Patricia Bean

Patricia Bean

Answered

2022-07-21

Probability mass function of sum of two independent geometric random variables
How could it be proved that the probability mass function of X + Y, where X and Y are independent random variables each geometrically distributed with parameter p; i.e. p X ( n ) = p Y ( n ) = { p ( 1 p ) n 1 n = 1 , 2 , . . . 0 o t h e r w i s e equas to P X + Y ( n ) = ( n 1 )   p 2 ( 1 p ) n 2

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Answer & Explanation

taguetzbo

taguetzbo

Expert

2022-07-22Added 16 answers

Step 1
Since X , Y 1, the summation should run over k = 1 , 2 , , n 1.
Step 2
Using this your convolution becomes
P ( X + Y = n ) = k = 1 n 1 p 2 ( 1 p ) n 2 = p 2 ( 1 p ) n 2 k = 1 n 1 1 = p 2 ( 1 p ) n 2 ( n 1 ) .

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Ruby Briggs

Ruby Briggs

Expert

2022-07-23Added 3 answers

Step 1
A geometric random variable is the count of Bernouli trial until a success. We measure the probability of obtaining n 1 failures and then 1 success.
P ( X = n ) = ( 1 p ) n 1 p : n { 1 , 2 , }
The sum of two such is the count of Bernouli trials until the second success. We measure the probability of obtaining 1 success and n 2 failures, in any arrangement of those n 1 trials, followed by the second success.
P ( X + Y = n ) = ( n 1 ) ( 1 p ) n 2 p 2 : n { 2 , 3 }
Step 2
This may also be counted by summing P ( X + Y = n ) = k = 1 n 1 P ( X = k , Y = n k ) note the range = k = 1 n 1 P ( X = k ) P ( Y = n k ) by independence = k = 1 n 1 ( 1 p ) k 1 p ( 1 p ) n k 1 p = ( 1 p ) n 2 p 2 k = 1 n 1 1 = ( n 1 ) ( 1 p ) n 2 p 2
Since X + Y must equal n and neither can be less than 1, then neither can be more than n 1. Hence this the range of X values we must sum over.

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