$X_i\sim\text{Exp}(1), S_n=\sum_{i=1}^nX_i.$ $\mathbb{E}(\frac{e^{S_{n+1}}(n+1)!}{(1+S_{n+1})^{n+2}}|\mathcal{F_n})=\frac{e^{S_n}n!}{(1+S_n)^{n+1}}$?

Question

$\newcommand{\e}{\operatorname{\mathbb E}}$ We need to show that $ \displaystyle \e \left( \frac{e^{S_{n+1}}(n+1)!}{(1+S_{n+1})^{n+2}} \mid \mathcal{F_n} \right) = \frac{e^{S_n}n!}{(1+S_n)^{n+1}}$.

The $X_i$'s are iid.

My only line of working so far is:

\begin{align} & \e \left( \frac{e^{S_{n+1}}(n+1)!}{(1+S_{n+1})^{n+2}} \mid \mathcal{F_n} \right) \\[8pt] = {} & e^{S_n}(n+1)!\e \left( \frac{e^{X_{n+1}}}{(1+S_n+X_{n+1})^{n+2}}\mid\mathcal{F_n} \right) \\[8pt] = {} & e^{S_n}(n+1)! \e \left(\frac{e^{X_{n+1}}}{(1+S_n+X_{n+1})^{n+2}}\right) \end{align}

At this point I'm not sure on how we can proceed? Maybe by using the definition of expectation we could integrate to get the expectation (but this looks long-winded - perhaps there's a neater way)?

Answer 1

$\newcommand{\e}{\operatorname E}$ \begin{align} & \e \left(\frac{e^{X_{n+1}}}{(1+S_n+X_{n+1})^{n+2}} \,\big|\, \mathcal F_n \right) \\[8pt] = {} & \int_0^\infty \frac{e^x}{(1+S_n + x)^{n+2}} \big( e^{-x}\,dx\big) \\[8pt] = {} & \int_0^\infty \frac{dx}{(1+S_n + x)^{n+2}} = \frac 1 {(n+1)(1+S_n)^{n+1}}. \end{align}