Interarrival times for Poisson process Let $(N_t)_{t\geq 0}$ be the Poisson process with rate $\lambda$ and denote by $T_n:=\inf\{t\geq 0 : N_t=n\}$ the first hitting time of $n$. I want to understand the interarrival times $Y_t:=T_{N_t+1}-T_{N_t}$.
I thought those interarrival times are exponentially distributed, in particular independent of $t$, but from what I read it depends on $t$ and converges to a $Gamma(2,\lambda)$ distribution as $t\rightarrow \infty$. More precisely, $Y_t$ has density:
$$f_{Y_t} = \begin{cases} \lambda^2xe^{-\lambda x}, 0<x\leq t \\ \lambda(1+\lambda t) e^{-\lambda x}, x > t\end{cases}$$
Where is my mistake? How does this distribution arise?
 A: 
You are confusing the interarrival times $(T_n)$ with the sum $Y_t=K_t+L_t$ of the time $K_t=T_{N_t+1}-t$ till the next event and the time $L_t=t-T_{N_t}$ since the last event. 

It is a general fact that, for i.i.d. interarrival times $(T_n)$ distributed like $T$, the length $Y_t=K_t+L_t$ of the interval around $t$ is distributed like $\min\{T^*,t\}$, where the distribution $T^*$ is the size-biased distribution of $T$. Furthermore, conditionally on $Y_t$, $K_t$ and $L_t$ are uniformly distributed in the interval $(0,Y_t)$, that is, $K_t=U_tY_t$ and $L_t=(1-U_t)Y_t$, where $U_t$ is uniform on $(0,1)$ and independent of $Y_t$.
Recall that, starting from an almost surely nonnegative integrable random variable $X$, its size-biased distribution is the distribution of a random variable $X^*$ such that, for every bounded measurable function $u$,
$$
E(u(X^*))=\frac{E(Xu(X))}{E(X)}.
$$
If $X$ has density $f$ and mean $m$, $X^*$ has density $f^*$ defined by $f^*(x)=m^{-1}xf(x)$ for every $x\gt0$.
When $T$ is exponential $\lambda$, $T^*$ is gamma $(2,\lambda)$ and, if $U$ is uniform on $(0,1)$ and independent of $T^*$, then $UT^*$ is again exponential $\lambda$. Thus, in the homogenous renewal process, $K_t$ is exponential $\lambda$, $L_t$ is distributed like $\min\{t,L\}$ where $L$ is exponential $\lambda$, and $Y_t$ is distributed like $\min\{t,Y\}$, where $Y$ is gamma $(2,\lambda)$.
