# Sum of linear combination of product of exponentials is exponential

This problem has arisen in my research: suppose that $V_i \sim \text{ED}$ are iid exponential distributions (ED) with mean $1$ and let $\lambda$ be a nonnegative number. Is it true that $$\sum_{k=0}^{\infty} \frac{\lambda^k e^{-\lambda}V_{0} \cdots V_k}{k!} \sim \text{ED}?$$ This passes the sanity check, as the expected value of both sides is equal to $1$, and if we let $\lambda = 0$, then the left hand side equals $V_0$, which is exponential. Other than that, I am not sure how to approach this problem, as I do not know how to deal with the product of ED's.

• how do you guarantee this is a true statement? – Zhanxiong Dec 19 '17 at 13:37
• @Zhanxiong I'm not entirely sure if it's true, which is why I am asking if anyone can supply a proof (or disprove it if it's false.) – Alex Dec 19 '17 at 17:49
• ok, then you should avoid using "prove that" – Zhanxiong Dec 19 '17 at 18:02
• My bad, I edited the question. – Alex Dec 19 '17 at 18:16
• Is $\lambda$ the same rate/mean parameter for the exp RVs? – AdamO Dec 19 '17 at 18:41

Not a complete answer, sorry, but a few ideas (to long for a comment). Note that what you have is a product of $K+1$ iid random variables, where $K$ is a random variable (rv) with a poisson distribution with parameter $\lambda$. That can be used for another "sanity check", a simulation (using exponentials of rate 1):

set.seed(7*11*13)
N <- 1000000

prods <- rep(0, N)
ks <- rpois(N, 1)+1

for (i in 1:N) {
k  <-  ks[i]
prods[i]  <-  prod( rexp(k, 1))
}

qqplot( qexp(ppoints(N)), prods)


The resulting qqplot (not shown here) is far from a straight line, so this do not look to be an exponential of rate 1. The mean is right, the variance to large, there is a right tail much longer than for an exponential. What can be done theoretically? The Mellin transform https://en.wikipedia.org/wiki/Mellin_transform is adapted to products of independent random variables. I will compute only for the exponential with rate 1. The Mellin transform of $V_0$ then is $$\DeclareMathOperator{\E}{\mathbb{E}} M_1(s) = \E V_0^s = \int_0^\infty x^s e^{-x}\; dx = \Gamma(s+1)$$ so the Mellin transform of a product of $k+1$ iid exponentials is $$M_{k+1}(s) = \Gamma(s+1)^{k+1}$$ Since $K$ has a poisson distribution with parameter $\lambda$, the Mellin transform of the random product of a random number $K+1$ factors, is $$M(s) = \E M_{K+1}(s) = \E \Gamma(s+1)^{K+1}= \Gamma(s+1)e^{-\lambda}\sum_{k=0}^\infty \frac{\lambda^k}{k!}\Gamma(s+1)^k=e^{-\lambda}\Gamma(s+1) e^{\lambda \Gamma(s+1)}$$ but I cannot find an inverse of this transform. But note that if $X$ is a nonnegative random variable with Mellin transform $M_X(t)$, then defining $Y=\log X$ we find that $$K_Y(t)=\E e^{tY} = \E e^{t\log X}=\E e^{\log (X^t)}=\E X^t =M_X(y)$$ so the Mellin transform of $X$ is the moment generating function of its logarithm $Y$. So, using that we can approximate the distribution of $X$ with saddlepoint approximation methods, How does saddlepoint approximation work? and search this site.

• (+1) Even the product of two exponentials does not have a closed form density. – Xi'an Dec 19 '17 at 21:41
• there was a post in SE showing that a product of three exponentials does not have moments or something along those lines – Aksakal Dec 19 '17 at 21:42
• Thanks kjetil. I was pretty sure the answer was no as well, but those are pretty good reasons for why. – Alex Dec 19 '17 at 22:04
• @Aksakal: the product of $k$ independent exponential has all moments finite. – Xi'an Dec 20 '17 at 5:15
• ...and the product $V_0\cdots V_k$ a.s. converges to zero. – Xi'an Dec 20 '17 at 6:59