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Let $N_t$ be a Poisson process with rate $\lambda$.
$T_k$ the inter arrival times of $N_t$.
$\{X_k\}$ a collection of i.i.d. random variables with mean $\mu$.
$X_k$ is independent of $N_t$.
Calculate the expectation of $$ S_t= \sum_{k=1}^{N_t} X_k e^{t-T_k}. $$

Given $N_t$, the inter arrival times are uniformly distributed on $[0,t]$.
Hence, $T_k \sim \text{Beta}(k,n-k+1)$ and $$ E\left( \left. e^{-T_k}\right| N_t=n \right)=\frac{1}{B(k,n-k+1)}\int_0^1 e^{-x}x^{k-1} (1-x)^{n-k} dx. $$ I don't see how to compute this integral.

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1 Answer 1

up vote 2 down vote accepted

Use $\frac{1}{\operatorname{B}(k,n-k+1)} = n \binom{n-1}{k-1}$: $$ \sum_{k=1}^n \frac{x^{k-1} (1-x)^{n-k}}{\operatorname{B}(k,n-k+1)} = n \sum_{k=1}^{n} \binom{n-1}{k-1} x^{k-1} (1-x)^{(n-1)-(k-1)} = n $$ Thus: $$ \mathbb{E}\left( \sum_{k=1}^{N_t} X_k \mathrm{e}^{t-T_k} \right) = \mathbb{E}\left( \mathbb{E}\left( \sum_{k=1}^{N_t} X_k \mathrm{e}^{t-T_k} \Big| N_t \right) \right) = \mathbb{E}(X) \mathbb{E}\left( N_t \int_0^1 \mathrm{e}^{t-t x} \mathrm{d} x \right) = \mathbb{E}(X) \mathbb{E}\left( N_t \right) \frac{\exp(t)-1}{t} = \lambda \left( \exp(t)-1 \right)\mathbb{E}(X) $$

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