I wonder if someone can help me with this problem:
Let $X_t= \exp\{ -\frac{1}{4c}(1-e^{-2ct}) + \int_{0}^t e^{-cs}dW_s\}$ where $W_s$ is a standard Brownian motion and $c>0$. I'm trying to show that $X_t$ is a martingale.
My attempt goes like this:
Let $T>t$.
$\mathbb{E}\big[ X_T | \mathcal{F}_t \big] = \mathbb{E}\big[ \exp\{
> -\frac{1}{4c}(1-e^{-2cT}) + \int_{0}^T e^{-cs}dW_s\} | \mathcal{F}_t \big]
> = \\ \mathbb{E}\big[ \exp\{ -\frac{1}{4c}(1-e^{-2cT}) + \int_{0}^t e^{-cs}dW_s +\int_{t}^T e^{-cs}dW_s \} | \mathcal{F}_t \big] $$\exp\{ \int_{0}^t e^{-cs}dW_s \}$ is $\mathcal{F}_t$-measurable so
one can take it out of the expectation while $\exp\{ \int_{t}^T
> e^{-cs}dW_s \}$ is independent of $\mathcal{F}_t$. So we get:$ \exp\{ -\frac{1}{4c}(1-e^{-2cT}) + \int_{0}^t e^{-cs}dW_s
> \}\mathbb{E}\big[ \exp\{\int_{t}^T e^{-cs}dW_s \} \big] $
Is this approach correct? How can I calculate the value of $\mathbb{E}\big[ \exp\{\int_{t}^T e^{-cs}dW_s \} \big]$? I tried using the Ito-formula but had no success.
Best Answer
Let us define:
$f(t):=-\frac{1}{4c}(1-e^{-2ct})$
$Y_{t}:=\int_{0}^{t}e^{-cs}dW_{s}$
so
$dY_{t}=e^{-ct}dW_{t}$
and
$X_{t}=e^{f(t)+Y_{t}}$
From Ito formula, we get:
$dX_{t}=d(e^{f(t)+Y_{t}})=e^{f(t)+Y_{t}}f'(t)dt+e^{f(t)+Y_{t}}dY_{t}+\frac{1}{2}e^{f(t)+Y_{t}}dY_{t}dY_{t}$
$=e^{f(t)+Y_{t}}*(-\frac{1}{2}e^{-2ct})dt+e^{f(t)+Y_{t}}*e^{-ct}dW_{t}+\frac{1}{2}e^{f(t)+Y_{t}}*e^{-2cs}dt$
$=e^{f(t)+Y_{t}}*e^{-ct}dW_{t}=X_{t}e^{-ct}dW_{t}$
Summing up, we have:
$dX_{t}=X_{t}e^{-ct}dW_{t}$,
so
$X_{t}=X_{0}+\int_{0}^{t}X_{s}e^{-cs}dW_{s}$
Now, to solve your problem fully, follow the solution (to be more precise point Ad. 2, for your process $X_{t}$) of this problem: Link
All the calculations I leave to you.