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Let $t_1,t_2,\dots$ be independent exponential($\lambda$) random variables and let $Y$ be an independent random variable with $P(Y=n) = p(1-p)^{n-1}$. What is the distribution of the random sum $T = t_1 + t_2 + ... + t_Y$?

I understand that, for a given $n$, $T$ is the gamma($n,\lambda$) distribution. But how do we account for randomness of $n$ in the distribution of $T$?

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  • $\begingroup$ I have a feeling $P(Y)$ is missing a factor of $p$. $\endgroup$
    – Patrick
    Oct 13, 2013 at 4:09
  • $\begingroup$ You are right. Will correct that. $\endgroup$ Oct 13, 2013 at 4:52

2 Answers 2

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For every $n\geqslant1$, the distribution of $t_1+\cdots+t_n$ has density $f_n:t\mapsto\lambda^nt^{n-1}\mathrm e^{-\lambda t}/(n-1)!$ on $t\geqslant0$ hence $T$ has density $f_T$ defined on $t\geqslant0$ by $$ f_T(t)=\sum_{n=1}^\infty P[Y=n]\,f_n(t)=\sum_{n=1}^\infty p(1-p)^{n-1}\frac{\lambda^nt^{n-1}}{(n-1)!}\mathrm e^{-\lambda t}, $$ that is, $$ f_T(t)=p\lambda\mathrm e^{-\lambda t}\sum_{n=0}^\infty \frac{(1-p)^{n}\lambda^nt^n}{n!}=p\lambda\mathrm e^{-\lambda t}\mathrm e^{(1-p)\lambda t}=p\lambda\mathrm e^{-p\lambda t}. $$ Thus, $T$ is exponential with parameter $p\lambda$. (It might be worth mentioning that this is a classical result in terms of thinning of Poisson processes.)

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Use the law of total probability:

$P\{T\leq t\}=\sum\limits_{n=1}^{\infty} P\{T \leq t |Y=n \}P\{Y=n\}=p\sum\limits_{n=1}^\infty \frac{\gamma\left(\lambda,\frac{t}{n}\right )}{\Gamma(\lambda)}(1-p)^{n-1}$

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