X and Y are geometric RV's with parameter p. A) P{X+Y=n}(n=1,2,...) ?

Jaxon Hamilton 2022-07-18 Answered
X and Y are geometric RV's with parameter p.
A) P { X + Y = n } ( n = 1 , 2 , . . . )?
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Answers (2)

Cael Cox
Answered 2022-07-19 Author has 11 answers
Step 1
I am assuming X and Y are independent. Without this assumption, it is not possible to answer your question without further information.
Step 2
So, you have to be a little more clever: Suppose we let Z = X + Y. We wish to find Pr [ Z = n ]. Note that we can write Pr [ Z = n ] = y = 0 n Pr [ ( X = n y ) ( Y = y ) ] = y = 0 n Pr [ X = n y ] Pr [ Y = y ] .
Now you know what each of these probabilities is, so write them out, and calculate the sum.

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Kade Reese
Answered 2022-07-20 Author has 3 answers
Step 1
For the shifted geometric distribution, of X failures before success on trial X + 1, X S G e o ( p ) Pr ( X = x ) = p ( 1 p ) x
Given:  X Y , X S G e o ( p ) , Y S G e o ( p ) Pr ( X + Y = n ) = Pr ( x = 0 n ( X = x Y = n x ) ) n { 0.. }   = x = 0 n Pr ( X = x ) Pr ( Y = n x )   = x = 0 n p ( 1 p ) x p ( 1 p ) n x   = ( n + 1 ) p 2 ( 1 p ) n     Pr ( X = k X + Y = n ) = Pr ( X = k X + Y = n ) Pr ( X + Y = n ) n { 0.. } , k { 0.. n }   = et cetera
Step 2
For the geometric distribution of X 1 failures before success on trial X,
X G e o ( p ) Pr ( X = x ) = p ( 1 p ) x 1
Given:  X Y , X G e o ( p ) , Y G e o ( p ) Pr ( X + Y = n ) = Pr ( x = 1 n 1 ( X = x Y = n x ) ) n { 2.. }   = x = 1 n 1 Pr ( X = x ) Pr ( Y = n x )   = x = 1 n 1 p ( 1 p ) x 1 p ( 1 p ) n x 1   = ( n 1 ) p 2 ( 1 p ) n 2     Pr ( X = k X + Y = n ) = Pr ( X = k X + Y = n ) Pr ( X + Y = n ) n { 2.. } , k { 1.. n 1 }   = et cetera

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