Confused about how to show independent random variable $Y$ has the Poisson distribution with parameter $t\lambda$

Question

Assume there are $N$ independent exponential random variable $(X_1, X_2,\ldots, X_N)$ with parameter $\lambda$. Fix a real number $t > 0$. Let $Y$ be the largest $N$ so that $X_1 + X_2 + \ldots + X_N \leqslant t$ ($Y = 0$ if $X_1 > t$). How to show independent random variable $Y$ has the Poisson distribution with parameter $t\lambda$?

Approach: I want to prove that $ P(Y \geqslant k) = 1 - \sum_{j=1}^{k-1} \frac{e^{-\lambda t}(\lambda t)^k}{k!}$, in order to do this, I wanted to use

\begin{align} & P(Y\geqslant k) = P(X_1 + X_2 + \cdots + X_k \leqslant t) \\[8pt] = {} & \int_{x_1 =0}^t P(X_2 + X_3 + \ldots + X_k \leqslant t - x_1)f_{X_1}(x_1) \, dx_1. \end{align}

Then I don't know how to keep going from here. please help me...

Answer 1

Let $S_k:=\sum_{i=1}^k X_i$ (note that $S_k\sim \Gamma(k,\lambda^{-1})$). Then

\begin{align} \mathsf{P}(Y=k)&=\mathsf{P}(S_k\le t, S_k+X_{k+1}>t)=\mathsf{E}\!\left[1\{S_k\le t\} \mathsf{P}(X_{k+1}>t-S_k\mid S_k)\right] \\ &=\mathsf{E}\!\left[1\{S_k\le t\}e^{-\lambda(t-S_k)}\right]=\int_0^t\frac{\lambda^k}{(k-1)!}x^{k-1}e^{-\lambda t}\,dx \\ &=\frac{(\lambda t)^ke^{-\lambda t}}{k!}. \end{align}

Answer 2

Suppose it has been shown that $$ f_{X_1+\cdots+X_k}(x) \, dx = \frac 1 {(k-1)!} (\lambda x)^{k-1} e^{-\lambda x} (\lambda \, dx) \text{ for } x>0. $$ Then \begin{align} & \Pr(X_1+\cdots+X_k< t) \\[8pt] = {} & \int_0^t \frac 1 {(k-1)!} (\lambda x)^{k-1} e^{-\lambda x} (\lambda \, dx) \\[8pt] = {} & \frac 1 {(k-1)!} \int_0^{\lambda t} u^{k-1} e^{-u} \, du \\[8pt] = {} & \frac 1 {(k-1)!} \left( \Big[ {-u^{k-1}}e^{-u} \Big]_0^{\lambda t} - \int_0^{\lambda t} (k-1)u^{k-2} (-e^{-u}) \, du \right) \\ & \text{(integration by parts)} \\[8pt] = {} & -\frac{(\lambda t)^{k-1} e^{-\lambda t}}{(k-1)!} + \frac 1{(k-2)!} \int_0^{\lambda t} u^{k-2} e^{-u} \, du \end{align} That last step is valid if $k-1\ne0.$ Keep going until you get to the point where it is $0.$ That should give you the sum that you're looking for.