Question

Let X and Y are independent negative binomial distribution. Find the conditional distribution of X|Y=y.​

Answer 1

Answer:

Correct option is A)

We have P(

X+Y=r+s

X=r

)=

P(X+Y=r+s)

P[(X=r)∩(X+Y=r+s)]

=

P(X+Y=r+s)

P[(X=r)∩(Y=s)]

=

P(X+Y=r+s)

P(X=r)P(Y=s)

P(X+Y=r+s)=

k=0

∑

r+s

P[(X=k)∩(Y=r+s−k)]

=

k=0

∑

r+s

(

n

C

k

.p

k

.q

n−k

)(

m

C

r+s−k

.p

r+s−k

.q

m−r−s+k

)

=p

r+s

.q

m+n−r−s

.

k=0

∑

r+s

(

n

C

k

)(

m

C

r+s−k

)

Now the last sum is the expression for the number of ways of choosing r+s persons out of n men and m women, which is

m+n

C

r+s

.

Therefore P(X+Y=r+s)=

m+n

C

r+s

.p

r+s

.q

m+n−r−s

so that

P(

X+Y=r+s

X=r

)=

m+n

C

r+s

.p

r+s

.q

m+n−r−s

(

m

C

r

.p

r

.q

n−r

)(

n

C

s

.p

s

.q

m−s

)

=

m+n

C

r+s

(

m

C

r

)(

n

C

s

)