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