2
$\begingroup$

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 $T$?

$\endgroup$
  • $\begingroup$ I have a feeling $P(Y)$ is missing a factor of $p$. $\endgroup$ – Patrick Oct 13 '13 at 4:09
  • $\begingroup$ You are right. Will correct that. $\endgroup$ – user2373713 Oct 13 '13 at 4:52
1
$\begingroup$

Use the law of total probability:

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

$\endgroup$
1
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.