The problem is:
Let $\tau_1,\tau_2,\dots$ be independent exponentially distributed with parameter $\lambda >0$. Define $N_t:=|${$ k \geq 1 | \tau_1+\dots + \tau_k\leq t$}$ |$.
Show that $N_t$ is Poisson distributed with parameter $\lambda t$.
This is shown by induction:
Firstly $\mathbb{P}(N_t=0)=e^{-\lambda t}$. Now suppose that $\mathbb{P}(N_t=k)=e^{-\lambda t}\frac{(\lambda t)^k}{k!}$ for all $t>0$. Then by conditioning on $\tau_1$ we find
$$ \mathbb{P}(N_t=k+1)=\int_0^t\lambda e^{-\lambda x}\mathbb{P}(N_{t-x}=k)\text{d}x=\dots $$
which then yields the result easily, but I completely don't understand that step.
I tried starting with something like this:
$$ \mathbb{P}(N_t=k+1)=\int_0^{\infty}\lambda e^{-\lambda x}\mathbb{P}(N_t=k+1 | \tau_1=x)\text{d}x $$
but I couldn't figure out how to obtain the result from the solution. Can you please help me with that?
Thanks!
Let $S_{n}:=\sum_{i=1}^{n}\tau_{i}$ and $R_{n}=\sum_{i=2}^{n+1}\tau_{i}$ and observe that $S_{n}$ and $R_{n}$ have the same distribution.
We have: $$N_{t}=\max\left\{ n\mid S_{n}\leq t\right\} =\max\left\{ n\mid\tau_{1}+R_{n-1}\leq t\right\} $$
Then under condition $\tau_{1}=x<t$ we have: $$N_{t}=\max\left\{ n\mid x+R_{n-1}\leq t\right\} =\max\left\{ n\mid R_{n-1}\leq t-x\right\} $$ where: $$\max\left\{ n\mid R_{n-1}\leq t-x\right\} \sim\max\left\{ n\mid S_{n-1}\leq t-x\right\} =$$$$1+\max\left\{ n-1\mid S_{n-1}\leq t-x\right\} =1+N_{t-x}$$(here $LHS\sim RHS$ means that both sides have the same distribution)
For $x<t$ this leads to: $$P\left(N_{t}=k+1\mid\tau_{1}=x\right)=P\left(1+N_{t-x}=k+1\right)=P\left(N_{t-x}=k\right)$$