Given two iid geometric random variables with p=1/5, what is the probability that variates from them equate?

benatudq 2022-10-16 Answered
Given two iid geometric random variables with p = 1 5 , what is the probability that variates from them equate?
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Answers (2)

Plutbantonavv
Answered 2022-10-17 Author has 15 answers
Step 1
No need for the joint mass function. If the variables count the number of failures before the first success:
- The probability that both variates are 0 is ( 1 5 ) 2
- The probability that both variates are 1 is ( 1 5 ) 2 ( 4 5 ) 2
- The probability that both variates are 2 is ( 1 5 ) 2 ( 4 5 ) 4 , etc.
Step 2
Thus the probability that both variates are equal is
n = 0 ( 1 5 ) 2 ( 4 5 ) 2 n
= 1 25 n = 0 ( 16 25 ) n
= 1 25 1 1 16 25 = 1 9
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Vincent Norman
Answered 2022-10-18 Author has 1 answers
Step 1
Think of it this way: You want to find
P ( X = Y ) = P ( X = Y = 1  or  X = Y = 2  or  X = Y = 3  or  . . . )
These are clearly disjoint cases, so we break apart into a sum of probabilities while also realizing that each such statement X = Y = n is equivalent to saying X = n AND Y = n. Thus we have P ( X = Y ) = P ( X = 1 , Y = 1 ) + P ( X = 2 , Y = 2 ) + P ( X = 3 , Y = 3 ) + . . .
Step 2
Now, we use the fact that the two variates were considered IID. As such, each term P ( X = n , Y = n ) = P 2 ( X = n ) and we ultimately end up with the infinite sum P ( X = Y ) = P 2 ( X = 1 ) + P 2 ( X = 2 ) + P 2 ( X = 3 ) + . . .
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