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

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- May 19th 2008, 12:44 PMstatsnewbieMGF Problem
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 - May 19th 2008, 07:59 PMmr fantastic
The mass density function is $\displaystyle f(x) = \frac{x + 1}{14}$ where x = 1, 2, 3, 4.

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

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

So for your distribution:

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

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

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

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

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

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

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

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

I've left the details for you to fill in. - May 20th 2008, 12:04 AMstatsnewbieMGF Problem
Thanks

One further question.

If t=0 does that mean that each element of (e) evaluates to zero or have I missed something. - May 20th 2008, 12:23 AMmr fantastic
- May 20th 2008, 12:29 AMstatsnewbieMGF Problem
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!!!!