Show that $\mathbb{E}[X_t]=X_0e^{-ct}$ if $X_t=X_0e^{-ct}+\sigma e^{-ct}\int_0^te^{cs}dW_s$, $X_0\in\mathbb{R}$

Question

so I know the result is trivially correct, but I am being asked to prove it. I tried using a theorem, but it seems rather contradictory. Thanks in advance!

Question: Show that $\mathbb{E}[X_t]=X_0e^{-ct}$ if $X_t=X_0e^{-ct}+\sigma e^{-ct}\int_0^te^{cs}dW_s$, $X_0\in\mathbb{R}$ and $c,\sigma>0$

Attempt:

Let $M_t=e^{cs} dWs$, then if $\mathbb{E}\int_0^t(M_t)^2<\infty$ for all $t\ge 0$, we will have that $\{M_t\}_{t\ge 0}$ is a martingale, so $\mathbb{E}(M_t)=\mathbb{E}(M_0)=0$.

However, I get $$\mathbb{E}\int_0^te^{2cs}ds=\left[\frac{e^{2cs}}{2c}\right]_0^t=\frac{e^{2ct}-1}{2c}$$ However, as far as I know that goes to $\infty$ as $t\to \infty$.

Is this the right approach?

Answer 1

If $M = \int h dX$ where $X$ is a continuous square integrable martingale, you don't need $E\int_0^\infty h_s^2d\langle X\rangle_s < \infty$ for $M$ to be a martingale, you just need $E\int_0^t h_s^2d\langle X\rangle_s < \infty$ for all $t$. You did show this, and hence your $M$ is a martingale and you can say $E M_t = E M_0$ to solve your problem.

Answer 2

To add onto nullUser's response. Of course $\int_0^t 1 dW_s=W_t$ is a martingale right? Well:

$$ \int_0^{\infty} 1 ds=\infty$$

This is just to give a solid counterexample, the rest was explained in his post.