# Sum of two independent geometric random variables

Let X and Y be independent random variables,

$P(X = k) = P(Y = k) = p(1 - p)^{k-1}$

How do you show that the pmf of $Z = X + Y$, is negative binomial, and how do you find

$P(X = Y)$?

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This is the probability that the $k$th trial is the first success. You would have to use convolutions. – PEV Nov 15 '10 at 0:23

$P(Z=r)=\Sigma_{k=1}^r P(X=k)P(Y=r-k)$

$P(X=Y)=\Sigma_{k=1}^{\infty} P(X=k)P(Y=k)$

I am assuming $k$ is a positive integer. Otherwise, the summations need to be modified accordingly.

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Thats great! One question, is P(Y = r - k) = p(1-p)^r-k-1? – MacMath2010 Nov 15 '10 at 0:34
Yes, precisely. – Timothy Wagner Nov 15 '10 at 1:09

While Timothy Wagner's answer is correct, I thought you might like to see another way to answer your first question.

Often the simplest way to prove that the sum of independent random variables has a particular distribution is to use moment-generating functions. This is because 1) if $X$ and $Y$ are independent with mgf's $M_X(t)$ and $M_Y(t)$, then $M_{X+Y}(t) = M_X(t) M_Y(t)$ and 2) moment-generating functions (when they exist) characterize distributions.

Applying this to your problem, a geometric $(p)$ random variable has mgf $$\frac{pe^t}{1 - (1-p)e^t}.$$

Thus $$M_{X+Y}(t) = \left(\frac{pe^t}{1 - (1-p)e^t}\right)^{2}.$$ Since this is the mgf of a negative binomial random variable, $X+Y$ must have a negative binomial distribution.

(There are different conventions for defining negative binomial and geometric random variables, so depending on the convention used in a particular reference the mgf's there may be slightly different from the ones I give here.)

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