Question : A Binomial variate $\displaystyle X$ has the mean $\displaystyle 6$ and standard deviation $\displaystyle \sqrt{2}$. Find its moment generating function. Hence obtain $\displaystyle E[e^{2X+3}]$. Also compute $\displaystyle E[X(X-1)]$

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- Dec 14th 2009, 01:39 AMzorroCompute E[X(X-1)]
Question : A Binomial variate $\displaystyle X$ has the mean $\displaystyle 6$ and standard deviation $\displaystyle \sqrt{2}$. Find its moment generating function. Hence obtain $\displaystyle E[e^{2X+3}]$. Also compute $\displaystyle E[X(X-1)]$

- Dec 14th 2009, 02:37 AMmr fantastic
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

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

$\displaystyle E[X(X-1)] = E[X^2 - X] = E[X^2] - E[X] = \left.\frac{d^2m}{dt^2}\right|_{t = 0} - \left.\frac{dm}{dt}\right|_{t = 0} = ....$. - Dec 18th 2009, 07:27 PMzorroThank u Mr fantastic
- Dec 18th 2009, 09:04 PMzorroIs this correct

http://www.mathhelpforum.com/math-he...4a5f9dd5-1.gif. = $\displaystyle \mu'_2 - 6$ = 2............Is this correct ? - Dec 19th 2009, 12:16 AMMoo
Hello,

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