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Probability - Conditional Expectation

My professor explained this concept absolutely horribly and I have no idea how to do these problems.

Let A and B be independent Poisson random variables with parameters α and β, respectively. Find the conditional expectation of A given A + B = c.
(Hint: For discrete random variables, there is no conditional density. Use the definition of conditional probability.)

Attempt:
Starting with the definition, f(A | A + B = c) = [f(A, A+B=c)] / [f(A+B=c)]

Not sure how to proceed.
 
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We will have to work with A+B. Do you know the probability distribution of this??
 

Ray Vickson

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My professor explained this concept absolutely horribly and I have no idea how to do these problems.

Let A and B be independent Poisson random variables with parameters α and β, respectively. Find the conditional expectation of A given A + B = c.
(Hint: For discrete random variables, there is no conditional density. Use the definition of conditional probability.)

Attempt:
Starting with the definition, f(A | A + B = c) = [f(A, A+B=c)] / [f(A+B=c)]

Not sure how to proceed.
It is best to be clear and to use correct notation: you want P{A=k|A+B=c} for the possible values of k in {0,1,2,...}. So, you need to compute P{A=k & A+B=c} in the numerator (and, of course, you need P{A+B=c} in the denominator).

Do you know the distribution of A+B? It should be in your textbook or course notes; if not, look on-line, or work it out for yourself from first principles, using the distributions of A and B and the formula for the distribution of a sum of independent random variables (really: it is not that hard!).

RGV
 
Last edited:
The distribution for a Poisson distribution is p(x) = [e^(-λ)*λ^x] / x!
 
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Yes, that is the distribution for A and B with [itex]\lambda=\alpha[/itex] and [itex]\lambda=\beta[/itex] respectively.

But we are asking about the distribution of A+B.
 
A + B also has a Poisson distribution with parameters Poisson(A+B), as the sum of independent Poisson random variables has a Poisson distribution.
 
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OK, that's good. Now we want to figure out (for fixed c)

[tex]f(A=x~|~A+B=c)[/tex]

In order to find to, we want to find

[tex]f(A=x,~A+B=c)[/tex]

Of course, this is equal to

[tex]f(A=x,~B=c-x)[/tex]

Can you find this?? This is just a two-dimensional pmf. Remember that A and B are independent, so you can find it easily.

Then we also nee to find

[tex]f(A+B=c)[/tex]

This should be easy since you just figured out the distribution of A+B.
 
Not sure how to go about finding f(A=x , B=c−x)
 

Ray Vickson

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Can anyone show me this problem step by step? I'm not picking up on any of this question, which is why I posted this.
 

Ray Vickson

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Can anyone show me this problem step by step? I'm not picking up on any of this question, which is why I posted this.
No, we can't. That is not how this forum works. However, I will give you a hint: if your professor did not explain things to your satisfaction, and if, for some reason you do not have access to course notes or to a textbook, then *look online*. Google 'independent + probability' to turn up hundreds of articles at various levels of sophistication, from step-by-step explanations to abstract discussions.

RGV
 
Last edited:
P(A|A+B=c)
= P(A|B=c-A)
= P(A and B=c-A) / P(B=c-A)

=P(α + β) / P(β) ?
 

Ray Vickson

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P(A|A+B=c)
= P(A|B=c-A)
= P(A and B=c-A) / P(B=c-A)

=P(α + β) / P(β) ?
I have no idea what you mean by P(α + β) or P(β). I know what α and β are, and I know what is meant by P(A=u) or P(B=v) and how to write them in terms of α, β, u and v, but I cannot figure out your P(α+β), etc. Anyway, I certainly would get something very different from what you wrote.

RGV
 

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