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

Answer 1

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

Answer 2

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

Answer 3

But then again i dont really know haha

Answer 4

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.