MGF of Normal Dist

I am not able to derive E(X^2) from the MGF of Normal Distribution.

I get that that E(X) = mu which is the co-efficient of t/1! in the second term of the series.

But how is E(X^2) = (sigma)^2 + (mu)^2 ??

Am i supposed to expand the third term of the series?

I feel as if I am missing some point here.

Thanks.

 

No what you quoted is the standard Var(X) = E(X^2) - [E(X)]^2 result .. so the coefficient of (t^2)/2! is the 2nd moment [EX^2], and you can re-arrange to find sigma using the standard identity. The third term would be used to find skewness.

 

Sorry but I still dont get it. Lets say, mean = m and variance = v^2, the series goes -

M(t) = 1 + [mt + 1/2*(vt)^2] + {[mt + 1/2*(vt)^2]^2}/2! + ..........

Can you help evaluate 1st and 2nd moment from this because the study material (Chp 5 Pg 18) says,

E[X^2] = co - efficient of t^2/2! = v^2 + m^2.

How do i re-arrange to get the above step?

Apologies if my mathematics is poor

 

hey, just calculate the second derivative of MGF, wrt t, and evaluate it at t = 0. This will give you the E(X^2).

Otherwise, just see, that as you take second derivate, the first term in series expansion will disappear, and you are left with series with second term onwards. Then set t = 0. All other terms except first will disappear. Remaining expression is you E(X^2)..

 

i have done all that .... i am just curious about the series method which i couldnt derive .... thanks anyway ....