How do I calculate the expectation of the process given by the SDE $$dX_t=-\tanh(X_t) dt + dW_t, \qquad X_0=x_0$$ and $W_t$ a Wiener process?

If I start with $$ d\left(e^{t/2}\sinh(X_t)\right) = e^{t/2}\cosh(X_t)dW_t\tag{1} $$

which gives

$$ \sinh(X_t) = e^{-t/2}\sinh(x_0)+\int_0^t e^{-(t-s)/2}\cosh(X_s)dW_s\label{a}\tag{2} $$

then, at first sight, the expectation would seem to be

$$ \mathbb{E} [\sinh(X_t)\,|\,X_0] = e^{-t/2}\sinh(x_0)\tag{3} $$

However, it seems that the Itô integral in (\ref{a}) may be not zero in expectation, that it is a local martingale but not a martingale? Perhaps also that

$$ \mathbb{E} [\sinh(X_t)\,|\,X_0] \le e^{-t/2}\sinh(x_0)\tag{4}\quad ? $$

In general, how can I calculate the expectation of a stochastic integral such as

$$ \int_0^t e^{-(t-s)} \cosh(X_s) dW_s\ \tag{5} $$

for a process $\{X_t\}$ such as above? Is there any way to do it explicitly?

I've been out of the game almost 20yrs, I'm a little rusty, so if what I have posted above is incorrect then please forgive me.

  • $\begingroup$ How do you get from (2) to (3)? Even if the stochastic integral is a martingale (and hence has expectation zero), then you get only $$\sinh^{-1}(\mathbb{E}(\sinh(X_t)) = \sinh^{-1}(e^{-t/2} \sinh(x_0))$$ which is not the same as $$\mathbb{E}(X_t) = \sinh^{-1}(e^{-t/2} \sinh(x_0))$$, right? $\endgroup$ – saz Mar 3 '19 at 17:36
  • $\begingroup$ Yes you are probably right, what an elementary mistake, clearly I can't match powers there, so I am likely incorrect already at that point. $\endgroup$ – duquesne Mar 3 '19 at 17:39

Step 1: Let's first construct a weak solution to the SDE $$dX_t = - \tanh(X_t) \, dt + dB_t, \qquad X_0 = x_0. \tag{1}$$

For a Brownian motion $(W_t)_{t \geq 0}$ on a probability space $(\Omega,\mathcal{A},\mathbb{P})$ we define

$$\tilde{X}_t :=- x_0 + W_t \quad \text{and} \quad \tilde{B}_t := W_t - \int_0^t \tanh(-x_0+W_s) \, ds.$$


$$d\tilde{B}_t = dW_t - \tanh(\tilde{X}_t) \, dt = d\tilde{X}_t - \tanh(\tilde{X}_t) \, dt. \tag{2}$$

By Girsanov's theorem, $(\tilde{B}_t)_{t \leq T}$ is a Brownian motion with respect to the probability measure $\mathbb{Q}_T$, $$d\mathbb{Q}_T := Z_T \, d\mathbb{P}, \tag{3}$$

where $T>0$ is fixed and

$$Z_T := \exp \left( \int_0^T \tanh(-x_0+W_s) \, dW_s - \frac{1}{2} \int_0^T \tanh^2(-x_0+W_s) \, ds \right).$$

Hence, by $(2)$, $(\tilde{X}_t)_{t \leq T}$ is a weak solution to $$d\tilde{X}_t = \tanh(\tilde{X}_t) \, dt + d\tilde{B}_t, \qquad \tilde{X}_0 = -x_0$$ on the probability space $(\Omega,\mathcal{A},\mathbb{Q}_T)$. Since $\tanh(x)=-\tanh(-x)$ we find that $X_t := -\tilde{X}_t$ is a weak solution to $$dX_t = - \tanh(X_t) \, dt + dB_t, \qquad X_0 = x_0$$ with $B_t := - \tilde{B}_t$.

Step 2: We need to compute $$\mathbb{E}_{\mathbb{Q}_T}(X_T) = - \mathbb{E}_{\mathbb{Q}_T}(\tilde{X}_T).$$

By Itô's formula, we have

$$d(\log(\cosh(-x_0+W_t)) = \tanh(-x_0+W_t) \, dW_t + \frac{1}{2} (1-\tanh^2(-x_0+W_t)) \, dt,$$

and so

\begin{align*}&\int_0^T \tanh(-x_0+W_t) \, dW_t - \frac{1}{2} \int_0^T \tanh^2(-x_0+W_t) \, dt \\ &= \log(\cosh(-x_0+W_T)) - \log(\cosh(-x_0))- \frac{T}{2}.\end{align*}

This implies

$$Z_T = \frac{\cosh(-x_0+W_T)}{\cosh(x_0)} \exp \left(- \frac{T}{2} \right). \tag{4}$$

Consequently, we conclude that

$$\mathbb{E}_{\mathbb{Q}_T}(\tilde{X}_T) \stackrel{(3)}{=} \mathbb{E}(Z_T \tilde{X}_T) \stackrel{(4)}{=} \exp \left(- \frac{T}{2} \right) \frac{1}{\cosh(x_0)} \mathbb{E}((-x_0+W_T) \cosh(-x_0+W_T)). \tag{5}$$

Since $W_T$ is Gaussian with mean zero and variance $T$, we have

\begin{align*} \mathbb{E}(W_T \cosh(-x_0+W_T)) &=\frac{1}{\sqrt{2\pi T}} \int_{\mathbb{R}} x \cosh(x-x_0) \exp \left(- \frac{x^2}{2T} \right) \, dx \\ &= - \frac{T}{\sqrt{2\pi T}} \int_{\mathbb{R}} \cosh(x-x_0) \frac{d}{dx} \exp \left( - \frac{x^2}{2T} \right) \, dx\end{align*}

and so, by the integration by parts formula,

\begin{align*} \mathbb{E}(W_T \cosh(-x_0+W_T)) &=- \frac{T}{\sqrt{2\pi T}} \int_{\mathbb{R}} \sinh(x-x_0) \exp \left( - \frac{x^2}{2T} \right) \, dx \\ &= -T \mathbb{E}(\sinh(-x_0+W_T)). \end{align*}

Consequently, we obtain from $(5)$ that

$$\mathbb{E}_{\mathbb{Q}_T}(\tilde{X}_T) = \exp \left(- \frac{T}{2} \right) \frac{1}{\cosh(x_0)} \left[ -x_0 \mathbb{E}(\cosh(x_0+W_T)) - T \mathbb{E}\sinh(-x_0+W_T) \right].$$

Writing $$\cosh(x) = \frac{e^x+e^{-x}}{2} \quad \text{and} \quad \sinh(x) = \frac{e^x-e^{-x}}{2}$$

and using the fact that the exponential moments of Gaussian random variables are known, we conclude that

$$\mathbb{E}_{\mathbb{Q}_T}(\tilde{X}_T) = \frac{1}{\cosh(x_0)} \left[ -x_0 \cosh(x_0) + T \sinh(x_0) \right] = -x_0 + T \tanh(x_0).$$


$$\mathbb{E}_{\mathbb{Q}_T}(X_T) = - \mathbb{E}_{\mathbb{Q}_T}(\tilde{X}_T) = x_0 - T \tanh(x_0).$$

  • $\begingroup$ Great, thank you so much saz, I was aware that this process (which I simplified from something slightly more complicated) has a $sech^2(x)$ distribution, but I could not calculate this expectation that you have now done. Everything I tried just collapsed it back into an OU process! I'll go through your solution precisely. I cannot upvote you yet until I get to a 15 reputation score it seems, but I'll be sure to come back and do that! $\endgroup$ – duquesne Mar 3 '19 at 19:55
  • $\begingroup$ @duquesne You are welcome; I'm glad that I could help you. $\endgroup$ – saz Mar 3 '19 at 20:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.