I have the following question. I have a random variabele that is the sum of the exponent of the jumping times $T_i$ of a Poisson process $N(\cdot)$ with parameter $\lambda$.
Say:
$$B(t) = \sum_{i=1}^{N(t)} e^{-r (t - \ T_i)} $$
I want to know $\mathbb{E}[B(t)]$. I can condition on $N(t)$ to see that $\mathbb{E}[B(t)] = \sum_{i=1}^\infty\mathbb{E}[B(t)|N(t)=k]\mathbb{P}[N(t)=k]$. Where $\mathbb{P}[N(t)=k] = e^{-\lambda t} \frac{(\lambda t)^k}{k!}$ is a Poisson distribution of parameter $\lambda t$.
To understand the problem I first calculated $\mathbb{E}[B(t)|N(t)=1]$, where since we condition on having one jump the jump is uniformly distributed:
\begin{align} \mathbb{E}[B(t)|N(t)=1]&= \int_0^t \mathbb{P}(T_1=s)e^{-r(t-s)}ds \\ &= \int_0^t \frac{1}{t} e^{-rt}e^{rs}ds \\ &= \frac{e^{-rt}}{t} \int_0^t e^{rs} ds \\ &= \frac{1}{e^{rt}} \left[ \frac{e^{rs}}{r} \right]_0^t \\ &= \frac{1}{te^{rt}} \left( \frac{e^{rt}}{r} - 1 \right) = \left( \frac{1}{tr} - \frac{e^{-rt}}{t} \right) \end{align}
Which looks like it makes some sense. I know that the joint distribution of the jump times. Have a joint density distribution of $f(t_1,...,t_n)= \frac{n!}{t^n} 1_{(0 \leq t_1 \leq ... \leq t_n \leq t)}$ so the expected value generalised to some n would look like:
\begin{align} \mathbb{E}[B(t)|N(t)=n] &= \int_0^t ... \int_0^t f(t_1,...,t_n) \sum_{i=1}^n e^{-r(t-t_i)} dt_1 ... dt_n \\ &= \int_0^{t_2} ... \int_{t_{n-2}}^t \frac{n!}{t^n} \sum_{i=1}^n e^{-r(t-t_i)} dt_1 ... dt_n \end{align}
This I am not so sure how to evaluate. Anyone has an idea?