Question

Problem 41.3 Let X and Y be independent random variables each geometrically distributed with parameter p, i.e. p(1- p otherwise. Find the probability mass function of X +Y

Answer 1

A geometric random variable is the count of Bernouli trial untill a success. we count the probability of obtaining n−1 failures and then 1 success.

P_X(n) = p (1−p)^n-1   :n=1,2,…

The sum of two such is the count of Bernouli trials untill the 2nd success. the probability of getting 1 success and (n−2) failures, in any arrangement of those (n−1) trials, followed by the second success. is given by:

P_x+y(n)=(n−1)(1−p)ⁿ⁻²p²   n=2,3….........

Now

$P(X+Y=n)=\sum_{k=1}^{n-1}P(X=k,Y=n%u2212k)$

$=\sum_{k=1}^{n-1}P(X=k)P(Y=n-k)$

due to independenc of X and Y

$=\sum_{k=1}^{n}(1-p)^{k-1}p(1-p)^{n-k-1}p$

$=(1-p)^{n-2}p^2\sum_{k=1}^{n-1}1 =(n-1)(1-p)^{n-2}p^{2}$

which is the required pmf of x+y. where n=2,3,4,........