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Conditional Probability Distribution

  1. May 31, 2012 #1
    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=47884&stc=1&d=1338504146.png

    3. The attempt at a solution

    I have that the joint probability mass function would be

    [itex]\Pi_{i=1}^{k} \frac{\lambda_{i}^{n_{i}}}{n_{i}!} e^{-\lambda_{i}}[/itex]

    How would I go about applying the conditional to get the conditional distribution?
     

    Attached Files:

  2. jcsd
  3. May 31, 2012 #2

    Ray Vickson

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    You need to compute
    [tex] P(N_1 = n_1, N_2 = n_2, \ldots, N_k = n_k | \sum_{i=1}^k N_i = n)
    = \frac{ P(N_1 = n_1, N_2 = n_2, \ldots, N_k = n_k \: \& \sum_{i=1}^k N_i = n)}{P(\sum_{i=1}^k N_i = n)}[/tex] for any k-tuple [itex] (n_1, n_2, \ldots, n_k)[/itex] that sums to n. Can you simplify the "event" in the numerator probability? Can you compute the denominator probability?

    RGV
     
  4. May 31, 2012 #3
    I don't understand how the [itex] \sum_{i=1}^k N_i = n[/itex] term constrains the joint distribution, given that each [itex]\lambda_{i}[/itex] can be distinct.

    Also, to evaluate [itex]P(\sum_{i=1}^k N_i = n)[/itex], wouldn't this be the joint distribution, evaluated at [itex]N_{1}=n-\sum_{i=2}^{k}N_{i}[/itex] and then [itex]N_{2}=n-\sum_{i=3}^{k}N_{i}[/itex] and so on? I feel that's wrong because it doesn't seem like it'd collapse down nicely.
     
  5. May 31, 2012 #4

    Ray Vickson

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    The [itex] \lambda_i[/itex] have absolutely nothing to do with the [itex] n_i[/itex]; if you think they do, then you seriously misunderstand the material. The formula you wrote for the joint pmf is [itex] P(N_1 = n_1, \ldots, N_k = n_k).[/itex] The only role of the [itex] \lambda_i[/itex] is to control the size of this probability, but that has nothing to do with the [itex] n_i[/itex] values themselves.

    RGV
     
  6. May 31, 2012 #5
    But because the joint PMF is a multiplication of the individual PMFs, the term becomes [itex]\lambda_{1}^{n_{1}}\lambda_{2}^{n_{2}}\lambda_{3}^{n_{3}}...[/itex] which means the term [itex]\sum_{i=1}^k N_i = n[/itex] doesn't appear in the joint PMF (although it would if the [itex]\lambda_{i}[/itex] were equal), so I don't understand how the condition [itex]\sum_{i=1}^k N_i = n[/itex] is applied to give [itex]P(N_1 = n_1, N_2 = n_2, \ldots, N_k = n_k \: \& \sum_{i=1}^k N_i = n)[/itex]
     
  7. May 31, 2012 #6

    Ray Vickson

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    Try first to figure out what happens when k = 2, then maybe k = 3. You should see a pattern emerging.

    RGV
     
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