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Is the sum of two independent geometric random variables with the same success probability parameter a geometric random variable? What is it's distribution?

My approach is as follows: $Z=X+Y$

$P(X+Y=z)=\sum\limits_{x} P(X=x)P(Y=z-x)$

$=\sum\limits_{x} p(1-p)^{(x-1)}p(1-p)^{(z-x-1)}$

$=\sum\limits_{x} p^{2}(1-p)^{z-2}$

$= p^{2}(1-p)^{z-2} \sum\limits_{x=0}^\infty 1$

I am not sure how to turn this into a distribution. It looks like binomial with n = z and x = 2, but I don't know how to get the coefficient from this.


marked as duplicate by symplectomorphic, user147263, Joel Reyes Noche, Ali Caglayan, Graham Kemp Mar 3 '15 at 10:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


If $X$ and $Y$ are geometric random variables, they are each a count of Bernoulii trials until the first success.

$Z$ then counts Bernoulii trials until the second success.

$$\begin{align}P_Z(z) & = \sum_{k=1}^{z-1} P_X(k)P_Y(z-k) & \text{Note: $k$ cannot be greater than $z-1$} \\ & \vdots \\ & = (z-1) (1-p)^{z-2} p^2 \end{align}$$

PS: this is called a negative binomial distribution.

  • $\begingroup$ Thank you, I believe that if I had the bounds correctly set as Brent Kerby stated below then I would have the sum going from 1 to (z-1) and then I would have the negative binomial distribution you have shown here. Am I wrong in saying that? $\endgroup$ – satellite_help Mar 3 '15 at 5:09

You've made a mistake in letting the bounds for $x$ go from 0 to $\infty$. If $x$ were larger than $z$, then in $P(Y=z-x)$ we would have a negative number for $z-x$, and resulting probability should be zero. Also, it appears that in the definition of geometric random variable that you're using, 1 is the smallest possible value, so $x$ should not start at 0.


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