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Let $N_t$ a Poisson process with parameter $\lambda$.

$T_k$ the arrival times of $N_t$

Let $H_t=t-T_{N_t}$

Show that $E(H_t|N_t=k)=\frac{t}{k+1}$

This is what i did :

$E(H_t|N_t=k)=t-E(T_k|N_t=k)$

I know that for any Poisson process, conditional on the event $N_t = k$, the joint distribution of $(T_1,...,T_k)$ is the same as the joint distribution of the order statistics of $k$ i.i.d. uniform on $(0;t)$ but I am stuck to compute $E(T_k|N_t=k)$

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  • $\begingroup$ So... $P(T_k\leqslant s\mid N_t=k)=(s/t)^k$, right? Can you (prove this and) finish? (Note that $T_k$ denotes the $k$th arrival time, not the $k$th interarrival time.) $\endgroup$ – Did Oct 10 '16 at 23:16
  • $\begingroup$ its the joint distribution of $(T_1;...;T_k)$ is the same as the joint distribution of the order statistics of $k$ i.i.d. uniform on $(0;t)$ not $T_k$ (I edited it). Iam stuck to find the distribution of $T_k$ (only) conditional on the event $N_t=k$ $\endgroup$ – John Oct 10 '16 at 23:23
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You stated that the distribution of $T_k\mid N_t=k$ is that of the highest order statistic for a sequence of $k$ independent samples of $U\sim\mathcal U(0;t)$.   Then the density function is $$\begin{align}f_{T_k\mid N_t=k}(s) ~=~& k\, F_{U}^{k-1}(s)\,f_{U}(s) \\[1ex]~=~& k\cdot\tfrac {s^{k-1}}{t^k}\mathbf 1_{s\in(0;t)}\end{align}$$

Hence the expectation is:

$$\mathsf E(T_k\mid N_t=k) ~=~ k\cdot t^{-k}\int_0^t s\cdot s^{k-1}\operatorname d s$$

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