How do I compute $\mathbb{E}\left[e^{\lambda X_t} X_s^2 \right]$ for $X_t= \mu t + W_t$ where $(W_t)_t$ is a Brownian motion

Question

I want to compute $\mathbb{E}\left[e^{\lambda X_t} X_s^2 \right]$ with $X_t= \mu t + W_t$. Also $0 0$ and $W_t$ is a standard Brownian Motion.

To do this I first computed \begin{align*} \mathbb{E}\left[e^{\lambda X_t}X_s^2\right] &= e^{\lambda \mu (t-s)} \mathbb{E}\left[e^{\lambda (W_t-W_s)} \right] \mathbb{E}\left[e^{\lambda X_s} X_s^2\right] \\ &=e^{\lambda \mu (t-s)} \mathbb{E}\left[e^{\lambda (W_t-W_s)} \right] \mathbb{E}\left[e^{\lambda X_s}\right] e^{\lambda \mu s} \mathbb{E}\left[ e^{\lambda W_s} \left(W_s^2 +2 \mu s W_s + \mu^2 s^2 \right) \right] \end{align*} Now since $W_t-W_s \sim N(0,t-s)$ we can substitute this in $\mathbb{E}\left[e^{\lambda (W_t-W_s)} \right]$ to compute this using integral manipulation.

Can I also do this for $\mathbb{E}\left[ e^{\lambda W_s} \left(W_s^2 +2 \mu s W_s + \mu^2 s^2 \right) \right]$? (I know I can split this into 3 parts by the linearity of the expectation)

Or is there another easier way of evaluating this expectation?

Answer 1

Yes, it's possible to do this without calculating all these integrals.

1. Show (or recall) that for any Gaussian random variable $Y \sim N(m,\sigma^2)$, we have $$\mathbb{E}(e^{\lambda Y}) = \exp \left( \lambda m+ \frac{1}{2} \lambda^2 \sigma^2 \right). \tag{1}$$
2. Differentiating $(1)$ two times with respect to $\lambda$ yields $$\mathbb{E}(Y^2 e^{\lambda Y}) = \exp \left( \lambda m+\frac{1}{2} \sigma^2 \lambda \right) ((m+\sigma^2 \lambda)^2+\sigma^2). \tag{2}$$
3. As $$\mathbb{E}(e^{\lambda X_t} X_s^2) = \mathbb{E}(X_s^2 e^{\lambda X_s}) \underbrace{\mathbb{E}(e^{\lambda (X_t-X_s)})}_{\mathbb{E}e^{\lambda X_{t-s}}}$$ and $X_r \sim N(\mu r,r)$ for any $r \geq 0$, we can use $(1)$ and $(2)$ to calculate $\mathbb{E}(e^{\lambda X_t} X_s^2)$.

Note that $(2)$ also gives $\mathbb{E}(W_s^2 e^{\lambda W_s})$, and therefore you can also use this to continue with your calculations.