MGF Problem

**statsnewbie** · May 19th 2008, 12:44 PM

Hi

Can someone help with the pointing me in right direction with the following

"Use the MGF to compute the mean and variance of the distribution of x which has the following probability distribution

f(x) = x+1/14 if x=1,2,3,4

Many thanks

**mr fantastic** · May 19th 2008, 07:59 PM

Originally Posted by statsnewbie

Hi

Can someone help with the pointing me in right direction with the following

"Use the MGF to compute the mean and variance of the distribution of x which has the following probability distribution

f(x) = x+1/14 if x=1,2,3,4

Many thanks

The mass density function is $f(x) = \frac{x + 1}{14}$ where x = 1, 2, 3, 4.

By definition, the moment generating function m(t) for a random variable X is

$E\left( e^{tX}\right) = \Pr(X = x_1) \, e^{tx_1} + \Pr(X = x_2) \, e^{tx_2} + \Pr(X = x_3) \, e^{tx_3} + ......$ .

So for your distribution:

$m(t) = \Pr(X = 1) \, e^{t} + \Pr(X = 2) \, e^{2 t} + \Pr(X = 3) \, e^{3 t} + \Pr(X = 4) \, e^{4 t}$

$= \frac{1}{14} \left( 2 \, e^{t} + 3 \, e^{2 t} + 4 \, e^{3 t} + 5 \, e^{4 t}\right)$ .

By definition: $\mu = E(X) = \frac{dm}{dt}$ evaluated at t = 0:

............

By definition: $E(X^2) = \frac{d^2 m}{d t^2}$ evaluated at t = 0:

.............

Formula: $Var(X) = E(X^2) - \mu^2$ :

.............

I've left the details for you to fill in.

**statsnewbie** · May 20th 2008, 12:04 AM

Thanks

One further question.

If t=0 does that mean that each element of (e) evaluates to zero or have I missed something.

**mr fantastic** · May 20th 2008, 12:23 AM

Originally Posted by statsnewbie

Thanks

One further question.

If t=0 does that mean that each element of (e) evaluates to zero or have I missed something.

?
You find the appropriate derivatives and then substitute t = 0. You do know that e^0 = 1 I hope .....

**statsnewbie** · May 20th 2008, 12:29 AM

Yr prompt reply is appreciated

e^0=1. Of course............ it's what I meant but not what I wrote!!! It's still too early!!!!

Thanks a lot!!!!

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