Let $X$ ~ $N(0,1)$ and $Y=e^X$. Find the moment generating function of Y.
I think I first need to find the cdf of Y. So I take:
$F_Y(y) = P(Y \le y) = P(e^X \le y) = P(X \le ln(y)) = F_X(ln(y))$
I think that part is correct. Now I get a little confused.
I think $M_Y(t) = Ee^{tX} = \int^{ln(y)}_{-\infty} \frac{1}{\sqrt{2\pi}}e^\frac{x^2}{2} e^{tx}\,dx $
Does that look right?
Best Answer
Hint: $Y$ has lognormal distribution $\ln\!\mathcal{N}(0,1)$. Write $X=\ln Y$, we get $$\frac{dx}{dy}=\frac{1}{y}.$$ It follows then \begin{align*} f_Y(y)&=f_X({\ln Y})\cdot\left|\frac{1}{y}\right|\\ &=\frac{1}{y\sqrt{2\pi}}\cdot\exp\left(-\frac{(\ln y)^2}{2}\right). \end{align*} Hint 2: If we change the variable back to $X$, then \begin{align*} M_Y(t)&=E[e^{te^x}]\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\exp\left(te^x-\frac{x^2}{2}\right)\,dx. \end{align*} which is divergent if $t>0$, why? Consider Taylor series of $e^x$: $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$ and the fact that $$\int_0^\infty e^{tx^3}\,dx$$ diverges for all $t>0$.