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, 2019 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, 2019 at 17:39

1 Answer 1


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, 2019 at 19:55
  • $\begingroup$ @duquesne You are welcome; I'm glad that I could help you. $\endgroup$
    – saz
    Mar 3, 2019 at 20:42

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