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Probability - Conditional Expectation

  1. Apr 23, 2012 #1
    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.
     
  2. jcsd
  3. Apr 23, 2012 #2

    micromass

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    We will have to work with A+B. Do you know the probability distribution of this??
     
  4. Apr 23, 2012 #3

    Ray Vickson

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    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: Apr 23, 2012
  5. Apr 23, 2012 #4
    The distribution for a Poisson distribution is p(x) = [e^(-λ)*λ^x] / x!
     
  6. Apr 23, 2012 #5

    micromass

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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.
     
  7. Apr 24, 2012 #6
    A + B also has a Poisson distribution with parameters Poisson(A+B), as the sum of independent Poisson random variables has a Poisson distribution.
     
  8. Apr 24, 2012 #7

    micromass

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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.
     
  9. Apr 24, 2012 #8
    Not sure how to go about finding f(A=x , B=c−x)
     
  10. Apr 24, 2012 #9

    Ray Vickson

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    You told us that A and B are independent. What do you think that means?

    RGV
     
  11. Apr 25, 2012 #10
    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.
     
  12. Apr 25, 2012 #11

    Ray Vickson

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    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: Apr 25, 2012
  13. Apr 25, 2012 #12
    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) ?
     
  14. Apr 25, 2012 #13

    Ray Vickson

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    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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