Let $B_t$ be a Brownian movement.
Let $X_t = \mu t + B_t $, where $\mu$ is a constant and $t$ is a variable, but not a Random Variable.
Let $s<t$ and $\lambda >0$.
I need to solve $\mathbb{E}[e^{X_t\cdot\lambda}X_s^{2}]$.
I am trying to use the independent increments of the Brownian Movement. Hence, I tried to define this equality: $X_t = X_s + (X_t - X_s)$.
I ended up with:
$\mathbb{E}[e^{X_s\cdot\lambda}\cdot e^{(X_t-X_s)\cdot \lambda}\cdot X_s^{2}]$
Now, I can split the products:
$\mathbb{E}[e^{X_s\cdot\lambda}\cdot X_s^{2}]\cdot \mathbb{E}[e^{(X_t-X_s)\cdot \lambda}]$
However, I am stucked again. How can I solve this?
Thanks in advance.
Note - I'm not providing full answer, rather hints.
Once certain regularity conditions* are fulfilled, you can write $$\frac{\partial}{\partial \lambda} \mathbb{E} e^{X_t \lambda} = \mathbb{E} \frac{\partial}{\partial \lambda} e^{X_t \lambda} = \mathbb{E} X_t e^{X_t \lambda}$$
Use this twice, with some minor tweaks, to get the result. You should combine this with the fact, that moment generating function for normal random variable has known, explicit form, i.e. $$\mathbb{E} e^{X_t \lambda} = \mathbb{E} e^{(B_t + \mu t) \lambda} = e^{\mu t \lambda} \mathbb{E} e^{ B_t \lambda} = e^{\mu t \lambda} e^{\frac{t\lambda}{2}}$$.
*you need to be able to interchange integral and derivative: https://en.wikipedia.org/wiki/Leibniz_integral_rule