Random variable $V$ has moment generating function $M(t)=e^{3e^{t}-1}$. What is $E(V)$? Note: you do not have to derive the mean.

Question

Random variable $V$ has moment generating function $M(t)=e^{3e^{t}-1}$. What is $E(V)$? Note: you do not have to derive the mean.

I have tried deriving the function , which i got $2e^2$ after subbing in $t=0$ That was no where close to the answer $3$ .

Answer 1

I don't think $M(t)=e^{3e^t-1}$ is a valid moment generating function:

$M(t) = E[e^{tV}]$ so

$M(0) = E[e^0]=E[1]=1$ necessarily, yet $e^{3e^0-1}=e^2$.

I think you meant $M(t)=e^{3(e^t-1)}$, in which case we recognize $V\sim Poisson(3)$. Thus $E[V]=3$.

Alternatively we could find $E[V]$ by evaluating the derivative of $M(t)$ at $0$.