This is a bit of a strange question, but suppose I have some random variables.

$$Y_i \sim Gamma(i,\lambda)$$

Where this comes from the fact that each $Y_i$ is defined as the sum of $i$ independent exponential random variables

$$X_i \sim Exp(\lambda)$$

So that, for example, $Y_{i+1} = Y_i + X_{i+1}$.

I am interested in this infinite summation:

$$\sum_{i=1}^\infty P(Y_i \leq t)$$

Clearly as $i$ increases so does the mean of the random variables $Y_i$ which is $E[Y_i] = \frac{i}{\lambda}$. I would expect that these probabilities go to zero sufficiently fast for this sum to converge to a finite number, which I would like to derive, but I am unsure if it even converges. I am also unsure if it's even possible to derive what it converges to since Gamma distributions do not have a closed form expression.

I tried substituting in the Gamma CDF form given here so that we can write it as a double summation:

$$\sum_{i=1}^\infty P(Y_i \leq t) = \sum_{i=1}^\infty e^{-\lambda t} \sum_{k=i}^\infty \frac{(\lambda t)^k}{k!}$$

But this doesn't lead anywhere useful. So I just have two questions:

  1. Is there anyway we can tell/show whether this converges at all?
  2. If we can, is it possible to actually derive what it converges to?

My gut instinct is that the answer to (1) is yes, but only by using inequalities to show it is bounded. What inequalities I can use, I'm unsure.

  • 1
    $\begingroup$ Using inequalities to show it is bounded would suffice to answer 1. so if you did that right it seems you've done that already. in relation to 2. ... one possibility might be to consider whether you can relate that to the expectation of a Poisson. $\endgroup$
    – Glen_b
    Commented Jun 30, 2017 at 1:46
  • $\begingroup$ Hey @glen_b I took your advice and looked at the Poisson (kind of ashamed I didn't think of that given the pmf is right there) and it seems to be a very simple result. Do you mind taking a second to check my answer for me? I wasn't expecting such a simple result! So I'm a little skeptical of my working. $\endgroup$
    – Patty
    Commented Jun 30, 2017 at 4:20
  • 1
    $\begingroup$ You can always check your result via a simulation of the original problem. $\endgroup$
    – Glen_b
    Commented Jun 30, 2017 at 5:42
  • $\begingroup$ Haha actually I did earlier and it seems to be correct. Very cool! Cheers for the tip on Poisson. $\endgroup$
    – Patty
    Commented Jun 30, 2017 at 9:08

1 Answer 1


It turns out you can relate this to the Poisson distribution very easily, and the result is actually really cool. Carrying on from where I left off:

$$\sum_{i=1}^\infty P(Y_i \leq t) = \sum_{i=1}^\infty \sum_{k=i}^\infty \frac{(\lambda t)^k}{k!} e^{-\lambda t}$$

$$= \sum_{i=1}^\infty P(X \geq i)$$ Where $X \sim Pois(\lambda t)$

Then it simply follows from the fact that if $X$ is a random variable that takes values $0,1,2...$ then it has expected value:

$$E[X] = \sum_{i=1}^\infty P(X \geq i)$$

Which gives us the nice tidy result of:

$$\sum_{i=1}^\infty P(Y_i \leq t) = \sum_{i=1}^\infty P(X \geq i) = E[X] = \lambda t$$


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