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Let $X_1, X_2,\dots$ be iid. random variables each with density $xe^{-x}$ for $x > 0$ and $0$ otherwise. Let $S_0 = 0$ and $S_n = X_1 +\cdots + X_n$, and $N(t) = \max\ \{n : S_n < t\}$.

I need to find the mass function of the random variable $N(t)$. I have worked out that the $X_i$ are distributed with a gamma distribution with $n = 2$ and $\lambda = 1$. Subsequently $S_n$ is distributed with $\gamma(1,2n)$. I realise that somehow the gamma distribution as a waiting time means that $N(t)$ is a Poisson, but I cannot show it. Any help would be appreciated.

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  • $\begingroup$ Inter-arrival times of a standard Poisson $P(t)$ are $\exp(1)$ and $X_i$ is distributed exactly as a sum of two i.i.d exponentials. Hence $N(t)=\lfloor P(t)/2\rfloor$ which is NOT Poisson. $\endgroup$
    – A.S.
    Commented Feb 18, 2016 at 13:15
  • $\begingroup$ so is Sn not distributed with a gamma distribution? $\endgroup$
    – user200632
    Commented Feb 18, 2016 at 14:22
  • $\begingroup$ $S_n\sim \gamma(1,2n)$ as you've written. It would be Poisson if it was $\gamma(1,n)$. $\endgroup$
    – A.S.
    Commented Feb 18, 2016 at 14:24
  • $\begingroup$ No. $N(t)=\lfloor P(t)/2\rfloor$. $N$ is a counting process, so it must be an integer. But think through exactly why the form I listed. $\endgroup$
    – A.S.
    Commented Feb 18, 2016 at 14:58
  • $\begingroup$ I understand how you've got to where you have but I am unsure on how to translate that to a distribution for N(t) $\endgroup$
    – user200632
    Commented Feb 18, 2016 at 15:18

1 Answer 1

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Note that $$\{N(t)=n\}=\{S_n\geqslant t, S_{n-1}<t\}=\{X_n\geqslant t-S_{n-1},S_{n-1}<t \}. $$ Hence \begin{align} \mathbb P(N(t)=n) &= \mathbb P(X_n\geqslant t-S_{n-1},S_{n-1}<t)\\ &=\int_0^t\mathbb P(X_n\geqslant t-S_{n-1}\mid S_{n-1}=s)f_{S_{n-1}}(s)\ \mathsf ds\\ &= \int_0^t e^{-(t-s)}(1+(t-s))\frac{s^{2n-1}e^{-s}}{(2n-1)!}\ \mathsf ds\\ &= \frac{e^{-t}t^{2n}(2n+1+t)}{(2n+1)!}. \end{align} Alternatively, we have $$\{N(t)=n\} = \{P(t)=2n\}\cup\{P(t)=2n+1\} $$ where $P(t)$ is a Poisson process with rate $1$, and so \begin{align} \mathbb P(N(t)=n) &= \mathbb P(P(t)=2n) + \mathbb P(P(t)=2n+1))\\ &= \frac{e^{-t}t^{2n}}{(2n)!} + \frac{e^{-t}t^{2n+1}}{(2n+1)!}\\ &= \frac{e^{-t}t^{2n}(2n+1+t)}{(2n+1)!}. \end{align}

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