1. ## 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 $E[X(X-1)]$

2. Originally Posted by zorro
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

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

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

3. ## Thank u Mr fantastic

Originally Posted by mr 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

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

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

Thank you Mr fantastic for helping me

4. ## Is this correct

Originally Posted by zorro
Thank you Mr fantastic for helping me

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

5. Hello,

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