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Thread: marginal variance of discrete joint distribution

  1. #1
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    marginal variance of discrete joint distribution

    \\\text{Given }Pr(X=i,\text{ }Y=j)=\frac{i+j-1}{8}\text{ for }i=1,\text{ }2\text{ and }j=1,\text{ }2.\\\\\text{Find }Var[Y].

    I know how to find the solution through finding E[Y] and E[Y^2], but my book gives an alternative solution that I haven't seen before.

    Alternative solution:

    \\\text{Note:}\\\\\text{Alternatively, since }Y\text{ can take only two values, 1 and 2, with probabilities }\frac{3}{8}\text{ and }\frac{5}{8}\text{ respectively, we can use the Bernoulli shortcut to find }Var[Y].\\\\Var[Y]=(1-2)^2\left(\frac{3}{8}\right) \left(\frac{5}{8}\right)=0.2344


    I know that the variance of the Bernoulli distribution is p(1-p), but if that's what the book is referring to, what's with the (1-2)^2 part?
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  2. #2
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    Re: marginal variance of discrete joint distribution

    I think you are confusing "Bernoulli distribution" with "Bernoulli shortcut". If your book says it is using the "Bernoulli shortcut", it must have an explanation of that. Have you tried looking it up in the index?
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  3. #3
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    Re: marginal variance of discrete joint distribution

    Thanks, there is no index
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  4. #4
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    Re: marginal variance of discrete joint distribution

    Nevermind, I figured it out. Since Y can only take on two values, you can find the variance the same way it's found in the Bernoulli distribution.
    Last edited by downthesun01; Aug 14th 2013 at 02:24 PM.
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  5. #5
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    Re: marginal variance of discrete joint distribution

    \\\text{Let } q=1-p\\\\f(x;p;q)=\begin{cases}p, & \text{for }x=a\\q, & \text{for }x=b \end{cases}

    \\E[X]=ap+bq\\\\E[X^2]=a^2p+b^2q\\\\Var[X]=a^2p+b^2q-(ap)^2-2abpq-(bq)^2

    =a^2p(1-p)+b^2q(1-q)-2abpq

    =a^2pq+b^2pq-2abpq

    =pq(a^2+b^2)-2abpq

    =pq(a^2-2abpq+b^2)

    =pq(a-b)^2
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