X and Y are both geometric distributions with success p. What it the Pr(X+Y=n). Would I use a convolution with a sum for this? Do I need to define a third random variable?

Melina Barber 2022-09-23 Answered
Finding the probability distribution of the sum of geometric distributions?
X and Y are both geometric distributions with success p. What it the P r ( X + Y = n )
Would I use a convolution with a sum for this? Do I need to define a third random variable?
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Answers (2)

asijikisi67
Answered 2022-09-24 Author has 10 answers
Explanation:
We have
  Pr ( X + Y = n ) =   i = 1 n 1 Pr ( X = i ) Pr ( Y = n i X = i ) =   i = 1 n 1 Pr ( X = i ) Pr ( Y = n i ) =   i = 1 n 1 p ( 1 p ) i 1 p ( 1 p ) n i 1 =   i = 1 n 1 p 2 ( 1 p ) n 2 = ( n 1 ) p 2 ( 1 p ) n 2
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Alan Sherman
Answered 2022-09-25 Author has 1 answers
Step 1
From the definition of Geometric Random Variables:
If X,Y are the independent counts of trials, from sequences of Bernoulli trials, until first success (with rate p), then X + Y would be the count of trials from a sequence of Bernoulli trials until the second success.
Step 2
We would then be seeking the probability of obtaining exactly one success somewhere among n 1 trials, then a second success.
P ( X + Y = n ) = ( n 1 ) p 2 ( 1 p ) n 2   1 n { 2 , 3 , }
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