Problem: Let X be a nonnegative random variable and $log X \sim N(0,1)$, prove that $\mathbb{E}[X^k] = e^{\frac{k^2}{2}}$.
I'm self learning probability in order to understand some theorems in number theory, and this problem came in a book I'm reading.
I know that $\mathbb{E}[X^k]=\mathbb{E}[e^{klog X }]=\int_{-\infty}^{\infty} e^{klog(x)}f_{X}(x)$.
Now, I'm pretty lost in the $f_{X}(x)$ part. I suppose the question is how can I link the logX which its distribution is known and the X but Im not sure what is allowed to use in these cases.
Any help would be really appreciated.
Thanks! 🙂
Best Answer
Let $Z = \log X$, so that $Z \sim \operatorname{Normal}(0,1)$ is a standard normal random variable. As you have already observed, $$\mathbb E [X^k] = \mathbb E [e^{k \log X}] = \mathbb E [e^{kZ}] = M_Z(k),$$ where $M_Z(k)$ is the moment generating function of $Z$. We find $$M_Z(k) = \int_{z=-\infty}^\infty e^{kz} \frac{e^{-z^2/2}}{\sqrt{2\pi}} \, dz = \int_{z=-\infty}^\infty \frac{e^{-(z^2/2 - kz)}}{\sqrt{2\pi}} \, dz.$$ Note your integral is problematic because first, we must have $x > 0$, and second, you have not established the value of $f_X(x)$, the density of the transformed variable $X$. Dealing with $Z$ directly avoids both of these issues. All that remains is to complete the square in the exponent: $$\frac{z^2}{2} - kz = \frac{1}{2}(z^2 - 2kz + k^2) - \frac{k^2}{2} = \frac{(z-k)^2}{2} - \frac{k^2}{2},$$ hence $$M_Z(k) = e^{k^2/2} \int_{z=-\infty}^\infty \frac{e^{-(z-k)^2/2}}{\sqrt{2\pi}} \, dz,$$ and observing that the integrand is the density of a normal random variable with mean $k$ and variance $1$, its integral over the reals is simply $1$, and the result is proven.
It is worth noting that $X$ is what we call a lognormal distribution, and that this result shows (in this case, but is easily generalized to other values for the location and scale parameters) that the $k^{\rm th}$ raw moment of the lognormal distribution corresponds to the MGF of the normal distribution evaluated at $k$.