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Math Help - Possible to find p(X=x) given E(X) and E(X^2)?

  1. #1
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    Possible to find p(X=x) given E(X) and E(X^2)?

    This practice problem gives us E(X) = 11 and E(X^2) = 121 and asks for us to find probability fn and cdf.

    Initially I was thinking we could use some identities of the probability generating function like P"(1) = E(X^2) - E(X) and P'(1) = E(X), but I'm running into a mental wall where I'm positive that you can't work it out unless you know how X is distributed (as in is it a poisson, binomial, etc).

    Any advice?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by drowned View Post
    This practice problem gives us E(X) = 11 and E(X^2) = 121 and asks for us to find probability fn and cdf.

    Initially I was thinking we could use some identities of the probability generating function like P"(1) = E(X^2) - E(X) and P'(1) = E(X), but I'm running into a mental wall where I'm positive that you can't work it out unless you know how X is distributed (as in is it a poisson, binomial, etc).

    Any advice?
    If \overline{x}=11 and \overline{x^2}=121 what is the variance of x

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    If \overline{x}=11 and \overline{x^2}=121 what is the variance of x

    CB
    I see that as well where V(X) = E(X^2) - E(X)^2 = 121 - 121 = 0. Getting the vibe that a variance = 0 means something significant but I'm a bit lost still. Does 0 variance mean we're limited to one value of x being 100% probable and all other values or 0%?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by drowned View Post
    I see that as well where V(X) = E(X^2) - E(X)^2 = 121 - 121 = 0. Getting the vibe that a variance = 0 means something significant but I'm a bit lost still. Does 0 variance mean we're limited to one value of x being 100% probable and all other values or 0%?
    Yes.

    CB
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  5. #5
    MHF Contributor matheagle's Avatar
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    since the mean is 11 and the second moment is 11 squared....

    P(X=11)=1
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