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Assume there are N independent exponential random variable $(X_1, X_2,..., X_N)$ 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 $P(Y\geqslant k) = P(X_1 + X_2 + ... + X_k \leqslant t) = \int_{x_1 =0}^{t} P(X_2 + X_3 + \ldots + X_k \leqslant t - x_1)f_{X_1}(x_1)dx$. Then I don't know how to keep going from here. please help me...

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

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  • $\begingroup$ Could you elaborate please? What is $S_k$ or S? What is equality of the first line based on? I don't really follow the step. $\endgroup$ – Chameleon_7 Mar 21 at 4:58
  • $\begingroup$ What equality are u asking about? $\endgroup$ – d.k.o. Mar 21 at 5:04
  • $\begingroup$ This one: $P(Y = k) = P(S_k \le t, S_k + X_(k+1) > t) = E[....]$. Also, I haven't learned gamma function yet... Sorry $\endgroup$ – Chameleon_7 Mar 21 at 5:37
  • $\begingroup$ The first equality is self-evident: $Y=k$ iff $S_k\le t$ and $S_{k+1}>t$. The second one is the law of iterated expectations: for random variables $X$ and $Y$, $\mathsf{E}[f(X)g(Y)]=\mathsf{E}[f(X)\mathsf{E}[g(Y)\mid X]]$. $\endgroup$ – d.k.o. Mar 21 at 5:40
  • $\begingroup$ $\Gamma(k,\lambda^{-1})$ is the gamma distribution. $\endgroup$ – d.k.o. Mar 21 at 5:43

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