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Math Help - Compute E[X(X-1)]

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    Compute E[X(X-1)]

    Question : A Binomial variate X has the mean 6 and standard deviation \sqrt{2}. Find its moment generating function. Hence obtain E[e^{2X+3}]. Also compute E[X(X-1)]
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    Quote Originally Posted by zorro View Post
    Question : A Binomial variate X has the mean 6 and standard deviation \sqrt{2}. Find its moment generating function. Hence obtain E[e^{2X+3}]. Also compute E[X(X-1)]
    First you need the value of n and p. Get them by solving the following equations:

    Mean: np = 6 .... (1)

    Variance: np(1-p) = 2 .... (2)

    Substitute the values of p and n into the moment generating function m(t), which is found here: Binomial distribution - Wikipedia, the free encyclopedia


    E[e^{2X+3}] = E[e^{2X}e^3] = e^3 E[e^{2X}] = e^3 m(2) = ....


    E[X(X-1)] = E[X^2 - X] = E[X^2] - E[X] = \left.\frac{d^2m}{dt^2}\right|_{t = 0} - \left.\frac{dm}{dt}\right|_{t = 0} = .....
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    Thank u Mr fantastic

    Quote Originally Posted by mr fantastic View Post
    First you need the value of n and p. Get them by solving the following equations:

    Mean: np = 6 .... (1)

    Variance: np(1-p) = 2 .... (2)

    Substitute the values of p and n into the moment generating function m(t), which is found here: Binomial distribution - Wikipedia, the free encyclopedia


    E[e^{2X+3}] = E[e^{2X}e^3] = e^3 E[e^{2X}] = e^3 m(2) = ....


    E[X(X-1)] = E[X^2 - X] = E[X^2] - E[X] = \left.\frac{d^2m}{dt^2}\right|_{t = 0} - \left.\frac{dm}{dt}\right|_{t = 0} = .....

    Thank you Mr fantastic for helping me
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    Is this correct

    Quote Originally Posted by zorro View Post
    Thank you Mr fantastic for helping me


    . = \mu'_2 - 6 = 2............Is this correct ?
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    Hello,

    You may also use the probability generating function (only available for discrete distributions), because its second derivative taken at t=1 is just E[X(X-1)]
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