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Math Help - Product of two discrete distributions is one of the same family?

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    Product of two discrete distributions is one of the same family?

    Hi, I have this problem with which I could do with some help.

    Given the marginal pdfs shown below, find the pdf of X + Y (given X and Y are independent).

    p_{X}(k) = \exp{(- \lambda)} \cdot \frac{\lambda^k}{k!}

    and

    p_{Y}(k) = \exp{(- \mu)} \cdot \frac{\mu^k}{k!}


    where k = 1, 2, 3,\ldots

    I know from the solution manual this should be a distribution of the same form, namely

    p_{X+Y}(z) = \exp{-(\lambda + \mu)} \cdot \frac{(\lambda + \mu)^z}{z!}

    where z = 1, 2, 3,\ldots

    But I don't see how this was arrived at. I'm especially miffed as to what z stands for here. Is z = k_Y + k_X or what?


    Many thanks in advance to anyone who can clarify this for me.
    Last edited by Mathsdog; May 14th 2012 at 11:40 PM.
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    Re: Product of two discrete distributions is one of the same family?

    Hello,

    This is the sum, not the product you're looking for.

    And k can be equal to 0. It's a Poisson distribution.

    Think about what values can X+Y take. The same : 0,1,2,...
    So for k=0,1,2,..., compute P(X+Y=k)=\sum_{j=0}^k P(X=j;Y=k-j) (try to think why it's this way, then use the independence)
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    Re: Product of two discrete distributions is one of the same family?

    Hi,
    I see what you are saying I think. I should have said the distribution of the sum of two independent random variables. But to find the distribution for the sum X+Y, i.e. the joint, one has to take the sum over the product of the two distributions (that's where independence comes in), right ? i.e. for Z = X +Y canonically

    p_{Z}(z) = \sum_{all x} p_{X}(x) \cdot p_{Y}(z-x)

    so does this translate for our example as

    p_{X + Y = Z}(z)= \sum_{all k} \exp{(-\mu) \cdot\frac{\mu^k}{k!} \cdot \exp{(-\lambda) \cdot\frac{\lambda^{(z-k)}}{(z-k)!}


    ?

    But I cant massage this form into the desired form. Am I setting things up incorrectly here?

    And here the notation implies that k is the same for both initial distributions. Should the notation not more correctly be k_Y = 1,2, \ldots and k_X = 1,2, \ldots

    So that z = k_Y + k_X

    Or am I missing something (as I guess is much more likely) ?

    Thanks again.
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    Re: Product of two discrete distributions is one of the same family?

    Quote Originally Posted by Mathsdog View Post
    Hi,
    I see what you are saying I think. I should have said the distribution of the sum of two independent random variables. But to find the distribution for the sum X+Y, i.e. the joint, one has to take the sum over the product of the two distributions (that's where independence comes in), right ? i.e. for Z = X +Y canonically

    p_{Z}(z) = \sum_{all x} p_{X}(x) \cdot p_{Y}(z-x)

    so does this translate for our example as

    p_{X + Y = Z}(z)= \sum_{all k} \exp{(-\mu) \cdot\frac{\mu^k}{k!} \cdot \exp{(-\lambda) \cdot\frac{\lambda^{(z-k)}}{(z-k)!}


    ?

    But I cant massage this form into the desired form. Am I setting things up incorrectly here?

    And here the notation implies that k is the same for both initial distributions. Should the notation not more correctly be k_Y = 1,2, \ldots and k_X = 1,2, \ldots

    So that z = k_Y + k_X

    Or am I missing something (as I guess is much more likely) ?

    Thanks again.
    you are on the right track.


    p_{X + Y = Z}(z)= \sum_{k=0}^z \exp{(-\mu) \cdot\frac{\mu^k}{k!} \cdot \exp{(-\lambda) \cdot\frac{\lambda^{(z-k)}}{(z-k)!}

    =e^{-(\mu+\lambda)} \sum_{k=0}^z \dfrac{z!}{z!}\cdot \dfrac{\mu^k \lambda^{z-k}}{k!(z-k)!}

    now use the binomial theorem
    Last edited by harish21; May 15th 2012 at 08:12 AM.
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    Re: Product of two discrete distributions is one of the same family?

    Hi, Thanks for your post again,


    If I multiply by e^{-(\mu+\lambda)}\sum_{k=0}^z\dfrac{z!}{z!} \cdot \dfrac{\mu^k%20\lambda^{z-k}}{k!(z-k)!} by  \frac{(1-\mu)^{z-k}}{(1-\mu)^{z-k}}


    then the sum over the binomial distribution with parameters (z, k) (equalling 1) falls out and I am left with

    e^{-(\mu+\lambda)} \cdot \frac{\lambda^{z-k}}{(1-\mu)^{z-k} \cdot z!}

    So I reckon I need to show that

    \frac{\lambda^{z-k}}{(1-\mu)^{z-k}} = (\lambda + \mu)^{z}


    in order to arrive at the final expression

    p_{X+Y}(z)= e^{-(\lambda%20+%20\mu)} \cdot \frac{(\lambda+\mu)^z}{z!}

    But I can't quite see how this final step works. Wanna put me out of my misery?

    Thanks, MD
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    MHF Contributor harish21's Avatar
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    Re: Product of two discrete distributions is one of the same family?

    Quote Originally Posted by Mathsdog View Post
    Hi, Thanks for your post again,


    If I multiply by e^{-(\mu+\lambda)}\sum_{k=0}^z\dfrac{z!}{z!} \cdot \dfrac{\mu^k \lambda^{z-k}}{k!(z-k)!} by  \frac{(1-\mu)^{z-k}}{(1-\mu)^{z-k}}


    then the sum over the binomial distribution with parameters (z, k) (equalling 1) falls out and I am left with

    e^{-(\mu+\lambda)} \cdot \frac{\lambda^{z-k}}{(1-\mu)^{z-k} \cdot z!}

    So I reckon I need to show that

    \frac{\lambda^{z-k}}{(1-\mu)^{z-k}} = (\lambda + \mu)^{z}


    in order to arrive at the final expression

    p_{X+Y}(z)= e^{-(\lambda + \mu)} \cdot \frac{(\lambda+\mu)^z}{z!}

    But I can't quite see how this final step works. Wanna put me out of my misery?

    Thanks, MD
    e^{-(\mu+\lambda)} \sum_{k=0}^z  \dfrac{\mu^k \lambda^{z-k}}{k!(z-k)!}

    multiplying by z!/z! gives:

    =e^{-(\mu+\lambda)} \sum_{k=0}^z \dfrac{z!}{z!}\cdot \dfrac{\mu^k \lambda^{z-k}}{k!(z-k)!}

    =\dfrac{e^{-(\mu+\lambda)}}{z!} \sum_{k=0}^z z!\cdot \dfrac{\mu^k \lambda^{z-k}}{k!(z-k)!}

    =\dfrac{e^{-(\mu+\lambda)}}{z!} \sum_{k=0}^z \dfrac{z!}{{k!(z-k)!}}\cdot \mu^k \lambda^{z-k}

    =\dfrac{e^{-(\mu+\lambda)}}{z!} \underbrace{\sum_{k=0}^z \dbinom{z}{k}\cdot \mu^k \lambda^{z-k}}_{\text{This by binomial theorem is equal to} (\lambda+\mu)^z}

    =\dfrac{e^{-(\mu+\lambda)}}{z!} (\lambda+\mu)^z}
    Last edited by harish21; May 16th 2012 at 10:35 AM.
    Thanks from Mathsdog
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