Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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) $?

share|improve this question
    
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

2 Answers 2

up vote 4 down vote accepted

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

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

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

share|improve this answer
    
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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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