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Math Help - Joint probability mass function

  1. #1
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    Joint probability mass function

    Suppose X and Y are independent Poisson random variables with parameters a and b respectively. Find the joint mass function

    P[X = k, Y = m]

    I got

    (exp(-a).a^k)/k! x (exp(-b).b^m)/m!

    Find P[X + Y = n]

    I got = P[X = k, Y = n - k]

    = sum over k to n (exp(-a).a^k)/k! x (exp(-b).b^n - k)/(n-k)!

    But I'm not sure how to complete

    is this right so far?

    thanks
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  2. #2
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    Quote Originally Posted by James0502 View Post
    Suppose X and Y are independent Poisson random variables with parameters a and b respectively. Find the joint mass function

    P[X = k, Y = m]

    I got

    (exp(-a).a^k)/k! x (exp(-b).b^m)/m!

    Find P[X + Y = n]

    I got = P[X = k, Y = n - k]

    = sum over k to n (exp(-a).a^k)/k! x (exp(-b).b^n - k)/(n-k)!

    But I'm not sure how to complete
    So you got (correctly) P(X+Y=n)=\sum_{k=0}^n e^{-a}\frac{a^k}{k!}e^{-b}\frac{b^{n-k}}{(n-k)!}=e^{-(a+b)}\sum_{k=0}^n \frac{1}{k!(n-k)!}a^k b^{n-k}. This should remind you the binomial formula : remember that {n\choose k}=\frac{n!}{k!(n-k)!}...
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  3. #3
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    Quote Originally Posted by James0502 View Post
    Suppose X and Y are independent Poisson random variables with parameters a and b respectively. Find the joint mass function

    P[X = k, Y = m]

    I got

    (exp(-a).a^k)/k! x (exp(-b).b^m)/m!

    Find P[X + Y = n]

    I got = P[X = k, Y = n - k]

    = sum over k to n (exp(-a).a^k)/k! x (exp(-b).b^n - k)/(n-k)! Isomorphism: Not from k to n, but k varying from 0 to n.

    But I'm not sure how to complete

    is this right so far?

    thanks
    First of all observe the correction in red.

    What you have done is not wrong... but it is not very insightful..

    Lets start with what you have written..

    P(X+Y = n) = \sum_{k = 0}^{k=n} P(X = k, Y = n - k) = \sum_{k = 0}^{k=n} P(X = k) P(Y = n - k) =  \sum_{k = 0}^{k=n}\left(\frac{e^{-a} a^k}{k!}\right)\left(\frac{e^{-b}.b^{n - k}}{(n-k)!}\right)<br />

    Lets continue with what you should have done:

    \sum_{k = 0}^{k=n}\left(\frac{e^{-a} a^k}{k!}\right)\left(\frac{e^{-b}.b^{n - k}}{(n-k)!}\right) = e^{-(a+b)} b^n \sum_{k = 0}^{k=n} \dfrac{\left(\frac{a}{b}\right)^k}{k! (n-k)!}

    Now we can write \frac1{k! (n-k)!} = \frac1{n!}\frac{n!}{k! (n-k)!} = \frac{\binom{n}{k}}{n!}

    So:

    e^{-(a+b)} b^n \sum_{k = 0}^{k=n} \dfrac{\left(\frac{a}{b}\right)^k}{k! (n-k)!} = e^{-(a+b)} b^n \sum_{k = 0}^{k=n} \dfrac{\binom{n}{k}\left(\frac{a}{b}\right)^k}{n!} =  e^{-(a+b)} b^n \dfrac{\sum_{k = 0}^{k=n} \binom{n}{k}\left(\frac{a}{b}\right)^k}{n!}

    Remember Binomial Theorem? It says \sum_{k = 0}^{k=n}  \binom{n}{k}\left(\frac{a}{b}\right)^k = \left(1 + \frac{a}{b}\right)^n

    This means that

    e^{-(a+b)} b^n \dfrac{\sum_{k = 0}^{k=n} \binom{n}{k}\left(\frac{a}{b}\right)^k}{n!} = e^{-(a+b)} b^n \dfrac{\left(1 + \frac{a}{b}\right)^n}{n!} =e^{-(a+b)} b^n\dfrac{(a+b)^n}{n! b^n} = \frac{e^{-(a+b)} (a+b)^n }{n!}

    But wait... \frac{e^{-(a+b)} (a+b)^n }{n!} is actually the distribution of a Poisson R.V with parameter a+b.

    Thus if X and Y are independent Poisson random variables with parameters a and b respectively, then X+Y is Poisson with parameter a+b !!

    P.S: There is a slick proof of this result using moment generating functions. Try it!
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  4. #4
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    ignore me - I misread what u'd written.. just gonna go over it in detail now.. thanks both! I'll let you know how I get on!

    wow.. that's awesome.. thankyou both so much! I have quite a bit of the rest of this question to go.. I'll give it a good go and get back here if there's any probs!
    Last edited by James0502; January 2nd 2009 at 09:59 AM.
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  5. #5
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    Ok, the next part asks for
    P[X = k | X + Y = n]

    I got

    [(n choose k)a^k . b^(n-k)]/[(a+b)^n]

    I am asked to state the distribution.. is this right and is it binomially distributed or does the [(a+b)^n] denominator affect this?

    many thanks
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  6. #6
    Lord of certain Rings
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    James,

    P( X = k | X+Y = n) = \frac{P(X = k, X+Y = n)}{P(X+Y = n)} = \frac{P(X = k) P(Y = n - k)}{P(X+Y = n)}

    And \frac{P(X = k) P(Y = n - k)}{P(X+Y = n)} = \dfrac{\frac{e^{-a} a^k}{k!}\frac{e^{-b} b^{(n-k)}}{(n-k)!}}{\frac{e^{-(a+b)} (a+b)^n}{n!}} = \frac{\binom{n}{k} a^k b^{(n-k)}}{(a+b)^n} = \binom{n}{k} \left(\frac{a}{a+b}\right)^k \left(\frac{b}{a+b}\right)^{(n-k)}

    Thus the R.V X conditioned on X+Y = n is \text{Binomial }\left(n, \frac{a}{a+b}\right) distributed
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