Question

Let X and Y be two independent random variables. Assume that X is Negative-
Binomial(2, θ) and Y is Negative-Binomial(3, θ) distributed. Let Z be another random

variable, Z = X + Y .
(a) Find the following probabilities: P(Z = 0), P(Z = 1) and P(Z = 2);
(b) Can you guess what is the distribution of Z?

Answer 1

Let random variables X and Y are independent random variables.

X ~ NB ( k1 =2, theta)

X denotes number of failures before the two successes.

The p.m.f. of X is given by

Y ~ NB ( k2 =3, theta)

Y denotes number of failures before the three successes.

The p.m.f. of Y is given by

a) Z = X + Y

i) Z = 0 when X= 0 and Y =0

P(Z =0) = P ( X=0, Y=0) = P(X=0) * P(Y=0) since X and Y are independent.

ii) Z = 1 when X= 0 and Y=1 or X=1 and Y=0

P ( Z=1) = P( X=0) * P(Y=1) + P(X=1) * P(Y=0)

iii) Z= 2 when X=0 and Y=1 or X=1 and Y=1 or X=2 and Y=0

P ( Z=2) = P ( X=0) * P(Y=2) + P(X=1) * P(Y=1) + P(X=2) * P(Y=0)

b) Additive property of Negative binomial distribution.

If X ~ NB ( K1, p) and Y ~ NB (k2, p) and X and Y are independent and prob. of success p is same for both X and Y, then

Z ~ NB ( k1+k2,p)

Hence in given problem distribution of Z is

.