I want to compute $\mathbb{E}[(B_n+n)^3\mathrm{exp}(-B_n-n/2)]$, where $B_n$ denotes the one dimensional discrete Brownian motion.
Clearly $\mathrm{exp}(-B_n-n/2)$ is a Girsanov transformation, thus it seems as it would natural to use this somehow. But I am very uncertain as to how to proceed. Any help is greatly appreciated!
Thank you!
Since $B_n\sim N(0,n)$, \begin{align} \mathsf{E}(B_n+n)^3e^{-B_n-n/2}&=\frac{1}{\sqrt{2\pi n}}\int_{-\infty}^{\infty}(x+n)^3e^{-x-\frac{n}{2}-\frac{x^2}{2n}} \,dx \\ &=\frac{1}{\sqrt{2\pi n}}\int_{-\infty}^{\infty}(x+n)^3e^{-\frac{(x+n)^2}{2n}} \,dx \\[1em] &=\mathsf{E}(Z+n)^3=\ldots, \end{align} where $Z\sim N(-n,n)$.