Let X and Y are independent random variables following geometric distribution with parameter p. Find the distribution of X given that X+Y=n.

trapskrumcu

trapskrumcu

Answered question

2022-09-14

Conditional probability distribution with geometric random variables
Let X and Y are independent random variables following geometric distribution with parameter p. Find the distribution of X given that X + Y = n.

Answer & Explanation

ahem37

ahem37

Beginner2022-09-15Added 15 answers

Step 1
I will assume that X, Y are the numbers of trials until the first success, where the probability of success on any trial is p. Then Pr ( X = j ) = Pr ( Y = j ) = ( 1 p ) j 1 p.
Let A be the event X = i and let B be the event X + Y = n. Then by the definition of conditional probability, we have Pr ( A | B ) = Pr ( A B ) Pr ( B ) .
The probability of A B is easy to compute. It is Pr ( X = i ) Pr ( Y = n i ). This is ( 1 p ) i 1 p ( 1 p ) n i 1 p, which simplifies to ( 1 p ) n 2 p 2 .
Step 2
For the probability that X + Y = n, we can use the fact that Pr ( X + Y = n ) = Pr ( X = 1 ) Pr ( Y = n 1 ) + Pr ( X = 2 ) Pr ( Y = n 2 ) + + Pr ( X = n 1 ) Pr ( Y = 1 ) ..
This simplifies to ( n 1 ) ( ( 1 p ) n 2 p 2 .
Divide. We get 1 n 1 . The conditional probability is discrete uniform. One can save a little time in the calculation of Pr ( X + Y = n ) by noting that X + Y has negative binomial distribution: it is the number of trials until the second success.

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