# Symmetry of the emergent inductance tensor exhibited by magnetic textures

Soju Furuta  Department of Physics, Tokyo Institute of Technology, Tokyo
152-8551, Japan  Wataru Koshibae  RIKEN Center for Emergent Matter Science
(CEMS), Wako 351-0198, Japan  Fumitaka Kagawa<EMAIL_ADDRESS>Department of Physics, Tokyo Institute of Technology, Tokyo 152-8551, Japan
RIKEN Center for Emergent Matter Science (CEMS), Wako 351-0198, Japan

###### Abstract

Metals hosting gradually varying spatial magnetic textures are attracting
attention as a new class of inductor. Under the application of an alternating
current, the spin-transfer-torque effect induces oscillating dynamics of the
magnetic texture, which subsequently yields the spin-motive force as a back
action, resulting in an inductive voltage response. In general, a second-order
tensor representing a material’s response can have an off-diagonal component.
However, it is unclear what symmetries the emergent inductance tensor has and
also which magnetic textures can exhibit a transverse inductance response.
Here we reveal both analytically and numerically that the emergent inductance
tensor should be a symmetric tensor in the so-called adiabatic limit. By
considering this symmetric tensor in terms of symmetry operations that a
magnetic texture has, we further characterize the magnetic textures in which
the transverse inductance response can appear. This finding provides a basis
for exploring the transverse response of emergent inductors, which has yet to
be discovered.

## Introduction

An inductor is a component that exhibits an inductive counter-electromotive
force, $V$, under a time-varying electric current, $I$, following

$V=L\frac{{\rm d}I}{{\rm d}t},$ (1)

where $L$ denotes the inductance. The electric work done by the external power
supply, $IV$, is hence

$\int{\rm d}t\>IV=\int{\rm d}t\>IL\frac{{\rm d}I}{{\rm d}t}=\int{\rm
d}\left(\frac{1}{2}LI(t)^{2}\right),$ (2)

which shows that the inductor stores an energy of $\frac{1}{2}LI^{2}$. Thus,
it can also be said that an inductor is a component that can store an energy
of $\Delta E=\frac{1}{2}LI^{2}$, under the application of an electric current.
A textbook example is a solenoidal inductor, which stores energy as a
magnetic-field energy Jackson . Other inductors possess similar energy-storing
properties. An established example is the so-called kinetic inductor, in which
the energy is stored as the kinetic energy of mobile charge carriers. When
considering the Drude model of conduction electrons, one can immediately find
that the inductance defined using the imaginary part of the angular-frequency
($\omega$)-dependent resistivity, $\rho(\omega)$, agrees with the inductance
defined using the total kinetic energy of electrons kinetic .

Recently, a new class of inductor, now referred to as emergent inductors, has
been proposed theoretically NagaosaJJAP  and confirmed experimentally
YokouchiNature ; KitaoriPNAS ; KitaoriPRB . In these inductors, the flowing
conduction electrons exert a spin-transfer torque (STT) STT1 ; STT2 ;
YamanouchiNature ; YamaguchiPRL  on the underlying magnetic texture; as a
result, the magnetic texture exhibits time-dependent elastic deformations
under an alternating current (AC) in the linear-response regime. Such current-
induced magnetic texture dynamics exert a back action on the flowing
conduction electrons, yielding the so-called spin-motive force or emergent
electric field (EEF) Volovik ; BarnesPRL2007 ; YangPRL ; JiadongPRL ;
Schulz_Nat.Phys. . This phenomenon can be derived microscopically in terms of
the so-called spin-Berry phase or the effective U(1) gauge field, and the
resulting EEF can be described by

$e_{i}(\bm{r},t)=\frac{\hbar}{2|e|}\bm{m}(\bm{r},t)\cdot[{\partial}_{i}\bm{m}(\bm{r},t)\times{\partial}_{t}\bm{m}(\bm{r},t)],$
(3)

where $e$ $(>0)$ is the elementary charge, $\bm{m}(\bm{r},t)$ is the unit
vector of the local magnetic moment at position $\bm{r}$ and time $t$, and
${\partial}_{i}$ ($i=x,y,z$) and ${\partial}_{t}$ denote spatial and time
derivatives, respectively (when the conduction-electron spins are not fully
polarized, the so-called spin-polarization factor $P$ is further multiplied on
the right-hand side of Eq. (3) Schulz_Nat.Phys. ; BarnesPRL2007 ). It has been
numerically demonstrated that in the so-called adiabatic limit (i.e.,
$\beta=0$; see the Methods section), the inductance value defined using the
EEF under an AC quantitatively agrees with that defined using the current-
induced magnetic-texture-deformation energy Furuta . Thus, in the adiabatic
limit, the emergent inductance is well defined, and both the electric and
energetic responses are correctly captured by Eq. (1). On the other hand, when
nonadiabaticity is concerned (i.e., $\beta\neq 0$), the inductance values
derived independently from the two definitions do not match, implying that the
system responses are beyond the framework of Eq. (1) and hence the inductance
interpretation does not apply.

An interesting aspect of emergent inductors is that the inductive electric
response is potentially not limited to the applied current direction but may
also appear along the perpendicular directions, as inferred from Eq. (3).
Thus, in general, the emergent inductance, when it is well defined, should be
represented by a tensor: $V_{i}=L_{ij}\frac{{\rm d}I_{j}}{{\rm d}t}(i,j=x,y)$
or

$\displaystyle\begin{pmatrix}V_{x}\\\
V_{y}\end{pmatrix}=\begin{pmatrix}L_{xx}&L_{xy}\\\
L_{yx}&L_{yy}\end{pmatrix}\frac{{\rm d}}{{\rm d}t}\begin{pmatrix}I_{x}\\\
I_{y}\end{pmatrix}.$ (4)

In classical electrodynamics, such an inductance tensor with $i,j=1,2$ may be
introduced to describe two mutually coupled coils, 1 and 2. It therefore
appears that an emergent inductor possesses a function similar to that of a
coupled classical inductor system. However, such an intuitive analogy requires
careful consideration because the microscopic mechanism is quite different
between classical and emergent inductors. For instance, in a coupled classical
inductor system, one can analytically express the mutual inductance and find
$L_{12}=L_{21}\equiv M$ Jackson ; moreover, the fact that the coupled system
stores a positive energy in the quadratic form of
$\frac{1}{2}L_{11}I_{1}^{2}+\frac{1}{2}L_{22}I_{2}^{2}+MI_{1}I_{2}$ for
arbitrary values of $I_{1}$ and $I_{2}$ leads to a constraint,
$L_{11}L_{22}\geq M^{2}$ Feynman , in addition to the obvious one,
$L_{11},L_{22}\geq 0$. Such classical electrodynamics considerations, however,
are not helpful for emergent inductors consisting of an arbitrary spin texture
including disorder, and thus, the relation between $L_{xy}$ and $L_{yx}$
appears to be nontrivial.

When considering the nature of $L_{ij}$, it is instructive to review the
resistivity tensor, $\rho_{ij}$, as a textbook example. Note that any second-
order tensor, $K_{ij}$, can always be decomposed into a symmetric part,
$K_{ij}^{\rm S}$, and an antisymmetric part, $K_{ij}^{\rm A}$; namely,
$K_{ij}=K_{ij}^{\rm S}+K_{ij}^{\rm A}$ with $K_{ij}^{\rm S}=(K_{ij}+K_{ji})/2$
and $K_{ij}^{\rm A}=(K_{ij}-K_{ji})/2$. In the case of $\rho_{ij}$, the
symmetric part represents dissipative transport, whereas the antisymmetric
part represents nondissipative transport, that is, the Hall resistivity. Thus,
the symmetric and antisymmetric parts of $\rho_{ij}$ have their own physical
meanings with quite different characteristics. Therefore, the symmetry of the
emergent inductance tensor is also an important issue in understanding the
underlying physics.

In this paper, focusing on the adiabatic limit, in which the emergent
inductance is well defined by Eq. (1) Furuta , we aim to reveal the symmetry
of the emergent inductance tensor and discuss the physical implications of the
revealed symmetry. Our approach is two-fold. First, we consider the tensor-
expressed circuit equation [Eq. (4)] in detail and draw a conclusion regarding
the symmetry of $L_{ij}$: this also enables us to discuss how the inductor
tensor should behave under the time-reversal operation. Second, we numerically
investigate $L_{ij}$ for various magnetic textures using micromagnetic
simulations. These two approaches consistently show that $L_{ij}$ is a
symmetric tensor (that is, $L_{xy}=L_{yx}$) and $L_{ij}$ is even under the
time-reversal operation. By combining the numerical results and symmetry
arguments, we also find what kinds of magnetic textures can or cannot exhibit
a transverse emergent inductance, $L_{yx}$. We note that the present
conclusion is for the case where the emergent inductance is well defined
(i.e., the adiabatic limit, $\beta=0$). The effect of nonadiabaticity (i.e.,
$\beta\neq 0$), which makes the emergent inductance ill-defined Furuta , is
discussed in the Supplementary Information.

## Results

### Considerations for the circuit equation

We discuss the consequences that are prescribed in the tensor-expressed
circuit equation, Eq. (4). Following the general arguments on a second-order
tensor, we decompose $L_{ij}$ into symmetric and antisymmetric parts:
$L_{ij}=L_{ij}^{\rm S}+L_{ij}^{\rm A}$, or explicitly,

$\begin{pmatrix}L_{xx}&L_{xy}\\\
L_{yx}&L_{yy}\end{pmatrix}=\begin{pmatrix}L_{xx}^{\rm S}&L_{xy}^{\rm S}\\\
L_{xy}^{\rm S}&L_{yy}^{\rm S}\end{pmatrix}\\\ +\begin{pmatrix}0&L_{xy}^{\rm
A}\\\ -L_{xy}^{\rm A}&0\end{pmatrix}.$ (5)

To gain insight into the physical meaning of the symmetric and antisymmetric
tensors, we consider the work done by the power source along a closed loop in
the $I_{x}$–$I_{y}$ plane, which is expressed as:

$\displaystyle\begin{split}\oint{\rm d}t\>(I_{x}V_{x}+I_{y}V_{y})=\oint{\rm
d}\left(\frac{1}{2}L_{xx}^{\rm S}I_{x}^{2}+\frac{1}{2}L_{yy}^{\rm
S}I_{y}^{2}+L_{xy}^{\rm S}I_{x}I_{y}\right)\\\ +\oint{\rm d}t\>L_{xy}^{\rm
A}\left(I_{x}\frac{{\rm d}}{{\rm d}t}I_{y}-I_{y}\frac{{\rm d}}{{\rm
d}t}I_{x}\right).\end{split}$ (6)

Note that the first term in the right-hand side consists of only the symmetric
tensor components and the integrand takes the form of a total derivative;
hence, the contour integral results in zero. The expression,
$\frac{1}{2}L_{xx}^{\rm S}I_{x}^{2}+\frac{1}{2}L_{yy}^{\rm
S}I_{y}^{2}+L_{xy}^{\rm S}I_{x}I_{y}$, is essentially the same as that derived
for mutually coupled classical inductors, representing an energy stored in the
emergent inductor under a current. In contrast, the second term consists of
only the antisymmetric-tensor components, and the integrand is not the form of
a total derivative, indicating that the second term is nonzero and dependent
on the path. These features imply that $L_{xy}^{\rm A}$ is associated with a
non-conserved quantity.

Figure 1:  A specific closed loop used to prove the absence of the
antisymmetric components of an inductance tensor.

To see the consequences of the antisymmetric component $L_{xy}^{\rm A}$ more
clearly, it is helpful to consider a specific closed path for the integral(s)
of Eq. (6). Suppose $L_{xy}^{\rm A}>0$; we consider a specific cycle $C$ that
consists of three paths, $C_{1}$, $C_{2}$ and $C_{3}$, as shown in Fig. 1:
$(I_{x},I_{y})=(0,0)\xrightarrow{C_{1}}(I_{0},I_{0})\xrightarrow{C_{2}}(I_{0},0)\xrightarrow{C_{3}}(0,0)$
with constraints of $I_{x}=I_{y}$ on $C_{1}$, $I_{x}=I_{0}$ on $C_{2}$, and
$I_{y}=0$ on $C_{3}$. Thus, taking the contour integral along the cycle in the
clockwise direction results in:

$\oint_{C}{\rm d}t\>(I_{x}V_{x}+I_{y}V_{y})=-L_{xy}^{\rm A}I_{0}^{2}<0.$ (7)

The result indicates that if a positive $L_{xy}^{\rm A}$ were present, the
power source could acquire energy by cycling the closed loop. Such behaviour
is obviously not allowed for a passive element, such as a stable material.
Similarly, one can consider the case of $L_{xy}^{\rm A}<0$, and the same
conclusion can be drawn by considering the same closed loop $C$ but in the
counterclockwise direction. Thus, Eq. (4) concludes that even for the case of
an emergent inductor, the inductance tensor cannot have an antisymmetric
component; that is, $L_{xy}^{\rm A}=0$, and an emergent inductor tensor should
be a symmetric tensor (below, we therefore omit the superscript, S),

$L_{xy}=L_{yx}\equiv L_{tr}.$ (8)

Hence, the energy stored in an emergent inductor under current is found to be
expressed by
$\frac{1}{2}L_{xx}I_{x}^{2}+\frac{1}{2}L_{yy}I_{y}^{2}+L_{tr}I_{x}I_{y}$, and
for this quadratic form to be nonnegative, $L_{ij}$ should satisfy

$L_{xx}L_{yy}\geq(L_{tr})^{2},$ (9)

in addition to $L_{xx},L_{yy}\geq 0$ Furuta .

Thus, although the microscopic mechanism is quite different between classical
and emergent inductors, it turns out that there is no difference in the
constraints that inductance tensors should satisfy. These characteristics are
implicitly prescribed by the relation between voltage and current, Eq. (4),
not depending on the microscopic mechanism for inductors.

Having established the symmetry of $L_{ij}$, we can discuss the behaviour of
$L_{ij}$ under the time-reversal operation. Since an emergent inductance
arises from a magnetic texture $\\{\bm{m}(\bm{r})\\}$, the behaviour of an
emergent inductance under the time-reversal operation is an interesting issue.
In fact, in experiments, the magnetic-field ($\bm{B}$)-dependence of an
emergent inductance has been frequently investigated YokouchiNature ;
KitaoriPNAS ; KitaoriPRB . To incorporate a case where $\\{\bm{m}(\bm{r})\\}$
shows hysteretic behaviour with respect to changes in $\bm{B}$, $L_{ij}$ may
be expressed as a function with $\\{\bm{m}(\bm{r})\\}$ and $\bm{B}$ as
variables. Note that regardless of the details of the variables, $L_{ij}$
should be a symmetric tensor as discussed above, and hence,
$L_{ij}(\bm{B},\\{\bm{m}(\bm{r})\\})=L_{ji}(\bm{B},\\{\bm{m}(\bm{r})\\})$
should always be satisfied. Moreover, with respect to the complex resistivity,
Onsager’s reciprocal theorem concludes
$\operatorname{Im}\rho_{ij}(\omega,\bm{B},\\{\bm{m}(\bm{r})\\})=\operatorname{Im}\rho_{ji}(\omega,-\bm{B},\\{-\bm{m}(\bm{r})\\})$
(the real part also satisfies the same relation) Kubo ; hence, the inductance
tensor should also satisfy
$L_{ji}(\bm{B},\\{\bm{m}(\bm{r})\\})=L_{ij}(-\bm{B},\\{-\bm{m}(\bm{r})\\})$.
By combining the two relations regarding $L_{ij}$, one can thus conclude

$L_{ij}(\bm{B},\\{\bm{m}(\bm{r})\\})=L_{ij}(-\bm{B},\\{-\bm{m}(\bm{r})\\}).$
(10)

This relation indicates that $L_{ij}$ is even under time-reversal, or
equivalently, $L_{ij}$ is a polar symmetric tensor. In particular, we note
$L_{yx}(\bm{B},\\{\bm{m}(\bm{r})\\})=L_{yx}(-\bm{B},\\{-\bm{m}(\bm{r})\\})$,
distinct from the Hall resistivity, which satisfies
$\rho_{yx}(\bm{B},\\{\bm{m}(\bm{r})\\})=-\rho_{yx}(-\bm{B},\\{-\bm{m}(\bm{r})\\})$.
For this reason, we call $L_{yx}$ the transverse inductance, not the Hall
inductance.

### Micromagnetic simulations

To observe the symmetry of the inductance tensor of emergent inductors, we
consider magnetic textures that slowly vary in space; for such magnetic
textures, the EEF can be calculated according to Eq. (3). We further consider
the pinned regime, in which a magnetic texture does not exhibit a steady flow
under a direct current Nattermann ; ChauvePRB ; Kleemann ; IntrinsicTatara ;
ThiavilleEPL ; ExtrinsicTatara ; IntrinsicOno ; TataraReview ;
ExtrinsicNatPhys ; Iwasaki_Nat.Commun. . The procedure for calculating the
emergent inductance arising from a slowly varying magnetic texture in the
pinned regime is detailed in the literature Furuta  and also in the Methods
section. We consider a spin Hamiltonian based on the continuum approximation
that can exhibit helical and skyrmion-lattice (SkL) Bogdanov1 ; Bogdanov2 ;
Muhlbauer_Science ; Yu_Nature  magnetic textures and calculate the current-
induced dynamics of a magnetic texture by numerically solving the Landau-
Lifshitz-Gilbert (LLG) equation LLG  (see the Methods section). To be more
specific, the magnetic texture dynamics under the application of an AC along
the $x$-direction are calculated by micromagnetic simulation; then, by
referring to Eq. (3), the time-dependent EEFs along both the $x$ and $y$
directions are further derived; and finally, by referring to Eq. (4), $L_{xx}$
and $L_{yx}$ are obtained. Similarly, we obtain $L_{yy}$ and $L_{xy}$ by
simulating the case of an AC along the $y$ direction. The emergent inductance
$L_{ij}$ depends on the system dimension in the form of
$L_{ij}=\tilde{L}_{ij}\frac{\ell}{S}$, where $\tilde{L}_{ij},\ell$, and $S$
represent the normalized inductance (we call it “inductivity”), system length,
and sample cross-section area. Below, we therefore present $\tilde{L}_{ij}$,
rather than the system-size-dependent $L_{ij}$. The inductivity tensor may be
defined by

$e_{i}=\tilde{L}_{ik}\frac{{\rm d}j_{k}}{{\rm d}t},$ (11)

where $j$ represents the current density. The following simulation results are
obtained for the case of $\beta=0$ (i.e., the adiabatic limit).

Figure 2:  Various metastable magnetic textures and corresponding inductance
tensors. a-e Helical magnetic textures with the helical ${\bm{q}}$-vector that
forms approximately an angle $\theta=0^{\circ}$ (a), $20^{\circ}$ (b),
$-20^{\circ}$ (c), $45^{\circ}$ (d), and $-45^{\circ}$ (e) with respect to the
$x$ direction. f Maze helix. g Skyrmion lattice. The corresponding fast-
Fourier-transform (FFT) images are also shown in each panel. Color wheels
specify the $x$-$y$ plane magnetization direction. The brightness of the color
represents the $z$ component of the magnetization, and white represents the
local magnetizations pointing toward the $z$ direction. The current-induced
magnetic texture dynamics are calculated under the application of a weak AC.
The parameters used for the simulation are tabulated in Table I; they are
chosen so that the resulting emergent voltage is in the linear-response and
low-frequency regimes (see the Methods section). The simulations were done for
$\beta=0$.

Figure 2 summarizes the magnetic textures investigated in this study and the
corresponding inductance tensors. We studied four examples of helical magnetic
textures, for which the helical ${\bm{q}}$-vector forms approximately an angle
$\theta=0^{\circ},\pm 20^{\circ}$, and $\pm 45^{\circ}$ with respect to the
$x$ direction (Fig. 2a–e, respectively); a maze-helix texture (Fig. 2f); and
an SkL (Fig. 2g). The intensity and concentration of disorder were minimized
as much as possible while confirming the linear response of the pinned
dynamics. As a result, more disorder had to be included when examining the
maze-helix and SkL, as summarized in Table I: Selecting a much lower current
density while keeping the disorder density as low as 0.3 $\%$ was not
appropriate in terms of the required numerical accuracy. As shown in Fig. 2,
we find that $\tilde{L}_{xy}=\tilde{L}_{yx}$ invariably holds within the
numerical error, consistent with the conclusion derived from the circuit
equations.

Table 1: Parameters used for the micromagnetic simulations displayed in Fig. 2. The disorder density was chosen as low as possible while confirming the linear response of the pinned magnetic textures under a given current density.  Magnetic texture | Magnetic field | $K_{\rm imp}$ | Disorder density | Current density | Frequency
---|---|---|---|---|---
| (T) |  $\times 10^{7}$ (J m-3) | ($\%$) |  $\times 10^{10}$ (A m-2) | (MHz)
Helix ($\theta=0,\pm 20^{\circ},\pm 45^{\circ}$) | 0 | 0.1 | 0.3 | 5.0 | 50
Maze helix | 0 | 2.0 | 3 | 2.0 | 50
SkL | 0.3 | 1.0 | 3 | 1.0 | 10
Figure 3:  Magnetic-field dependence of the transverse inductivity. H, Sk, and
F denote the single ${\bm{q}}$-helix with $\theta=45^{\circ}$, skyrmion
lattice, and ferromagnetic state, respectively. The disorder density and
strength $K_{\rm imp}$ (see Methods) is fixed to 3 $\%$ and 1.0$\times$107 J
m-3. The simulations were done for $\beta=0$. When $\beta$ is non-zero, the
effective inductivity tensor $\tilde{L}^{\rm eff}_{ij}$ should be discussed
($\tilde{L}^{\rm
eff}_{ij}=\operatorname{Im}[\rho_{ij}(\omega)-\rho_{ij}(0)]/\omega$): The
magnetic-field dependence of the transverse components of $\tilde{L}^{\rm
eff}_{ij}$ at finite $\beta$ is shown in Fig. S1 and discussed in
Supplementary Note 2. Note that $\tilde{L}^{\rm eff}_{ij}$ is a different
quantity from the inductivity tensor defined by Eq. (11).

The diagonal components of the inductance tensor are invariably positive,
whereas the off-diagonal components can be either positive or negative.
Nevertheless, we emphasize that the inductance tensor retains energetic
interpretations; that is, the energy increase in the magnetic system, $\Delta
E(I_{x},I_{y})$, caused by the application of an electric current agrees with
$\frac{1}{2}L_{xx}I_{x}^{2}Fig.~{}3+\frac{1}{2}L_{yy}I_{y}^{2}+L_{tr}I_{x}I_{y}$.
As an example, we discuss the results for the helical texture with
$\theta=-20^{\circ}$, in which the off-diagonal components of $\tilde{L}_{ij}$
are negative. The $\Delta E(I_{x},I_{y})$ is calculated for the following
three cases independently: (i) $(I_{x}\neq 0,I_{y}=0)$, (ii)
$(I_{x}=0,I_{y}\neq 0)$, and (iii) $(I_{x}\neq 0,I_{y}\neq 0)$. Then, by
solving the three simultaneous equations regarding $\Delta
E(I_{x},I_{y})=\frac{1}{2}L_{xx}I_{x}^{2}+\frac{1}{2}L_{yy}I_{y}^{2}+L_{tr}I_{x}I_{y}$,
we can obtain:
$(\tilde{L}_{xx},\tilde{L}_{yy},\tilde{L}_{tr})=(2.67,0.63,-1.00)\times
10^{-21}$ H m. These values are in quantitative agreement with
$\tilde{L}_{ij}$ calculated from the EEF (Fig. 2c), indicating that the
emergent inductivity is well defined by Eq. (11). We also confirmed
$\tilde{L}_{xy}(\bm{B},{\bm{m}(\bm{r}_{k})})=\tilde{L}_{xy}(-\bm{B},{-\bm{m}(\bm{r}_{k})})$
numerically (Fig. 3), in which the disorder density and strength $K_{\rm imp}$
(see Methods) are fixed to 3 $\%$ and 1.0$\times$107 J m-3, respectively, and
the single-${\bm{q}}$ helix with $\theta=45^{\circ}$ was considered. Thus, our
numerical study confirms that $\tilde{L}_{ij}$ for an emergent inductor is a
polar symmetric tensor.

When comparing the three helical textures quantitatively, one can find that as
the $\theta$ increases from 0∘ to 45∘, $\tilde{L}_{xx}$ decreases, whereas
$\tilde{L}_{xy}$ increases. We also note that in the maze-helix and SkL
textures, the transverse component, $\tilde{L}_{xy}$, is more than one order
of magnitude smaller than the longitudinal components, $\tilde{L}_{xx}$ and
$\tilde{L}_{yy}$. As discussed below, these observations can be explained by
considering an orthogonal transformation of $\tilde{L}_{ij}$ and the
rotational symmetry that each magnetic texture has.

## Discussion

In the following, we aim to categorize inductance tensors of a magnetic
texture origin and consider how our numerical results obtained in the
adiabatic limit can be explained in terms of the symmetry operations that each
magnetic system has. Note that because an inductance tensor is real and
symmetric, it can be diagonalized by performing an appropriate orthogonal
transformation, $R$, or equivalently by choosing appropriate Cartesian
coordinates:

$\begin{pmatrix}\tilde{L}_{xx}&\tilde{L}_{tr}\\\
\tilde{L}_{tr}&\tilde{L}_{yy}\end{pmatrix}\xrightarrow{R}\begin{pmatrix}\lambda_{1}&0\\\
0&\lambda_{2}\end{pmatrix},$ (12)

where $\lambda_{1}$ and $\lambda_{2}$ ($\lambda_{1},\lambda_{2}\geq 0$)
represent the eigenvalues of the inductance tensor. Hence, to classify an
emergent inductance tensor, it is sufficient to consider the diagonalized
form. This approach does not lose generality because a representation in
different Cartesian coordinates can be immediately obtained by performing the
corresponding orthogonal transformation. Following the group theory arguments
for a polar symmetric tensor, one can conclude that: (i) when the system has
three-fold or higher rotational symmetry with respect to the $z$ axis (i.e.,
$C_{3z},C_{4z},C_{6z}$ or $C_{\infty z}$), $\lambda_{1}$ and $\lambda_{2}$
should be equal, whereas (ii) when the system has only two-fold with respect
to the $z$ axis ($C_{2z}$) or no rotational symmetry, $\lambda_{1}$ and
$\lambda_{2}$ should be inequivalent: For the details of the derivation, see
Supplementary Note 3. Thus, the diagonalized two-by-two tensor can be
classified as one of the two categories, which are characterized by
$\lambda_{1}=\lambda_{2}$ and $\lambda_{1}\neq\lambda_{2}$, respectively.

The first category, $\lambda_{1}=\lambda_{2}$, is represented by an isotropic
tensor $\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}$, and thus, the off-diagonal
components are always zero for arbitrarily chosen Cartesian coordinates; that
is, the transverse inductance response does not appear. From group theory, a
magnetic texture that has $C_{3z},C_{4z}$, $C_{6z}$ or $C_{\infty z}$ symmetry
should belong to this category. Note that our numerical calculations deal with
finite-size systems including randomly distributed disorder, and therefore,
the simulated magnetic textures do not have any rotational symmetry in a
strict sense. Nevertheless, we numerically find that the inductance tensors of
the maze-helix and SkL textures satisfy $\tilde{L}_{xx}\approx\tilde{L}_{yy}$
and $\tilde{L}_{xy},\tilde{L}_{yx}\ll\tilde{L}_{xx},\tilde{L}_{yy}$ (Fig. 2f
and g, respectively), indicating that the obtained tensors are close to
isotropic. These results appear reasonable, considering that in a macroscopic
system, the maze-helix and SkL textures have global approximate $C_{\infty z}$
and $C_{6z}$ symmetries, respectively. The symmetry of the macroscopic systems
can be imagined by looking at the corresponding fast-Fourier-transform (FFT)
images. To be precise, the rotation symmetry of the SkL confined in the
finite-size system is $C_{2z}$, rather than $C_{6z}$, as indicated by the FFT
image (Fig. 2g): This perturbative symmetry lowering from $C_{6z}$ to $C_{2z}$
explains the small but finite symmetric off-diagonal component, which is
originally prohibited under $C_{6z}$ symmetry. When nonadiabaticity is not
negligible, the effective inductivity tensor defined by $\tilde{L}^{\rm
eff}_{ij}=\operatorname{Im}[\rho_{ij}(\omega)-\rho_{ij}(0)]/\omega$ is
discussed, but it should be noted that $\tilde{L}^{\rm eff}_{ij}$ is a
different quantity from the inductivity tensor in Eq. (11). For instance, the
$\tilde{L}^{\rm eff}_{ij}$ of the SkL has antisymmetric off-diagonal
components when $\beta\neq 0$, although the antisymmetric component in
$\tilde{L}_{ij}$ is energetically prohibited; for more details, see
Supplementary Notes 1 and 3.

Figure 4:  Numerically obtained inductivity tensors of helical magnetic
textures for various ${\bm{q}}$-directions. The $\theta=0$ represents the
${\bm{q}}$-direction parallel to the $x$ axis. The solid symbols are the data
obtained by the micromagnetic simulations, and the solid curves represent the
corresponding trigonometric functions multiplied by $\lambda_{1}=3.11\times
10^{-21}$ H m. The simulations were done for $\beta=0$.

The second category consists of tensors that have two inequivalent components,
$\begin{pmatrix}\lambda_{1}&0\\\ 0&\lambda_{2}\end{pmatrix}$, and thus, off-
diagonal components can appear if arbitrary Cartesian coordinates are chosen.
For instance, the matrix $R(\theta)$ that rotates Cartesian coordinates
clockwise by $\theta$ transforms the diagonalized tensor into a nondiagonal
form:

$\displaystyle\begin{pmatrix}\lambda_{1}&0\\\
0&\lambda_{2}\end{pmatrix}\xrightarrow{R(\theta)}\begin{pmatrix}\lambda_{1}\cos^{2}\theta+\lambda_{2}\sin^{2}\theta&(\lambda_{1}-\lambda_{2})\sin\theta\cos\theta\\\
(\lambda_{1}-\lambda_{2})\sin\theta\cos\theta&\lambda_{1}\sin^{2}\theta+\lambda_{2}\cos^{2}\theta\end{pmatrix}.$
(13)

Thus, $\tilde{L}_{xy}=\tilde{L}_{yx}$ can be either positive or negative
depending on the selection of Cartesian coordinates. An example of this
category is a single-${\bm{q}}$ helix, in which the inductance tensor is
diagonalized, for instance, when the $x$-axis is chosen parallel to the
helical-${\bm{q}}$ vector. An important feature of an ideal single-${\bm{q}}$
helix is that the local magnetic moments show no modulation along the
direction perpendicular to ${\bm{q}}$. Hence, no STT effect is expected for
the current along the $y$-axis, resulting in $\lambda_{2}=0$. Thus, for the
case of an ideal single-${\bm{q}}$ helix, the diagonalized form and its
orthogonal transformation are given as:

$\displaystyle\begin{pmatrix}1&0\\\
0&0\end{pmatrix}\xrightarrow{R(\theta)}\begin{pmatrix}\cos^{2}\theta&\sin\theta\cos\theta\\\
\sin\theta\cos\theta&\sin^{2}\theta\end{pmatrix}.$ (14)

Figure 4 displays the comparison between the numerically obtained inductivity
tensors of various ${\bm{q}}$-direction helices and the orthogonal
transformation of $\lambda_{1}=3.11\times 10^{-21}$ H m and $\lambda_{2}=0$,
which is an approximate inductivity tensor of the single-${\bm{q}}$ helix with
$\theta=0^{\circ}$ (Fig. 2a). Although the simulated single-${\bm{q}}$ helices
are more or less affected by random disorder and the open boundaries of the
system, the overall tendency is well reproduced by orthogonal transformation.
This observation demonstrates that the above arguments based on orthogonal
transformation of the symmetric inductivity tensor are helpful when
considering a single-${\bm{q}}$ helix with arbitrary ${\bm{q}}$ direction.

For a more complicated magnetic texture, positive $\lambda_{1}$ and
$\lambda_{2}$ with $\lambda_{1}\neq\lambda_{2}$ may be expected. For instance,
a mutidomain of single-$\bm{q}$-helices obviously belongs to this category. In
contrast, a long-range ordered state belonging to this category may not be so
clear. A candidate of this category is likely a magnetic texture that has
multiple ${\bm{q}}$ vectors with different wavenumbers, such as those observed
in EuAg4As2 EuAg4As2  and EuAl4 OrthoSk ; however, such an anisotropic
magnetic texture is beyond the scope of our model Hamiltonian based on the
continuum approximation [see Eq. (15) in the Methods section].

To conclude, we have revealed analytically and numerically the symmetry of the
emergent inductance tensor exhibited by pinned magnetic textures. We focused
on the adiabatic limit, where the inductance tensor is well defined by
$e_{i}=\tilde{L}_{ik}\frac{{\rm d}j_{k}}{{\rm d}t}$. We thus found that the
inductance tensor is a real symmetric tensor, and hence, the presence and
magnitude of the transverse component are determined by the degree to which
the measurement axis is tilted from the principal axis that generates the
diagonalized tensor. As a natural consequence of the real symmetric tensor,
the transverse component does not change sign with respect to the magnetic
field reversal. These fundamental aspects of the emergent inductance tensor
will be useful when exploring the transverse inductive response in a magnetic
texture. However, it must also be noted that when nonadiabaticity is not
negligible, the electric response produced by magnetic textures may not be
described by an inductance tensor in the strict sense defined by Eq. (4).

## Methods

### Numerical model

In this study, we consider long-period helical magnetic textures that are
stabilized by the Dzyaloshinskii-Moriya (DM) interaction Dzyaloshinskii ;
Moriya . We consider both a clean system without any disorder and dirty
systems including randomly distributed disorder. Our model Hamiltonian is:

$\displaystyle\begin{split}\mathscr{H}=&\int\frac{{\rm
d}^{3}r}{a^{3}}\left[\frac{J}{2}(\nabla\bm{m})^{2}+D\bm{m}\cdot(\nabla\times\bm{m})\right]\\\
&-\sum_{k\in\Lambda}\int_{V_{k}}{\rm d}^{3}r\>K_{{\rm
imp}}(\bm{m}_{k}\cdot\bm{n}_{{\rm imp},k})^{2}\end{split}$ (15)

where $J$ is the Heisenberg exchange energy, $D$ is the DM interaction and $a$
is the lattice constant. The extrinsic pinning effect is controlled by the
last term of Eq. (15), which is introduced to randomly selected cells to break
the translational symmetry: $K_{{\rm imp}}(>0)$ represents the magnetic-easy-
axis anisotropy along a randomly chosen direction, $\bm{n}_{{\rm imp},k}$, at
the $k$-th cell (the cell volume $V_{k}$ is $3^{3}$ nm3), and $\Lambda$ is a
set of random numbers. The disorder density displayed in Table I represents
the ratio of the number of cells with finite $K_{{\rm imp}}$ to the total
number of cells (243$\times$243).

When simulating the current-induced dynamics of a given helical magnetic
structure, we insert the spin Hamiltonian into the following Landau-Lifshitz-
Gilbert (LLG) equation LLG :

$\displaystyle\begin{split}\frac{{\rm d}\bm{m_{r}}(t)}{{\rm
d}t}=-\frac{|\gamma|}{1+\alpha^{2}}\frac{{\rm d}\mathscr{H}}{{\rm
d}\bm{m_{r}}}&\times\bm{m_{r}}-\frac{\alpha|\gamma|}{1+\alpha^{2}}\left[\bm{m_{r}}\times\left(\frac{{\rm
d}\mathscr{H}}{{\rm d}\bm{m_{r}}}\times\bm{m_{r}}\right)\right]\\\
+\frac{1}{1+\alpha^{2}}\\{(1&+\beta\alpha)\bm{m_{r}}\times[\bm{m_{r}}\times(\bm{u}\cdot\bm{\nabla})\bm{m_{r}}]\\\
&+(\beta-\alpha)[\bm{m_{r}}\times(\bm{u}\cdot\bm{\nabla})\bm{m_{r}}]\\},\end{split}$
(16)

where $\bm{u}$ represents the spin drift velocity, $\alpha$ is the Gilbert
damping constant, $\beta$ is a dimensionless constant that characterizes the
nonadiabatic electron spin dynamics, and $\gamma$ $(>0)$ is the gyromagnetic
ratio; $\bm{u}$ is related to the electric current density $\bm{j}$ by
$\bm{u}=\frac{P\mu_{\rm B}}{2|e|M_{\rm s}(1+\beta^{2})}\bm{j}$, where
$\mu_{\rm B}$ is the Bohr magneton and $M_{\rm s}$ is the saturation
magnetization. When implementing the micromagnetic simulation, we use the open
software MuMax3 Mumax1 ; Mumax2 . We choose the following parameter set:
$J/(2a^{3})=1.8\times 10^{-11}$ J m-1, $D/a^{3}=2.8\times 10^{-3}$ J m-2,
$M_{\rm s}=2.45\times 10^{5}$ A m-1, $P=1$, and $\alpha=0.04$.

In the simulation, we apply a current density of a sufficiently small
magnitude so that the magnetic system is certainly in the linear-response
regime; that is, with respect to the input alternating electric current along
the $x$ or $y$ direction, $j_{i}(t)=j_{0,i}\sin\omega t$ ($i=x,y$), the
magnetic system is in the pinned regime, and the output AC emergent voltage,
$V_{e,i}(t)$, is $\propto j_{0,k}\omega\cos\omega t$ $(i,k=x,y)$. Based on
these observations, $L_{ij}$ is derived from the following equations:

$\displaystyle V_{e,i}(t)=\langle e_{i}(t)\rangle\ell=L_{ij}\frac{{\rm
d}(I_{j}(t))}{{\rm d}t},$ (17)

where $\langle\cdots\rangle$ denotes a spatially averaged value, the system
length $\ell$ is $243\times 3$ nm, and $I=jS$ with the cross-section area
$S=243\times 1\times 3^{2}$ nm2. In the present frequency range ($\leq$ 100
MHz), it is confirmed that the inductivity is independent of $\omega$ (i.e.,
$\langle e_{i}\rangle\propto\omega$) (Fig. S2) and the $\alpha$ dependence of
the numerical results is negligibly small (Fig. S3) (see also Supplementary
Note 4). The numerical accuracy of MuMax3 is $\Delta{\bm{m}}/|{\bm{m}}|\sim
10^{-7}$, and the typical increment in one time step (4 ps) is $\sim$$10^{-5}$
under the current application of $\sim$1010 A m-2. This finite accuracy
eventually gives rise to an uncertainty of $\sim$$10^{-23}$ H m in the
calculated inductivity.

In the numerical simulation, a uniform current density is considered to
understand fundamental aspects of the inductivity tensor. On the other hand,
the local $\rho_{xx}$ and $\rho_{yx}$ may be non-uniform in real material,
reflecting spatial variations in magnetic textures. Nevertheless, the uniform
current is a good approximation as long as
$\langle\rho_{xx}\rangle\gg\delta\rho_{xx},\delta\rho_{yx}$, where
$\delta\rho_{xx}$ and $\delta\rho_{yx}$ represent magnitude of the spatial
variations. For instance, in the chiral magnet MnSi at 10 K, the presence or
absence of the metastable skyrmion lattice changes $\rho_{xx}$ and $\rho_{yx}$
by $\approx$50 n$\Omega$ cm and $\approx$30 n$\Omega$ cm, respectively,
whereas $\rho_{xx}\approx 5$ $\mu$$\Omega$ cm Oike . Such magnetic-texture-
dependent $\rho_{xx}$ and $\rho_{yx}$ imply that
$\langle\rho_{xx}\rangle\gg\delta\rho_{xx},\delta\rho_{yx}$ holds, although
the precise estimation of the spatial variations is experimentally difficult;
thus, the current uniformity is well expected. If $\delta\rho_{xx}$ and
$\delta\rho_{yx}$ are significant, the current distribution should be
determined self-consistently; for instance, see inhomo .

### Initial-state preparation

To obtain various metastable magnetic textures, a pristine helical texture
with a different oblique angle of the helical ${\bm{q}}$-vector, a random spin
configuration, or an SkL is prepared as an initial state and then relaxed
under zero current. Note that imposing the open-boundary condition and
introducing impurity sites are key in obtaining the intended magnetic
textures.

## Data availability

The data used in this work are available from the corresponding author upon
reasonable request.

## Acknowledgments

The authors thank N. Nagaosa and Y. Fujishiro for their valuable discussions.
This work was partially supported by JSPS KAKENHI (Grants No. 20K03810, No.
18H05225, No. 23K03291 and No. 21H04442), JST CREST (Grants No. JPMJCR1874 and
No. JPMJCR20T1).

## Competing interests

The authors declare no competing interests.

## Author contributions

S.F. conducted the calculations and analyzed the data. F.K. conceived the
project and wrote the draft with S.F. and W.K. All the authors discussed the
results and commented on the manuscript.

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  * (1)
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## Supplementary Note 1: Impact of nonadiabaticity

In the literature [16], it has been numerically demonstrated that when
nonadiabaticity is considered (i.e., $\beta\neq 0$), the inductivity value
defined using the emergent electric field under an AC current does not agree
with that defined using the current-induced energy increase. This observation
contradicts the conclusion derived from $e=\tilde{L}\frac{{\rm d}j}{{\rm d}t}$
and therefore implies that a simple equivalent circuit, such as a resistance
($R$)–inductance ($L$) series or $R$$\parallel$$L$ parallel circuit, can not
consistently explain the electric and energetic responses of the system under
the current. This fact also implies that the procedure for deriving the
inductance from the observed impedance is not well defined. Therefore, the
inductivity value of a magnetic texture is no longer well-defined when
$\beta\neq 0$, at least in the present framework. For convenience, however,
adopting a $R$–$L$ series circuit, one may define an effective inductivity in
the low-frequency regime by $\tilde{L}^{\rm
eff}_{ij}=\operatorname{Im}[\rho_{ij}(\omega)-\rho_{ij}(0)]/\omega$ and use
this to describe the electric response. However, note that this effective
inductivity or inductance ($L^{\rm eff}_{ij}=\tilde{L}^{\rm
eff}_{ij}\frac{\ell}{S}$) no longer represents the genuine inductivity or
inductance; i.e., $\tilde{L}^{\rm eff}_{ij}\neq\tilde{L}_{ij}$ Thus, there is
no reason why the constraints associated with $\tilde{L}_{ij}$ [i.e., the
diagonal components should be positive; the tensor should be symmetric; and
$\tilde{L}_{xx}\tilde{L}_{yy}\geq(\tilde{L}_{xy})^{2}$] should also apply to
$\tilde{L}^{\rm eff}_{ij}$. In fact, the diagonal $\tilde{L}^{\rm eff}_{ii}$
may be negative, whereas the diagonal $\tilde{L}_{ii}$ is invariably positive.
For instance, when $R$–capacitance ($C$) parallel circuit is considered,
$L^{\rm eff}_{xx}$ at low frequencies is negative, $-R^{2}C$, whereas the
energy stored under a direct current (DC) is positive,
$\frac{1}{2}(R^{2}C)I^{2}$. Thus, the effective inductivity or inductance
contradicts the physics that $V=L\frac{{\rm d}j}{{\rm d}t}$ encompasses, and
therefore, $L^{\rm eff}_{ij}$ should be viewed merely as a value that
characterizes the electric response. The previous results [16] suggest that
although $\tilde{L}^{\rm eff}_{ij}\neq\tilde{L}_{ij}$ in general,
$\tilde{L}^{\rm eff}_{ij}$ becomes identical to $\tilde{L}_{ij}$ when
$\beta=0$ (the adiabatic limit).

With this in mind, here we show how the effective inductivity $\tilde{L}^{\rm
eff}_{ij}$ changes as $\beta$ increases. In this context, we note that several
theoretical studies [42–45] have revealed that when $\beta$ is finite, the
expression of the emergent electric field should be modified as follows:

$\displaystyle e_{i}$
$\displaystyle=\frac{P\hbar}{2|e|}\bm{m}\cdot(\partial_{i}\bm{m}\times\partial_{t}\bm{m})-\beta\frac{P\hbar}{2|e|}(\partial_{i}\bm{m}\cdot\partial_{t}\bm{m}).$
(S1)

Thus, using this equation, we numerically obtain $e_{i}$ and thus
$\tilde{L}^{\rm eff}_{ij}$ at low frequencies for a single-${\bm{q}}$ helix
with $\theta=0^{\circ}$, a maze helix, and a skyrmion lattice: the results are
summarized in Supplementary Table SI-SIII, respectively.

In the single-${\bm{q}}$ helix with $\theta=0^{\circ}$ (Table SI), finite
$\beta$ leads exclusively to a decrease in $\tilde{L}^{\rm eff}_{xx}$, as
discussed in the literature [16, 46]. For even larger $\beta$, $\tilde{L}^{\rm
eff}_{xx}$ can be negative [46], but this conclusion relates to the effective
inductivity and does not necessarily contradict the fact that the diagonal
components of $\tilde{L}_{ij}$ must be positive. In contrast, no clear $\beta$
dependence is seen in the off-diagonal components of $\tilde{L}^{\rm
eff}_{ij}$. The absence of the off-diagonal components will be corroborated
from a symmetry point of view in Supplementary Note 3. The effective
inductivity of a single-${\bm{q}}$ helix with arbitrary $\theta$ at finite
$\beta$ is obtained immediately by applying the corresponding orthogonal
transformation to the results of $\theta=0^{\circ}$, and thus the results are
omitted here. For the maze helix (Table SII), the effect of nonadiabaticity is
weak.

The impact of nonadiabaticity on the skyrmion lattice (Table SIII) is
different from that on the single-${\bm{q}}$ helix. We find that as seen in
$\tilde{L}^{\rm eff}_{ij}(\beta)-\tilde{L}_{ij}$, the antisymmetric off-
diagonal components become pronounced as $\beta$ increases. As demonstrated in
the main text and Fig. 1, $e_{i}=\tilde{L}_{ik}\frac{{\rm d}j_{k}}{{\rm d}t}$
must result in a symmetric inductivity tensor,
$\tilde{L}_{ik}=\tilde{L}_{ki}$. Thus, the emergence of the antisymmetric
components in $\tilde{L}^{\rm eff}_{ij}(\beta)$ again indicates that at finite
$\beta$, $\tilde{L}^{\rm eff}_{ij}$ is a different quantity from the
inductivity tensor $\tilde{L}_{ij}$; only when $\beta=0$ (the adiabatic
limit), $\tilde{L}^{\rm eff}_{ij}=\tilde{L}_{ij}$, and hence, $\tilde{L}^{\rm
eff}_{ij}$ is symmetric. The symmetry arguments on the SkL also support the
emergence of the antisymmetric components in $\tilde{L}^{\rm eff}_{ij}$; for
more details, see Supplementary Note 3.

Table S 1: Impact of nonadiabaticity on the effective inductivity tensor in the single-${\bm{q}}$ helix with $\theta=0^{\circ}$. The unit is 10-21 H m. $\beta$ |  $\tilde{L}_{ij}$ or $\tilde{L}^{\rm eff}_{ij}(\beta)$ |  $\tilde{L}^{\rm eff}_{ij}(\beta)-\tilde{L}_{ij}$
---|---|---
0 |  $\tilde{L}_{ij}=\tilde{L}^{\rm eff}_{ij}=$ $\begin{pmatrix}3.11&-0.07\\\ -0.07&0.07\end{pmatrix}$ |
0.02 |  $\tilde{L}^{\rm eff}_{ij}=$ $\begin{pmatrix}2.91&-0.07\\\ -0.06&0.07\end{pmatrix}$ |  $\begin{pmatrix}-0.20&0.00\\\ 0.01&0.00\end{pmatrix}$
0.04 |  $\tilde{L}^{\rm eff}_{ij}=$ $\begin{pmatrix}2.32&-0.04\\\ -0.04&0.06\end{pmatrix}$ |  $\begin{pmatrix}-0.79&0.03\\\ 0.03&-0.01\end{pmatrix}$
Table S 2: Impact of nonadiabaticity on the effective inductivity tensor in the maze helix. The unit is 10-21 H m. $\beta$ |  $\tilde{L}_{ij}$ or $\tilde{L}^{\rm eff}_{ij}(\beta)$ |  $\tilde{L}^{\rm eff}_{ij}(\beta)-\tilde{L}_{ij}$
---|---|---
0 |  $\tilde{L}_{ij}=\tilde{L}^{\rm eff}_{ij}=$ $\begin{pmatrix}2.23&-0.07\\\ -0.08&2.67\end{pmatrix}$ |
0.02 |  $\tilde{L}^{\rm eff}_{ij}=$ $\begin{pmatrix}2.25&-0.09\\\ -0.05&2.66\end{pmatrix}$ |  $\begin{pmatrix}0.02&-0.02\\\ 0.03&-0.01\end{pmatrix}$
0.04 |  $\tilde{L}^{\rm eff}_{ij}=$ $\begin{pmatrix}2.26&-0.09\\\ -0.05&2.63\end{pmatrix}$ |  $\begin{pmatrix}0.03&-0.02\\\ 0.03&-0.04\end{pmatrix}$
Table S 3: Impact of nonadiabaticity on the effective inductivity tensor in the skyrmion lattice at 0.3 T. The unit is 10-21 H m. $\beta$ |  $\tilde{L}_{ij}$ or $\tilde{L}^{\rm eff}_{ij}(\beta)$ |  $\tilde{L}^{\rm eff}_{ij}(\beta)-\tilde{L}_{ij}$
---|---|---
0 |  $\tilde{L}_{ij}=\tilde{L}^{\rm eff}_{ij}=$ $\begin{pmatrix}7.40&-0.47\\\ -0.47&8.75\end{pmatrix}$ |
0.02 |  $\tilde{L}^{\rm eff}_{ij}=$ $\begin{pmatrix}7.39&-0.92\\\ -0.02&8.74\end{pmatrix}$ |  $\begin{pmatrix}-0.01&-0.45\\\ 0.45&-0.01\end{pmatrix}$
0.04 |  $\tilde{L}^{\rm eff}_{ij}=$ $\begin{pmatrix}7.36&-1.37\\\ 0.43&8.71\end{pmatrix}$ |  $\begin{pmatrix}-0.04&-0.90\\\ 0.90&-0.04\end{pmatrix}$

## Supplementary Note 2: Effective inductivity tensor under the time-reversal
operation

When the nonadiabaticity $\beta$ is finite, antisymmetric components may
appear in the effective inductivity tensor, $\tilde{L}_{ij}^{\rm
eff}(\bm{B},\\{\bm{m}(\bm{r})\\})$, in addition to symmetric components, as
most clearly seen in Table SIII. For clarity, we decompose
$\tilde{L}_{ij}^{\rm eff}(\bm{B},\\{\bm{m}(\bm{r})\\})$ into symmetric and
antisymmetric parts:

$\tilde{L}_{ij}^{\rm
eff}(\bm{B},\\{\bm{m}(\bm{r})\\})=\tilde{L}_{ij}^{\rm{eff},S}(\bm{B},\\{\bm{m}(\bm{r})\\})+\tilde{L}_{ij}^{\rm{eff},A}(\bm{B},\\{\bm{m}(\bm{r})\\}),$

where $\tilde{L}_{ij}^{\rm{eff},S}$ and $\tilde{L}_{ij}^{\rm{eff},A}$
represent the symmetric and antisymmetric components, respectively. By
definition, $\tilde{L}_{ij}^{\rm{eff},S}$ and $\tilde{L}_{ij}^{\rm{eff},A}$
satisfy

$\begin{split}\tilde{L}_{ij}^{\rm{eff},S}&=\tilde{L}_{ji}^{\rm{eff},S},\\\
\tilde{L}_{ij}^{\rm{eff},A}&=-\tilde{L}_{ji}^{\rm{eff},A}.\end{split}$

On the other hand, as discussed in the main text, Onsager’s reciprocal theorem
concludes:

$\tilde{L}_{ji}^{\rm{eff}}(\bm{B},\\{\bm{m}(\bm{r})\\})=\tilde{L}_{ij}^{\rm{eff}}(-\bm{B},\\{-\bm{m}(\bm{r})\\}).$

By combining the two relations, one can thus conclude:

$\displaystyle\tilde{L}_{ij}^{\rm{eff},S}(\bm{B},\\{\bm{m}(\bm{r})\\})$
$\displaystyle=\tilde{L}_{ij}^{\rm{eff},S}(-\bm{B},\\{-\bm{m}(\bm{r})\\}),$
(S2) $\displaystyle\tilde{L}_{ij}^{\rm{eff},A}(\bm{B},\\{\bm{m}(\bm{r})\\})$
$\displaystyle=-\tilde{L}_{ij}^{\rm{eff},A}(-\bm{B},\\{-\bm{m}(\bm{r})\\}).$
(S3)

Thus, Onsager’s reciprocal theorem concludes that the symmetric and
antisymmetric components of $\tilde{L}_{ij}^{\rm{eff}}$ are even and odd under
time-reversal, respectively.

This analytic conclusion is also confirmed numerically. Figure S1 shows the
magnetic-field dependence of the transverse effective inductivity,
$\tilde{L}_{xy}^{\rm{eff}}$ and $\tilde{L}_{yx}^{\rm{eff}}$, at a finite
$\beta$, 0.04. For the single ${\bm{q}}$-helix with $\theta=45^{\circ}$, the
$\tilde{L}_{ij}^{\rm{eff}}$ is approximately symmetric
($\tilde{L}_{xy}^{\rm{eff}}=\tilde{L}_{yx}^{\rm{eff}}$) and satisfies Eq. (S2)
under time-reversal. For the SkL, in contrast, the $\tilde{L}_{ij}^{\rm{eff}}$
consists of both symmetric and antisymmetric components, which satisfy Eqs.
(S2) and (S3) under time-reversal, respectively.

Fig. S 1:  Magnetic-field dependence of the transverse effective inductivity,
$\tilde{L}_{xy}^{\rm{eff}}$ and $\tilde{L}_{yx}^{\rm{eff}}$, at $\beta=0.04$.
H, Sk, and F denote the single $\bm{q}$-helix with $\theta=45^{\circ}$,
skyrmion lattice, and ferromagnetic state, respectively. Note that the
transverse components of $\tilde{L}_{ij}^{\rm{eff}}$ of the SkL includes both
symmetric and antisymmetric components. The simulation parameters are the same
as those used in Fig. 3 in the main text, except for $\beta$.

## Supplementary Note 3: Symmetry arguments on a two-by-two polar tensor

In this section, we discuss how the symmetry of the magnetic texture places
constrains on the allowed form of a two-by-two polar tensor. Note that such a
symmetry argument holds whether or not the system is in the adiabatic limit
($\beta=0$).

When the Cartesian coordinates are transformed by a clockwise rotation of
angle $\theta$ with respect to the $z$ axis, the representation of an
arbitrary two-by-two tensor, $\begin{pmatrix}a&b\\\ c&d\end{pmatrix}$, changes
accordingly. The new representation in the new Cartesian coordinates is given
by

$\begin{split}&\begin{pmatrix}\cos\theta&-\sin\theta\\\
\sin\theta&\cos\theta\end{pmatrix}\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\begin{pmatrix}\cos\theta&-\sin\theta\\\
\sin\theta&\cos\theta\end{pmatrix}^{-1}\\\
&=\begin{pmatrix}a\cos^{2}\theta-b\sin\theta\cos\theta-c\sin\theta\cos\theta+d\sin^{2}\theta&a\sin\theta\cos\theta+b\cos^{2}\theta-c\sin^{2}\theta-d\sin\theta\cos\theta\\\
a\sin\theta\cos\theta-b\sin^{2}\theta+c\cos^{2}\theta-d\sin\theta\cos\theta&a\sin^{2}\theta+b\sin\theta\cos\theta+c\sin\theta\cos\theta+d\cos^{2}\theta.\end{pmatrix}\end{split}$
(S4)

If the rotation by $\theta$ is a symmetry operation of the system, the old and
new representations should be identical (Neumann’s principle). Similarly, when
the Cartesian coordinates are transformed by a two-fold rotation with respect
to the $x$ axis (denoted by $C_{2x}$), $\begin{pmatrix}1&0\\\
0&-1\end{pmatrix}$, the new representation of an arbitrary tensor is given by

$\begin{split}\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\begin{pmatrix}1&0\\\
0&-1\end{pmatrix}^{-1}=\begin{pmatrix}a&-b\\\ -c&d\end{pmatrix}.\end{split}$
(S5)

If $C_{2x}$, which may be followed by translation, is a symmetry operation of
the system, Neumann’s principle concludes

$b=0,\hskip 28.45274ptc=0.$ (S6)

The same conclusion is obtained when $C_{2y}$ (a two-fold rotation with
respect to the $y$ axis), which may be followed by translation, is a symmetry
operation of the system. Thus, the symmetry that the magnetic texture has
places strong constraints on the allowed form of a two-by-two tensor.

Below, we shall first discuss the symmetry constraints on the symmetric
component of an arbitrary two-by-two tensor,
$\begin{pmatrix}a&\frac{b+c}{2}\\\ \frac{b+c}{2}&d\end{pmatrix}$. When the
system has, for instance, three-fold ($C_{3}$) symmetry
$(\theta=\pm\frac{2\pi}{3})$, Neumann’s principle leads to

$\begin{pmatrix}a&\frac{b+c}{2}\\\
\frac{b+c}{2}&d\end{pmatrix}=\begin{pmatrix}\frac{1}{4}a+\frac{\sqrt{3}}{2}\frac{b+c}{2}+\frac{3}{4}d&-\frac{\sqrt{3}}{4}a-\frac{1}{2}\frac{b+c}{2}+\frac{\sqrt{3}}{4}d\\\
-\frac{\sqrt{3}}{4}a-\frac{1}{2}\frac{b+c}{2}+\frac{\sqrt{3}}{4}d&\frac{3}{4}a-\frac{\sqrt{3}}{2}\frac{b+c}{2}+\frac{1}{4}d\end{pmatrix}.$

Solving this equation, one obtains

$a=d,\hskip 28.45274pt\frac{b+c}{2}=0.$

The symmetry arguments thus conclude that a two-by-two symmetric tensor in the
presence of $C_{3z}$ symmetry should be in the form of $\begin{pmatrix}1&0\\\
0&1\end{pmatrix}$, regardless of whether the system has $C_{2x}$, which may be
followed by translation, as a symmetry operation. The same conclusion is
obtained when the system has four-, six-, or infinite-fold symmetry with
respect to the $z$ axis ($C_{4z},C_{6z}$ or $C_{\infty z}$). In contrast, when
the system has only two-fold symmetry with respect to the $z$ axis ($C_{2z}$),
Neumann’s principle places no constraint because an arbitrary two-by-two
tensor is invariant under $C_{2z}$ [see Eq. (S4)].

Next, we shall consider the antisymmetric component of an arbitrary two-by-two
tensor, $\begin{pmatrix}0&\frac{b-c}{2}\\\ -\frac{b-c}{2}&0\end{pmatrix}$. As
demonstrated in Table SIII, when nonadiabaticity is finite ($\beta\neq 0$), an
antisymmetric component may appear in $\tilde{L}^{\rm eff}_{ij}$ (not in
$\tilde{L}_{ij}$, of which derivation from the observed impedance is not well
defined when $\beta$ is finite). Note that $\begin{pmatrix}0&1\\\
-1&0\end{pmatrix}$ is invariant under arbitrary rotation with respect to the
$z$ axis [Eq. (S4)], and hence, the rotational symmetry with respect to the
$z$ axis does not place any constraint on the antisymmetric component. In
contrast, when the system has $C_{2x}$ or $C_{2y}$, which may be followed by
translation, a finite antisymmetric component is prohibited, as derived in
Eqs. (S5) and (S6).

The above symmetry consideration is consistent with the numerical observations
on $\tilde{L}^{\rm eff}_{ij}$ of a single-$\bm{q}$ helix with
$\bm{q}$$\parallel$$x$ (Table SI). The magnetic texture has $C_{2x}$ followed
by translation as a symmetry operation, and hence, a finite antisymmetric
component is prohibited, regardless of $\beta$. As discussed in the main text,
by further considering the spin-transfer torque mechanism, one can conclude
that a single-$\bm{q}$ helix with $\bm{q}$$\parallel$$x$ exhibits
$\tilde{L}^{\rm eff}_{ij}$ in the form of $\begin{pmatrix}1&0\\\
0&0\end{pmatrix}$, regardless of $\beta$. The $\tilde{L}^{\rm eff}_{ij}$ of a
single-$\bm{q}$ helix with arbitrary $\bm{q}$-direction can be obtained by the
orthogonal transformation of the $\tilde{L}^{\rm eff}_{ij}$ with
$\bm{q}$$\parallel$$x$ [Eq. (14) in the main text], and this transformation of
the symmetric tensor does not add any antisymmetric component. One can thus
conclude that $\tilde{L}^{\rm eff}_{ij}$ of a single-$\bm{q}$ helix should be
symmetric, regardless of $\beta$ and the $\bm{q}$-direction. Nevertheless, the
application of a magnetic field along the $z$ direction breaks $C_{2x}$ and
$C_{2y}$ symmetry, and hence a magnetic-field-induced antisymmetric component
is not prohibited, at least, by symmetry.

For a maze helix, the magnetic texture has a global approximate $C_{\infty z}$
symmetry, and hence, the symmetric part of the off-diagonal components should
be zero, as discussed in the main text. Furthermore, because a single-$\bm{q}$
helix with $\theta$ is transformed into that with $-\theta$ by $C_{2x}$ and
$C_{2y}$, a maze-helix texture, which is a collection of randomly oriented
single-$\bm{q}$ helices, has global approximate $C_{2x}$ and $C_{2y}$
symmetries. Hence, the antisymmetric part should also be zero. In this way,
$\tilde{L}^{\rm eff}_{ij}$ of a maze helix is expected to be also symmetric
with no off-diagonal components, regardless of $\beta$. This symmetry argument
is well consistent with the numerical observations shown in Table SII, given
that the calculations are performed for a system of finite size.

For the case of a skyrmion magnetic texture, the system has neither $C_{2x}$
nor $C_{2y}$ symmetry operations. Hence, there is no symmetry reason to
prohibit a finite antisymmetric component. This symmetry argument is
consistent with the numerical observations shown in Table SIII.

## Supplementary Note 4: Frequency and $\alpha$ dependences of the inductivity
tensor

In this section, we shall consider the frequency and $\alpha$ dependences of
the inductivity tensor at $\beta=0$. In this paper, we deal with the low-
frequency regime in which the magnetic texture can sufficiently track the
steady-state magnetic texture corresponding to each instantaneous current
value. In other words, during AC application, the magnetic texture at a given
instant is determined by the instantaneous current value at that instant, and
it does not depend on the immediately preceding state. In fact, we found that
below 100 MHz, the emergent electric field is proportional to $\omega$ (Fig.
S2), or equivalently, the inductivity tensor is independent of $\omega$. This
observation confirms that our numerical results are in the low-frequency
regime. The $\alpha$ dependence of the inductivity tensor in the low-frequency
regime is therefore expected to be dictated by that of the steady-state
magnetic texture.

To consider the $\alpha$ dependence of the steady-state magnetic texture, it
is useful to refer to the LLG equation expressed in an equivalent form
different from Eq. (16) (for example, see [23]):

$\frac{{\rm d}\bm{m_{r}}(t)}{{\rm d}t}=-|\gamma|\frac{{\rm d}\mathscr{H}}{{\rm
d}\bm{m_{r}}}\times\bm{m_{r}}+\alpha\bm{m_{r}}\times\frac{{\rm
d}\bm{m_{r}}}{{\rm
d}t}-(\bm{u}\cdot\bm{\nabla})\bm{m_{r}}+\beta\bm{m_{r}}\times[(\bm{u}\cdot\bm{\nabla})\bm{m_{r}}].$

This expression implies that the steady-state configurations of magnetic
moments, in which $\frac{{\rm d}\bm{m_{r}}(t)}{{\rm d}t}=\bm{0}$, do not
depend on $\alpha$. Thus, the inductivity tensor in the low-frequency regime
is expected to be also insensitive to $\alpha$. Figure S3 displays the
inductivity tensor for the case of a single-$\bm{q}$ helix with
$\theta=20^{\circ}$ with $\alpha=0.04$, 0.08 and 0.12. As expected, the
results are the same within the numerical error, demonstrating that the choice
of $\alpha$ does not make a difference in the inductivity tensor.

Fig. S 2:  Frequency dependence of the amplitude of the emergent electric
field. The simulations were done for the single-$\bm{q}$ helix of
$\theta=0^{\circ}$ at $\beta=0$. Fig. S 3:  $\alpha$ dependence of the
inductivity tensor. The simulations were done for the single-$\bm{q}$ helix of
$\theta=20^{\circ}$ at $\beta=0$.


# Exponential tilting of subweibull distributions

F. William Townes
Department of Statistics and Data Science
Carnegie Mellon University
<EMAIL_ADDRESS>
###### Contents

  1. 1 Introduction
  2. 2 Laplace transforms
  3. 3 Subweibull random variables
    1. 3.1 Subweibull properties of the Poisson distribution
  4. 4 Exponential tilting
  5. 5 Discussion

##  1 Introduction

Subexponential and subgaussian distributions are of fundamental importance in
the application of high dimensional probability to machine learning [10, 12].
Recently it has been shown that the subweibull class unifies the
subexponential and subgaussian families, while also incorporating
distributions with heavier tails [11, 3]. Informally, a q-subweibull ($q>0$)
random variable has a survival function that decays at least as fast as
$\exp(-\lambda x^{q})$ for some $\lambda>0$. For example, the exponential
distribution is 1-subweibull and the Gaussian distribution is 2-subweibull.
Here, we provide two alternative characterizations of the subweibull class and
introduce a distinction between strictly and broadly subweibull distributions.
As an example, the Poisson distribution is shown to be strictly subexponential
($q=1$) but not subweibull for any $q>1$. Finally, we detail the conditions
under which the subweibull property is preserved after exponential tilting.

##  2 Laplace transforms

######  Definition 2.1.

We define the two-sided Laplace transform of a random variable $X$ with
distribution function $F$ as

$\mathcal{L}_{X}(t)=\operatorname{\mathbb{E}}[\exp(-tX)]=\int_{-\infty}^{\infty}\exp(-tx)dF(x)$

We do not restrict $X$ to be nonnegative or to have a density function. In the
special case that $\mathcal{L}_{X}(t)<\infty$ for all $t$ in an open interval
around $t=0$, then $X$ has a moment generating function (MGF) which is
$M_{X}(t)=\operatorname{\mathbb{E}}[\exp(tX)]=\mathcal{L}_{X}(-t)$. The
Laplace transform can characterize the distribution even if the MGF does not
exist.

######  Lemma 2.0.1.

If the Laplace transforms of random variables $X$ and $Y$ satisfy
$\mathcal{L}_{X}(t)=\mathcal{L}_{Y}(t)$ for all $t$ in any nonempty open
interval $(a,b)\subset\mathbb{R}$, not necessarily containing zero, then
$X\overset{d}{=}Y$.

For a proof refer to [8]. A random variable $X$ is considered subexponential
iff the MGF exists [10]. If $\mathcal{L}_{X}(t)=\infty$ for all $t>0$
(respectively $t<0$), $X$ is said to have a heavy left (respectively, right)
tail [9]. If a tail is not heavy it is said to be light. It is well known that
the one-sided Laplace transform characterizes nonnegative distributions [1].
Lemma 2.0.1 shows that the two-sided Laplace transform characterizes any
distribution with at least one light tail.

##  3 Subweibull random variables

######  Definition 3.1.

A random variable $X$ is q-subweibull if
$\operatorname{\mathbb{E}}[\exp(\lambda^{q}|X|^{q})]<\infty$ for some
$\lambda>0$. $X$ is strictly q-subweibull if the condition is satisfied for
all $\lambda>0$. If $X$ is q-subweibull but not strictly so, we refer to it as
broadly q-subweibull.

The first part of this definition was also proposed by [3] and by [11] using a
parameterization equivalent to $1/q$. Clearly $X$ is (strictly) q-subweibull
if and only if $|X|^{q}$ is (strictly) subexponential. As an example, the
Laplace distribution is broadly 1-subweibull (ie broadly subexponential).

######  Definition 3.2.

The radius of convergence of a q-subweibull random variable $X$ is defined by

$R_{q}=\sup\left\\{\lambda>0:\operatorname{\mathbb{E}}[\exp(\lambda^{q}|X|^{q})]<\infty\right\\}$

and if no such $\lambda>0$ exists we adopt the convention that $R_{q}=0$.

In the case of strictly q-subweibull distributions, $R_{q}=\infty$. $X$ has
“heavy tails” (in the sense of [9]) iff it is not subexponential ($R_{1}=0$).

######  Lemma 3.0.1.

Random variable $X$ with $\Pr(X<0)\notin\\{0,1\\}$ is q-subweibull if and only
if the nonnegative random variables $A=[-X\operatorname{\left.\right|}X<0]$
and $B=[X\operatorname{\left.\right|}X\geq 0]$ are q-subweibull. Let $R_{qx}$,
$R_{qa}$, and $R_{qb}$ denote the radii of convergence for $X$, $A$, and $B$,
respectively. Then $R_{qx}=\min\\{R_{qa},R_{qb}\\}$.

###### Proof.

Let $p=\Pr(X<0)$ and define nonnegative random variables
$A=[-X\operatorname{\left.\right|}X<0]$ and
$B=[X\operatorname{\left.\right|}X\geq 0]$.

$\displaystyle\operatorname{\mathbb{E}}[\exp(\lambda^{q}|X|^{q})]$
$\displaystyle=\operatorname{\mathbb{E}}[\exp(\lambda^{q}(-X)^{q})\operatorname{\left.\right|}X<0]p+\operatorname{\mathbb{E}}[\exp(\lambda^{q}X^{q})\operatorname{\left.\right|}X\geq
0](1-p)$
$\displaystyle=\operatorname{\mathbb{E}}[\exp(\lambda^{q}A^{q})]p+\operatorname{\mathbb{E}}[\exp(\lambda^{q}B^{q})](1-p)$

The left hand side is finite if and only if both terms on the right hand side
are finite. If $R_{qx}$ is the radius of convergence for $X$ then
$\operatorname{\mathbb{E}}[\exp(\lambda^{q}|X|^{q})]<\infty$ for all
$\lambda\in[0,R_{qx})$. Clearly
$\operatorname{\mathbb{E}}[\exp(\lambda^{q}A^{q})]<\infty$ and
$\operatorname{\mathbb{E}}[\exp(\lambda^{q}B^{q})]<\infty$ for all
$\lambda\in[0,R_{qx})$ also, implying $\min\\{R_{qa},R_{qb}\\}\geq R_{qx}$.
However, if $\min\\{R_{qa},R_{qb}\\}>R_{qx}$ then there exists some
$\lambda>R_{qx}$ such that
$\operatorname{\mathbb{E}}[\exp(\lambda^{q}|X|^{q})]<\infty$, which by
Definition 3.2 means $R_{qx}$ is not the radius of convergence of $X$, a
contradiction. Therefore, $\min\\{R_{qa},R_{qb}\\}=R_{qx}$. ∎

######  Proposition 3.1.

(Theorem 1 of [11]) The following are equivalent characterizations of a
q-subweibull random variable $X$ for $q>0$.

  1. 1.

Tail bound: $\exists~{}K_{1a}>0$ such that $\forall~{}t\geq 0$,

$\Pr(|X|>t)\leq 2\exp\big{(}-(t/K_{1a})^{q}\big{)}$

  2. 2.

Growth rate of absolute moments: $\exists~{}K_{2}>0$ such that
$\forall~{}p\geq 1$,

$\big{(}\operatorname{\mathbb{E}}[|X|^{p}]\big{)}^{1/p}\leq K_{2}p^{1/q}$

  3. 3.

MGF of $|X|^{q}$ finite in interval of zero: $\exists~{}K_{3}>0$ such that
$\forall~{}0<\lambda\leq 1/K_{3}$

$\operatorname{\mathbb{E}}\left[\exp(\lambda^{q}|X|^{q})\right]\leq\exp(K_{3}^{q}\lambda^{q})$

When $q\geq 1$, Condition 3 is equivalent to requiring the Orlicz norm
$\|X\|_{\psi_{q}}<\infty$ where $\psi_{q}(x)=\exp(x^{q})-1$. The case of $q<1$
(heavy tails) is further discussed in [3]. It is sometimes convenient to use
the following asymptotic alternatives to Proposition 3.1 (1 and 2).

######  Corollary 3.0.1.

A random variable $X$ is q-subweibull if and only if either of the following
hold

  1. 1.

Tail bound: $\exists~{}K_{1b}>0$ such that

$\limsup_{t\to\infty}\Pr(|X|>t)\exp\big{(}(t/K_{1b})^{q}\big{)}<\infty$

  2. 2.

Growth rate of absolute moments:

$\limsup_{p\to\infty}\frac{\big{(}\operatorname{\mathbb{E}}[|X|^{p}]\big{)}^{1/p}}{p^{1/q}}<\infty$

###### Proof.

Proposition 3.1 $(1)\implies(1)$ :

$\limsup_{t\to\infty}\Pr(|X|>t)\exp\big{(}(t/K_{1a})^{q}\big{)}\leq\sup_{t\geq
0}\Pr(|X|>t)\exp\big{(}(t/K_{1a})^{q}\big{)}\leq 2<\infty$

So we can simply set $K_{1b}=K_{1a}$.

$(1b)\implies$ Proposition 3.1 $(1a)$: Assume

$\limsup_{t\to\infty}\Pr(|X|>t)\exp\big{(}(t/K_{1b})^{q}\big{)}=K$

Then, for every $C>K$, there exists some $T$ such that for all $t>T$,

$\Pr(|X|>t)\exp\big{(}(t/K_{1b})^{q}\big{)}<C$

For all $t\in[0,T]$, $\Pr(|X|>t)\leq 1$ and
$\exp\big{(}(t/K_{1b})^{q}\big{)}\leq\exp\big{(}(T/K_{1b})^{q}\big{)}$.
Therefore

$\sup_{t\geq
0}\Pr(|X|>t)\exp\big{(}(t/K_{1b})^{q}\big{)}\leq\max\left\\{C,~{}\exp\big{(}(T/K_{1b})^{q}\big{)}\right\\}$

Let $U=\max\left\\{C,~{}\exp\big{(}(T/K_{1b})^{q}\big{)}\right\\}$. If $U\leq
2$ this directly implies $(1a)$ with $K_{1a}=K_{1b}$. In the case that $U>2$,
set

$K_{1a}=K_{1b}\left(\frac{\log U}{\log 2}\right)^{1/q}>K_{1b}$

Let $f(t)=U\exp\big{(}-(t/K_{1b})^{q}\big{)}$,
$g(t)=2\exp\big{(}-(t/K_{1a})^{q}\big{)}$, and $T^{\star}=K_{1b}(\log
U)^{1/q}$. Since $f(T^{\star})=g(T^{\star})=1$ and $g(t)$ is a strictly
decreasing function, this implies that $\Pr(|X|>t)\leq 1\leq g(t)$ for
$t\in[0,T^{\star}]$. For $t\geq T^{\star}$, $g(t)\geq f(t)$ since
$K_{1a}>K_{1b}$, and $f(t)\geq\Pr(|X|>t)$ by assumption therefore
$2\exp\big{(}-(t/K_{1a})^{q}\big{)}\geq\Pr(|X|>t)$ for all $t\geq 0$.

Proposition 3.1 $(2)\implies(2)$:

$\limsup_{p\to\infty}\frac{\big{(}\operatorname{\mathbb{E}}[|X|^{p}]\big{)}^{1/p}}{p^{1/q}}\leq\sup_{p\geq
1}\frac{\big{(}\operatorname{\mathbb{E}}[|X|^{p}]\big{)}^{1/p}}{p^{1/q}}\leq
K_{2}<\infty$

$(2)\implies$ Proposition 3.1 $(2)$: Assume

$\limsup_{p\to\infty}\frac{\big{(}\operatorname{\mathbb{E}}[|X|^{p}]\big{)}^{1/p}}{p^{1/q}}=K<\infty$

Then for every $C>K$, there exists some $p^{\star}$ such that for all
$p>p^{\star}$,

$\frac{\big{(}\operatorname{\mathbb{E}}[|X|^{p}]\big{)}^{1/p}}{p^{1/q}}<C$

The $L_{p}$ norm is increasing in $p$, so for $p\in[1,p^{\star}]$,
$\big{(}\operatorname{\mathbb{E}}[|X|^{p}]\big{)}^{1/p}\leq\big{(}\operatorname{\mathbb{E}}[|X|^{p^{\star}}]\big{)}^{1/p^{\star}}$
and $p^{1/q}\geq 1$, which establishes

$\sup_{p\geq
1}\frac{\big{(}\operatorname{\mathbb{E}}[|X|^{p}]\big{)}^{1/p}}{p^{1/q}}\leq\max\left\\{\big{(}\operatorname{\mathbb{E}}[|X|^{p^{\star}}]\big{)}^{1/p^{\star}},~{}C\right\\}$

∎

It was shown by [11] that a q-subweibull distribution is also r-subweibull for
all $r<q$. We now show that this also implies it is strictly r-subweibull.

######  Corollary 3.0.2.

If $X$ is q-subweibull then it is strictly r-subweibull for all $r\in(0,q)$.

###### Proof.

If $X$ is q-subweibull, by Proposition 3.1 we may assume there exists $K>0$
such that $\forall~{}p\geq 1$,

$\big{(}\operatorname{\mathbb{E}}[|X|^{p}]\big{)}^{1/p}\leq Kp^{1/q}$

Let $r\in(0,q)$. The MGF of $|X|^{r}$ is given by

$\displaystyle\operatorname{\mathbb{E}}\left[\exp(\lambda^{r}|X|^{r})\right]$
$\displaystyle=1+\sum_{p=1}^{\infty}\frac{\lambda^{pr}E[|X|^{pr}]}{p!}$
$\displaystyle\leq
1+\sum_{p=1}^{\infty}\frac{\lambda^{pr}K^{pr}(pr)^{pr/q}}{(p/e)^{p}}$
$\displaystyle=1+\sum_{p=1}^{\infty}\big{(}\lambda^{r}K^{r}er^{r/q}\big{)}^{p}p^{p(r/q-1)}$

Apply the root test to the series to determine convergence.

$\displaystyle R(p)=\lambda^{r}K^{r}er^{r/q}p^{r/q-1}$

Since $r<q$, then $\lim_{p\to\infty}R(p)=0$ and the series converges
regardless of the value of $\lambda$, which shows $X$ is strictly
r-subweibull. ∎

######  Corollary 3.0.3.

Every bounded random variable is strictly q-subweibull for all $q>0$.

###### Proof.

If $X$ is bounded then there exists $M\geq 0$ such that $|X|\leq M$. Then
$\operatorname{\mathbb{E}}[\exp(\lambda^{q}|X|^{q})]\leq\exp(\lambda^{q}M^{q})<\infty$
for all $\lambda>0$ and $q>0$. ∎

######  Corollary 3.0.4.

If $X$ is not strictly q-subweibull with $q\geq 1$ then it is not r-subweibull
for any $r>q$.

###### Proof.

From Corollary 3.0.1 we may assume $\exists~{}\lambda>0$ such that

$\limsup_{t\to\infty}\Pr(|X|>t)\exp(\lambda t^{q})=\infty$

which implies there is an infinite sequence $t_{n}\to\infty$ such that

$\lim_{n\to\infty}\Pr(|X|>t_{n})\exp(\lambda t_{n}^{q})=\infty$

Now let $\rho>0$ and $r>q$. Whenever $t\geq
t^{\star}=(\lambda/\rho)^{1/(r-q)}$, we have $\exp(\rho t^{r})\geq\exp(\lambda
t^{q})$. Let $\\{t_{m}\\}$ be the infinite subsequence of $\\{t_{n}\\}$
excluding the elements less than $t^{\star}$. Clearly $t_{m}\to\infty$ as
well. Then

$\lim_{m\to\infty}\Pr(|X|>t_{m})\exp(\rho
t_{m}^{r})\geq\lim_{m\to\infty}\Pr(|X|>t_{m})\exp(\lambda t_{m}^{q})=\infty$

which implies $X$ cannot be r-subweibull. ∎

###  3.1 Subweibull properties of the Poisson distribution

Corollaries 3.0.2 and 3.0.4 suggest a hierarchy of distributions based on the
heaviness of the tails. Broadly q-subweibull distributions, which have a
finite but nonzero radius of convergence ($R_{q}$), serve as “critical points”
in the transition between the strictly r-subweibull regime ($r<q$), with
$R_{q}=\infty$ and the not r-subweibull regime ($r>q$) with $R_{q}=0$.
However, the transition from strictly subweibull to not subweibull can be
immediate, without passing through the stage of broadly subweibull. A
sophisticated example using the moment sequence was suggested by [13]. Here we
provide a simple example: the Poisson tail is lighter than any exponential
tail, but heavier than any weibull tail with $q>1$.

######  Proposition 3.2.

The Poisson distribution is strictly q-subweibull for $q\leq 1$ but not
q-subweibull for any $q>1$.

###### Proof.

Since the Poisson distribution has a finite MGF with infinite radius of
convergence, it is strictly subexponential and by Corollary 3.0.2 strictly
q-subweibull for all $q\leq 1$. Let $X\sim Poi(\mu)$. Without loss of
generality assume $t>1$ and let $n=\lfloor t\rfloor+1$ with $t<n\leq t+1$.

$\Pr(X>t)=\sum_{j=n}^{\infty}\Pr(X=j)\geq\Pr(X=n)=\frac{\mu^{n}\exp(-\mu)}{n!}=\frac{\mu^{n}\exp(-\mu)}{n\Gamma(n)}$

Since $t\Gamma(t)$ is increasing for $t\geq 1$, we have
$n\Gamma(n)\leq(t+1)\Gamma(t+1)$. Also, $\Gamma(n)\leq n^{n}$ for $n\geq 1$.
For the $\mu^{n}$ term, it is increasing for $\mu\geq 1$ and decreasing for
$\mu<1$, so $\mu^{n}\geq\min\\{\mu^{t+1},\mu^{t}\\}=\mu^{t}\min\\{\mu,1\\}$.
Combining these we obtain

$\Pr(X>t)\geq\frac{\mu^{t}\min\\{\mu,1\\}e^{-\mu}}{(t+1)\Gamma(t+1)}=\frac{\mu^{t}\min\\{\mu,1\\}e^{-\mu}}{(t+1)(t)\Gamma(t)}\geq\frac{\mu^{t}\min\\{\mu,1\\}e^{-\mu}}{(t+1)(t)t^{t}}$

To assess whether the tail follows a subweibull rate of decay, choose any
$\lambda>0$ and $q>1$, then

$\displaystyle\limsup_{t\to\infty}$ $\displaystyle\Pr(X>t)\exp(\lambda
t^{q})\geq\min\\{\mu,1\\}e^{-\mu}\lim_{t\to\infty}\frac{\mu^{t}}{(t+1)(t)t^{t}}\exp(\lambda
t^{q})$
$\displaystyle=\min\\{\mu,1\\}e^{-\mu}\exp\left[\lim_{t\to\infty}t\log\mu-\log(t+1)-\log
t-t\log t+\lambda t^{q}\right]$

The expression inside brackets is of the form $\infty-\infty$ so we rearrange
terms and apply L’Hopital’s rule. Define

$\displaystyle\lim_{t\to\infty}$ $\displaystyle t\log\mu-\log(t+1)-\log
t-t\log t+\lambda t^{q}$ $\displaystyle=\lim_{t\to\infty}(t\log
t)\left[\lim_{t\to\infty}\frac{\log\mu}{\log t}-\frac{\log(t+1)}{t\log
t}-\frac{1}{t}+\frac{\lambda t^{q}}{t\log t}\right]$
$\displaystyle=\lim_{t\to\infty}(t\log
t)\left[\lim_{t\to\infty}0-\frac{1/(t+1)}{1+\log t}-(0)+\frac{\lambda
qt^{q-1}}{1+\log t}\right]$ $\displaystyle=\lim_{t\to\infty}(t\log
t)\left[\lim_{t\to\infty}\frac{\lambda
q(q-1)t^{q-2}}{1/t}\right]=\infty\cdot\infty=\infty$

Therefore

$\limsup_{t\to\infty}\Pr(X>t)\exp(\lambda t^{q})=\infty$

Since this holds for all $\lambda>0$, $X$ cannot satisfy Corollary 3.0.1 and
therefore is not q-subweibull for any $q>1$. ∎

##  4 Exponential tilting

######  Definition 4.1.

Let $X$ be a random variable with distribution function $F$. If the Laplace
transform satisfies $\mathcal{L}_{X}(-\theta)<\infty$ for some $\theta\neq 0$,
then the exponentially tilted distribution is given by

$F_{\theta}(x)=\int_{-\infty}^{x}\frac{\exp(\theta
t)}{\mathcal{L}_{X}(-\theta)}dF(t)$

We adopt the convention of using $-\theta$ instead of $\theta$ so that the
interpretation of the tilting parameter is consistent with other works that
assume $X$ has an MGF, in which case one could equivalently require
$M_{X}(\theta)<\infty$.

From the Radon-Nikodym theorem, $F_{\theta}$ is absolutely continuous with
respect to $F$. Since the density function $e^{\theta
x}/\mathcal{L}_{X}(-\theta)$ is also strictly positive, exponential tilting
does not change the support. Generally speaking it is possible to produce a
subexponential distribution by exponential tilting of any distribution with at
least one light tail.

######  Proposition 4.1.

If $X\sim F$ is a random variable having at least one light tail then
exponential tilting is possible for all $\theta$ in some open interval
$(-S,T)$ with $S,T\geq 0$ and $S+T>0$. The resulting tilted distribution
$F_{\theta}$ is subexponential with MGF
$M_{Z}(t)=\mathcal{L}_{X}(-\theta-t)/\mathcal{L}_{X}(-\theta)$ finite for all
$t\in(-S-\theta,T-\theta)$.

###### Proof.

Without loss of generality assume the right tail is light so
$\mathcal{L}_{X}(-\theta)<\infty$ for some $\theta>0$. For all
$\theta^{\prime}\in[0,\theta)$,

$\mathcal{L}_{X}(-\theta^{\prime})=\operatorname{\mathbb{E}}[\exp(\theta^{\prime}X)]\leq\operatorname{\mathbb{E}}[\exp(\theta
X)]<\infty$

Set $T=\sup\\{\theta:~{}\mathcal{L}_{X}(-\theta)<\infty\\}>0$. If $X$ has a
heavy left tail then $\mathcal{L}_{X}(-\theta)=\infty$ for all $\theta<0$, so
the interval of convergence is $(-S,T)$ with $S=0$. If $X$ has a light left
tail then we can set $S=-\inf\\{\theta:\mathcal{L}_{X}(-\theta)<\infty\\}>0$.
This establishes the interval is $(-S,T)$ with $S,T\geq 0$ and $S+T>0$. Let
$Z\sim F_{\theta}$ follow the tilted distribution with $\theta\in(-S,T)$. Its
Laplace transform is

$\displaystyle\mathcal{L}_{Z}(t)$
$\displaystyle=\operatorname{\mathbb{E}}[\exp(-tZ)]=\int_{-\infty}^{\infty}\exp(-tz)dF_{\theta}(z)=\int_{-\infty}^{\infty}\exp(-tx)\frac{\exp(\theta
x)}{\mathcal{L}_{X}(-\theta)}dF(x)$
$\displaystyle=\operatorname{\mathbb{E}}[\exp(-(t-\theta)X)]/\mathcal{L}_{X}(-\theta)=\mathcal{L}_{X}(-(\theta-t))/\mathcal{L}_{X}(-\theta)$

This is finite when $\theta-t\in(-S,T)$ or equivalently
$t\in(-T+\theta,S+\theta)$. Since $\theta\in(-S,T)$, the interval of
convergence for $\mathcal{L}_{Z}(t)$ is an open interval containing zero,
which proves $Z$ is subexponential and has the MGF

$M_{Z}(t)=\mathcal{L}_{Z}(-t)=\mathcal{L}_{X}(-\theta-t)/\mathcal{L}_{X}(-\theta)$

which is finite on the interval $t\in(-S-\theta,T-\theta)$. ∎

As an example, if $X\sim F$ is a nonnegative, heavy tailed random variable
($T=0$), its left tail is strictly subexponential ($S=\infty$) so exponential
tilting is possible for all $\theta<0$. By Proposition 4.1 the resulting
tilted distribution is subexponential and hence has lighter tails than the
original distribution. On the other hand, if $X$ is broadly subexponential,
exponential tilting produces another broadly subexponential distribution, with
a shifted interval of convergence.

While exponential tilting can alter the tail behavior of heavy tailed and
broadly subexponential distributions, it does not affect the tail behavior of
q-subweibull distributions with lighter than exponential tails (i.e., $q>1$).

######  Lemma 4.0.1.

Preservation of nonnegative subweibull tails under exponential tilting. Let
$\theta$ be any real number. If $X\sim F$ is nonnegative and q-subweibull
($q>1$), then the exponentially tilted variable $Z\sim F_{\theta}$ is also
nonnegative and q-subweibull with the same radius of convergence.

  1. 1.

$\operatorname{\mathbb{E}}[\exp(\lambda^{q}X^{q})]<\infty$ for all
$\lambda\in[0,R_{q})$ implies
$\operatorname{\mathbb{E}}[\exp(\lambda^{q}Z^{q})]<\infty$ for all
$\lambda\in[0,R_{q})$.

  2. 2.

$\operatorname{\mathbb{E}}[\exp(\lambda^{q}X^{q})]=\infty$ for all
$\lambda>R_{q}$ implies
$\operatorname{\mathbb{E}}[\exp(\lambda^{q}Z^{q})]=\infty$ for all
$\lambda>R_{q}$.

###### Proof.

If $X$ is q-subweibull with $q>1$ then by Corollary 3.0.2 it is strictly
subexponential and $\mathcal{L}_{X}(-\theta)<\infty$ for all
$\theta\in\mathbb{R}$. Let $Z\sim F_{\theta}$. The MGF of $Z^{q}$ is

$\operatorname{\mathbb{E}}[\exp(\lambda^{q}Z^{q})]=\frac{\int\exp(\lambda^{q}x^{q}+\theta
x)dF(x)}{\mathcal{L}_{X}(-\theta)}$

$(1)$ case of $\lambda<R_{q}$. If $\theta\leq 0$ then

$\int\exp(\lambda^{q}x^{q}+\theta
x)dF(x)\leq\int\exp(\lambda^{q}x^{q}+0)dF(x)=\operatorname{\mathbb{E}}[\exp(\lambda^{q}X^{q})]<\infty$

If $\theta>0$, choose $\rho\in(\lambda,R_{q})$ and define

$x^{\star}=\left(\frac{\theta}{\rho^{q}-\lambda^{q}}\right)^{\frac{1}{q-1}}$

Then for $x>x^{\star}$, $\lambda^{q}x^{q}+\theta x\leq\rho^{q}x^{q}$.
Therefore,

$\displaystyle\int\exp(\lambda^{q}x^{q}+\theta x)dF(x)$
$\displaystyle=\int_{0}^{x^{\star}}\exp(\lambda^{q}x^{q}+\theta
x)dF(x)+\int_{x^{\star}}^{\infty}\exp(\lambda^{q}x^{q}+\theta x)dF(x)$
$\displaystyle\leq\int_{0}^{x^{\star}}\exp\big{(}\theta
x^{\star}+\lambda^{q}(x^{\star})^{q}\big{)}dF(x)+\int_{x^{\star}}^{\infty}\exp(\rho^{q}x^{q})dF(x)$
$\displaystyle\leq\exp\big{(}\theta
x^{\star}+\lambda^{q}(x^{\star})^{q}\big{)}\Pr(X\leq
x^{\star})+\int_{0}^{\infty}\exp(\rho^{q}x^{q})dF(x)$ $\displaystyle<\infty$

$(2)$ case of $\lambda>R_{q}$. If $\theta\geq 0$ then

$\int\exp(\lambda^{q}x^{q}+\theta
x)dF(x)\geq\int\exp(\lambda^{q}x^{q}+0)dF(x)=\operatorname{\mathbb{E}}[\exp(\lambda^{q}X^{q})]=\infty$

If $\theta<0$. Choose $\rho\in(R_{q},\lambda)$ and define

$x^{\star}=\left(\frac{-\theta}{\lambda^{q}-\rho^{q}}\right)^{\frac{1}{q-1}}$

Then for $x>x^{\star}$, $\lambda^{q}x^{q}+\theta x\geq\rho^{q}x^{q}$.
Therefore,

$\displaystyle\int\exp(\lambda^{q}x^{q}+\theta x)dF(x)$
$\displaystyle=\int_{0}^{x^{\star}}\exp(\lambda^{q}x^{q}+\theta
x)dF(x)+\int_{x^{\star}}^{\infty}\exp(\lambda^{q}x^{q}+\theta x)dF(x)$
$\displaystyle\geq\int_{0}^{x^{\star}}\exp(\lambda^{q}x^{q}+\theta
x)dF(x)+\int_{x^{\star}}^{\infty}\exp(\rho^{q}x^{q})dF(x)$

The first term is finite. We will show the second term is infinite. By
assumption,

$\displaystyle\int_{0}^{\infty}\exp(\rho^{q}x^{q})dF(x)$
$\displaystyle=\infty$
$\displaystyle=\int_{0}^{x^{\star}}\exp(\rho^{q}x^{q})dF(x)+\int_{x^{\star}}^{\infty}\exp(\rho^{q}x^{q})dF(x)$

But

$\int_{0}^{x^{\star}}\exp(\rho^{q}x^{q})dF(x)\leq\exp\big{(}\rho^{q}(x^{\star})^{q}\big{)}\Pr(X\leq
x^{\star})<\infty$

Therefore

$\int_{x^{\star}}^{\infty}\exp(\rho^{q}x^{q})dF(x)=\infty$

implying

$\int\exp(\lambda^{q}x^{q}+\theta x)dF(x)=\infty$

as well.

∎

We now extend Lemma 4.0.1 to general random variables.

######  Theorem 4.1.

Preservation of subweibull tails under exponential tilting. Let $\theta$ be
any real number.

  1. 1.

If $X\sim F$ is q-subweibull ($q>1$) with radius of convergence $R_{q}$, then
the exponentially tilted variable $Z\sim F_{\theta}$ is also q-subweibull and
has the same radius of convergence.

  2. 2.

If $X\sim F$ is strictly q-subweibull ($q\geq 1$), the exponentially tilted
variable $Z\sim F_{\theta}$ is also strictly q-subweibull.

  3. 3.

If $X\sim F$ is not q-subweibull ($q>1$), then $Z\sim F_{\theta}$ is also not
q-subweibull.

###### Proof.

$(1)$ By Corollary 3.0.2, $X$ is strictly subexponential so
$\mathcal{L}_{X}(-\theta)<\infty$ for all $\theta\in\mathbb{R}$. Choose any
arbitrary $\theta$ and set $M_{1}=\mathcal{L}_{X}(-\theta)$. Define
nonnegative random variables $A=[X\operatorname{\left.\right|}X<0]$ and
$B=[X\operatorname{\left.\right|}X\geq 0]$ with distributions $F^{-}$ and
$F^{+}$, respectively. By Lemma 3.0.1 both $A$ and $B$ are q-subweibull and
strictly subexponential. Let $R_{qa}$ and $R_{qb}$ be the radii of convergence
of $A$ and $B$, respectively. Let $p=\Pr(X<0)$ and assume $p\notin\\{0,1\\}$
(otherwise simply apply Lemma 4.0.1 to $X$ or $-X$). Note that

$M_{1}=\mathcal{L}_{A}(\theta)p+\mathcal{L}_{B}(-\theta)(1-p)$ (1)

We have

$\displaystyle\operatorname{\mathbb{E}}[\exp(\lambda^{q}|Z|^{q})]$
$\displaystyle=\int_{-\infty}^{\infty}\exp(\lambda^{q}|z|^{q})dF_{\theta}(z)=\int_{-\infty}^{\infty}\exp(\lambda^{q}|x|^{q})\frac{\exp(\theta
x)}{M_{1}}dF(x)$
$\displaystyle=\frac{\operatorname{\mathbb{E}}[\exp(\lambda^{q}|X|^{q}+\theta
X)]}{M_{1}}$
$\displaystyle=\frac{p}{M_{1}}\operatorname{\mathbb{E}}[\exp(\lambda^{q}A^{q}-\theta
A)]+\frac{(1-p)}{M_{1}}\operatorname{\mathbb{E}}[\exp(\lambda^{q}B^{q}+\theta
B)]$
$\displaystyle=\frac{p\mathcal{L}_{A}(\theta)}{M_{1}}\int_{0}^{\infty}\exp(\lambda^{q}x^{q})\frac{\exp(-\theta
x)}{\mathcal{L}_{A}(\theta)}dF^{-}(x)\ldots$
$\displaystyle~{}\ldots+\frac{(1-p)\mathcal{L}_{B}(-\theta)}{\mathcal{L}_{X}(-\theta)}\int_{0}^{\infty}\exp(\lambda^{q}x^{q})\frac{\exp(\theta
x)}{\mathcal{L}_{B}(-\theta)}dF^{+}(x)$
$\displaystyle=(\tilde{p})\int_{0}^{\infty}\exp(\lambda^{q}z^{q})dF_{(-\theta)}^{-}(z)+(1-\tilde{p})\int_{0}^{\infty}\exp(\lambda^{q}z^{q})dF_{\theta}^{+}(z)$
$\displaystyle=\tilde{p}\operatorname{\mathbb{E}}[\exp(\lambda^{q}U^{q})]+(1-\tilde{p})\operatorname{\mathbb{E}}[\exp(\lambda^{q}V^{q})]$

where (see Equation 1), $\tilde{p}=p\mathcal{L}_{A}(\theta)/M_{1}$ so that
$\tilde{p}\in(0,1)$. The nonnegative random variable $U$ is distributed as
$F_{(-\theta)}^{-}$, which is the exponentially tilting of $A~{}F^{-}$ by
$-\theta$ and $V\sim F_{\theta}^{+}$ is similarly defined as the exponential
tilting of $B\sim F^{+}$. By Lemma 4.0.1, this implies $U$ and $V$ are
q-subweibull with radii of convergence $R_{qa}$ and $R_{qb}$, respectively.
Let $R_{qz}$ be the radius of convergence of $Z$. Note that

$\displaystyle\Pr(Z\geq 0)$
$\displaystyle=\int_{0}^{\infty}dF_{\theta}(z)=\int_{0}^{\infty}\frac{\exp(\theta
x)}{\mathcal{L}_{X}(-\theta)}dF(x)=\frac{\operatorname{\mathbb{E}}[\exp(\theta
X)\operatorname{\left.\right|}X\geq 0]\Pr(X\geq 0)}{M_{1}}$
$\displaystyle=\frac{\operatorname{\mathbb{E}}[\exp(\theta
B)](1-p)}{M_{1}}=\frac{\mathcal{L}_{B}(-\theta)(1-p)}{M_{1}}$
$\displaystyle=1-\tilde{p}$

So $\Pr(Z<0)=\tilde{p}$. By Lemma 3.0.1 this implies $Z$ is q-subweibull with
radius of convergence $\min\\{R_{qa},R_{qb}\\}$, which is also the radius of
convergence of $X$.

$(2)$ For $q=1$ apply Proposition 4.1 with $S=\infty$ and $T=\infty$. For
$q>1$, apply $(1)$ with $R_{q}=\infty$.

$(3)$ This is a direct corollary of $(1)$ obtained in the case of $R_{q}=0$. ∎

##  5 Discussion

The theory of subexponential and subgaussian distributions is a key
prerequisite to many results in nonparametric and nonasymptotic statistical
inference such as concentration inequalities. A comprehensive overview with
applications to high-dimensional machine learning problems is provided by [3].
Exponential tilting is used in a variety of statistical areas such as causal
inference [6] and Monte Carlo sampling [2]. If $F_{\theta}$ is a tilted
distribution, it is a natural exponential family with parameter $\theta$. The
exponential families are building blocks for generalized linear models [7].
For applications to machine learning see [4, 5].

Here, we have provided a brief overview of subweibull distributions. We showed
that the Poisson distribution is strictly 1-subweibull but not q-subweibull
for any $q>1$. Finally, we detailed the conditions under which the subweibull
property is perserved under exponential tilting.

## Acknowledgements

Thanks to Arun Kumar Kuchibhotla, Sam Power, Patrick Staples, Valerie Ventura,
Larry Wasserman, and Matt Werenski for helpful comments and suggestions.

## References

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