How can you find P(X/(Y-X)<0) if X∼Geometric(p) and Y∼Bernoulli(p)

Pierre Holmes 2022-07-19 Answered
How can you find P ( X Y X < 0 ) if X G e o m e t r i c ( p ) and Y B e r n o u l l i ( p )
Let the independent random variables X G e o m e t r i c ( p ) and Y B e r n o u l l i ( p ), I want to prove that P ( X Y X < 0 ) = ( p 1 ) 2 ( p + 1 )
Do I need the joint probability mass function for this or should it be proven in some other way?
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Answers (1)

Octavio Barr
Answered 2022-07-20 Author has 11 answers
Step 1
Given that Y is Bernoulli distributed, it will have probability mass function P ( Y = 0 ) = ( 1 p ) and P ( Y = 1 ) = p. As Frank has helpfully commented, we can condition on Y.
When Y = 0, we have P ( X Y X < 0 | Y = 0 ) = P ( X < 0 | Y = 0 ) = P ( X > 0 | Y = 0 )
When Y = 1, we P ( X Y X < 0 | Y = 1 ) = P ( X 1 X < 0 | Y = 1 ) = P ( X > 1 | Y = 1 )
Step 2
Thus, we have the following (the third line being a consequence of the independence between X and Y)
P ( X Y X < 0 ) = P ( X Y X < 0 | Y = 0 ) P ( Y = 0 ) + P ( X Y X < 0 | Y = 1 ) P ( Y = 1 ) = P ( X > 0 | Y = 0 ) P ( Y = 0 ) + P ( X > 1 | Y = 1 ) P ( Y = 1 ) = P ( X > 0 ) P ( Y = 0 ) + P ( X > 1 ) P ( Y = 1 ) = ( k = 1 p ( 1 p ) k ) ( 1 p ) + ( k = 2 p ( 1 p ) k ) ( p ) = ( 1 p ) 2 + p ( 1 p ) 2 = ( p 1 ) 2 ( 1 + p )
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