K
kartik_newpro
Member
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.
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.