find the mean and variance of the normal distribution of statistics using parts (a) and (b) with m(t) = e^(μt + (σ^2t^2)/2)

A) Mean = m'(0)
B) Variance = m"(0) - (m'(0))^2

Question

find the mean and variance of the normal distribution of statistics using parts (a) and (b) with m(t) = e^(μt + (σ^2t^2)/2)

A) Mean = m'(0)
B) Variance = m"(0) - (m'(0))^2

anonymous · Answer

is μt the mean of t, and σ^2 variance of t^2?

anonymous · Answer

I dont believe so.  I think that mu and sigma are constants.

anonymous · Answer

But then again i dont really know haha

anonymous · Answer

Solve the integral for the nth moment, the mean is when n=1. 

integral -inf to inf of ( t*e^(μt + (σ^2t^2)/2)) dt.

I dont know what mu and sigma is, so sorry.