_Monotonicity_in_volatility.ipynb" target="_parent"></a>

- __Suppose $f$ is convex and $X$ is submartingale, prove that 
$g(t) = \mathbb E[f(X_t)]$ is increasing.__
 
__Pf:__
Assuming that $X_n$ is a submartingale with respect to the filtration $\{\mathcal{F_n}\}_{n\in\mathbb{N}}$, which means that $$\mathbb E[X_t] \ge \mathbb E[X_s], \forall t \ge s$$

Now, assuming that $X_n \ge 0$ for all n and $f(x) = -x$, which is a convex function. Then we will have
\begin{equation}
\begin{aligned}
\mathbb E[f(X_t) - f(X_s) | \mathcal{F_s}] &= \mathbb E[-X_t - (-X_s)| \mathcal{F_s}] \\
&= X_s - \mathbb E[X_t| \mathcal{F_s}] \\
&\le 0, \forall t \ge s.
\end{aligned}
\end{equation}

This shows that $g(t) = \mathbb E[f(X_t)]$ is strictly decreasing instead of increasing, which means that this is a counter example of the proof.

On the other hand, if the function $f$ is given by the condition that it is increasing, then we will have the inequality as follows by Jensen's Inequality,

$$\mathbb E[f(X_t)|\mathcal{F_s}] \ge f(\mathbb E[X_t|\mathcal{F_s}]) \ge f(X_s), \forall t \ge s$$

then by taking expectation of both sides, we will have the following inequality by the property of conditional expectation:
$$\mathbb E[f(X_t)] = \mathbb E[\mathbb E[f(X_t)|\mathcal{F_s}]] \ge \mathbb E[f(X_s)], \forall t \ge s$$
Therefore, $g(t) = \mathbb E[f(X_t)]$ is increasing.

__Q.E.D__


- __Let $t \mapsto e^{-rt}S_t$ be a martingale, 
then prove that $C(t) = \mathbb E[e^{-rt}(S_t - K)^+]$ is increasing.__

__Pf:__
Assuming that $e^{-rt}S_t$ is a martingale with respect to the filtration $\{\mathcal{F_n}\}_{n\in\mathbb{N}}$, which means
$$\mathbb E[e^{-rt}S_t|\mathcal{F_s}] = e^{-rs}S_s, \forall t \ge s.$$

Since $f(x) = x^+$ is convex function, then

\begin{equation}
\begin{split}
\mathbb E[e^{-rt}(S_t - K)^+|\mathcal{F_s}] &\ge (\mathbb E[e^{-rt}(S_t - K)|\mathcal{F_s}])^+ \\
&= e^{-rs}(S_s-K)^+ \\
\end{split}
\end{equation}

##Can not solve by this since $S_t$ is a random variable
When $S_t \leq K$, then $$\mathbb E[e^{-rt}(S_t - K)^+|\mathcal{F_s}] = 0.$$
when $S_t > K$, then 
\begin{equation}
\begin{split}
\mathbb E[e^{-rt}(S_t - K)^+|\mathcal{F_s}] &= \mathbb E[e^{-rt}(S_t-K)|\mathcal{F_s}]\\
&= \mathbb E[e^{-rt}S_t - K(e^{-rt}-e^{-rs}+e^{-rs})|\mathcal{F_s}] \\
&= e^{-rs}(S_s - K) - K \mathbb E[e^{-rt} - e^{-rs}|\mathcal{F_s}] \\
&\ge e^{-rs}(S_s - K) \\
&\ge e^{-rs}(S_s - K)^+.
\end{split}
\end{equation}
Thus,$$\mathbb E[e^{-rt}(S_t - K)^+|\mathcal{F_s}] \ge e^{-rs}(S_s - K)^+, \forall t \ge s.$$
Therefore, by taking expectation of both sides, we will have
$$\mathbb E[e^{-rt}(S_t - K)^+] = \mathbb E[ \mathbb E[e^{-rt}(S_t - K)^+|\mathcal{F_s}]] \ge \mathbb E[e^{-rs}(S_s - K)^+], \forall t \ge s,$$
which shows that $C(t) = \mathbb E[e^{-rt}(S_t - K)^+]$ is increasing.

__Q.E.D__

- __Suppose $r = 0$ and $S$ is martingale, prove that
$P(t) = \mathbb E [(S_t - K)^-]$ is increasing.__

__Pf:__
Assuming that $S_t$ is a martingale with respect to the filtration $\{\mathcal{F_n}\}_{n\in\mathbb{N}}$, which means
$$\mathbb E[S_t|\mathcal{F_s}] = S_s, \forall t \ge s.$$
When $S_t \ge K$, then $$\mathbb E[(S_t - K)^-|\mathcal{F_s}] = 0.$$
when $S_t < K$, then 
\begin{equation}
\begin{split}
\mathbb E[(S_t - K)^-|\mathcal{F_s}] &= \mathbb E[K - S_t|\mathcal{F_s}]\\
&= K - S_s \\
&\ge (S_s - K)^-.
\end{split}
\end{equation}
Thus,$$\mathbb E[(S_t - K)^-|\mathcal{F_s}] \ge (S_s - K)^-, \forall t \ge s.$$
Therefore, by taking expectation of both sides, we will have
$$\mathbb E[(S_t - K)^-] = \mathbb E[ \mathbb E[(S_t - K)^-|\mathcal{F_s}]] \ge \mathbb E[(S_s - K)^-], \forall t \ge s,$$
which shows that $P(t) = \mathbb E[(S_t - K)^-]$ is increasing.

__Q.E.D__

d645512c3fa2fe1f975ef21f6d146f4b43b1bcb5
ipynb
Jupyter Notebook
src/Hw4_Monotonicity_in_volatility.ipynb
Jun-629/20MA573
addad663d2dede0422ae690e49b230815aea4c70
MIT
src/Hw4_Monotonicity_in_volatility.ipynb
Jun-629/20MA573
addad663d2dede0422ae690e49b230815aea4c70
MIT
src/Hw4_Monotonicity_in_volatility.ipynb
Jun-629/20MA573
addad663d2dede0422ae690e49b230815aea4c70
MIT
2020-02-05T21:42:08.000Z
2020-02-05T21:42:08.000Z
Qwen/Qwen-72B
1. YES
2. YES