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Customers arrive at a facility according to a Poisson process $N(t)$ of rate $\lambda$, with arrival times $S_1$, $S_2$, ...... Each customer pays \$1 on arrival. At time $t$, the discounted value of the total sum collected so far, discounted back to time zero, is $\sum_{i=1}^{N(t)}e^{- \beta S_i}$ , where $\beta> 0$ is the discount rate. Show that the discounted value at time $t$ has expectation 􀀀$\lambda \beta^{-1}(1-\exp(-\beta t))$.

My Attempt:

The sum $\sum_{i=1}^{N(t)}e^{- \beta S_i}=e^{- \beta S_1}+e^{- \beta S_2}+....+e^{- \beta S_{N(t)}}$

If want to take the expected value, then we can just do:

$n*E[e^{- \beta S_i}]$, which is easy to calculate. However, since $N(t) $ itself is also a random variable, I don't know how to calculate $n$.

What about condition on $N(t)=k$?

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  • $\begingroup$ Is the term in the sum supposed to be $e^{-\beta S_i}$? As written it is $e^{-\beta>S_i}$ which doesn't make sense. $\endgroup$
    – Math1000
    Commented Mar 17, 2016 at 8:43
  • $\begingroup$ yes. It is $e^{-\beta S_i}$ $\endgroup$
    – randy
    Commented Mar 17, 2016 at 18:44

1 Answer 1

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We apply the total expectation theorem as follows: $$ \mathbf{E} \sum_{i=1}^{N(t)} e^{-\beta S_i}=\sum_{k=0}^{\infty} \mathbf{E} \left[\sum_{i=1}^{N(t)} e^{-\beta S_i}| N(t)=k\right]e^{-\lambda t}\frac{(\lambda t)^k}{k!}.$$ Then we need to compute $$ \mathbf{E} \left[\sum_{i=1}^{N(t)} e^{-\beta S_i}| N(t)=k\right] = \mathbf{E} \left[\sum_{i=1}^{k} e^{-\beta S_i}| N(t)=k\right].$$ Given that $N(t)=k$, the arrival times $S_1, \cdots, S_k$ are chosen randomly in the interval $[0,t]$ to satisfy the only restriction $S_1\leq \cdots \leq S_k$. In fact, the $k$-tuples $(S_1, \cdots , S_k)$ are uniformly distributed on $k$-dimentional polygon $$ \{(x_1, \cdots , x_k) | 0\leq x_1 \leq \cdots \leq x_k \leq t \},$$ which has volume $t^k/k!$.

We now fix $1\leq i\leq k$, and compute $$ \mathbf{E}\left[e^{-\beta S_i} | N(t)=k\right].$$
This can be found from an integral: $$ \frac{k!}{t^k}\int_0^t \frac{u^{i-1}}{(i-1)!} e^{-\beta u} \frac{(t-u)^{k-i}}{(k-i)!}du.$$ Summing over $1\leq i\leq k$, we obtain by binomial theorem that $$ \begin{align} \mathbf{E} \left[\sum_{i=1}^{k} e^{-\beta S_i}| N(t)=k\right]&=\sum_{i=1}^k \mathbf{E}\left[e^{-\beta S_i} | N(t)=k\right]\\ &=\sum_{i=1}^k \frac{k!}{t^k}\int_0^t \frac{u^{i-1}}{(i-1)!} e^{-\beta u} \frac{(t-u)^{k-i}}{(k-i)!}du\\ &=\frac{k}{t^k}\int_0^t \sum_{i=1}^k \frac{(k-1)!}{(i-1)!(k-i)!} u^{i-1}(t-u)^{k-i} e^{-\beta u}du\\ &=\frac{k}{t^k}\int_0^t t^{k-1} e^{-\beta u} du =\frac k {t\beta} (1-e^{-\beta t}). \end{align} $$ Putting this back into the sum over $k$, the sum becomes $$ \frac1{t\beta}(1-e^{-\beta t}) \mathbf{E}N(t).$$ Since $N(t)$ is Poisson distribution of parameter $\lambda t$, we have $\mathbf{E}N(t)=\lambda t$, thus we conclude that $$ \mathbf{E} \sum_{i=1}^{N(t)} e^{-\beta S_i}=\frac{\lambda}{\beta}(1-e^{-\beta t}).$$

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  • $\begingroup$ Just noticed that this could be done easier if we ignore the order between $S_i$'s and regard them as the independent uniform random variables on $[0,t]$. $\endgroup$ Commented Mar 19, 2016 at 0:16

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