Let $X$ be the strong solution to the SDE $$ dX_t = \tanh X_t \,dt + dW_t, $$ where $W$ is a scalar Brownian motion defined on a probability space $(\Omega, \mathcal{F} ,\mathbb{P})$. (Such solution exists since $\tanh$ is Lipschitz.)

We want to prove that

$ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \bigg\{ \frac{e^{t/2}}{\cosh X_t} \bigg\}_{t \geq 0}$ is a positive martingale.

It is tedious to apply Ito's formula to obtain that $$ d \bigg(\frac{e^{t/2}}{\cosh X_t} \bigg) = - \frac{e^{t/2} \sinh X_t}{(\cosh X_t)^2} dW_t.$$ Hence, it is a local martingale. But to prove that it is a martingale, one way is to show that $\mathbb{E} \bigg[ \bigg\langle \frac{e^{\cdot/2}}{\cosh X} \bigg\rangle_t\bigg] < + \infty$, for all $t \geq 0$. But this doesn't seem to work.

Another method is to note that $\bigg\{ \frac{e^{t/2}}{\cosh X_t} \bigg\}_{t \geq 0}$ is nonnegative, and hence is a supermartingale. If we can prove that the expectation remains the same, then we are also done. But computing the expectation of $\frac{1}{\cosh X_t}$ is difficult.

Any suggestions?


1 Answer 1


Since $\cosh(x) \ge 1$ for all real $x$ you have $\frac{e^{t/2}}{\cosh X_t} \le e^{t/2}$. So on any finite time interval you have a bounded local martingale, which is a martingale.

Alternatively, note that $\frac{\sinh(x)}{(\cosh(x))^2} \le \frac{1}{2}$ for all $x$. So for any $T$, $$\int_0^T E\left[\left|-\frac{e^{t/2} \sinh X_t}{(\cosh(X_t))^2}\right|^2\right]\,dt \le \int_0^T \frac{1}{4} e^{t}\,dt = \frac{1}{4} (e^T-1) < \infty$$ so that $\displaystyle \frac{e^{t/2}}{\cosh(X_t)} = \frac{1}{\cosh(X_0)} -\int_0^t \frac{e^{s/2} \sinh X_s}{(\cosh(X_s))^2}\,dW_s$ is a martingale.


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