Finding Median for continuous distribution

Question

I have a homework problem asking me to find the median of $Y$, where $Y=e^Z$ and $Z \sim N(0,1)$.

I found that the density of $Y=e^Z$ equals $f(Y)=\frac{1}{y \sqrt{2\pi}}e^{-\frac{(ln(y))^2}{2}}$; $y>0$.

To find the median, $f(m)=\int_{-\infty}^m f(y)dy = \int_0^m \frac{1}{y\sqrt{2\pi}}e^{-\frac{(ln(y))^2}{2}}= \frac{1}{2}$.

However I am having a hard time solving the integral. Before solving it myself I went to Wolfram Alpha and it said that there was an error. Is there an easier way to do this computation or did I do something wrong previously?

Any help is appreciated, thank you!

Answer 1

The median of $Y=e^Z$ is $1$. Since $Z \sim N(0,1)$, half of $Z$ is less than or equal to $0$. So half of $Y$ is less than or equal to $e^0=1$.

Answer 2

Monotonic-increasing transforms commute with calculation of quantiles. i.e. $\text{median}(e^Z)$ is $e^{\text{median}(Z)}$

Intuitively it's obvious -- if half the distribution is at or below it before you transform, half the distribution will remain at or below what you transform it to after you transform.

Proof is elementary and along the same lines. For your particular transform:

Let $Y=e^Z$

$F_Y(y) = P(Y\leq y) = P(e^Z\leq y) = P(Z\leq \log(y)) = F_Z(\log(y))$.

Let the median of $Z$ be $m$. Then setting $m=\log(y)$ we have $F_Z(\log(y))=\frac12$ but we see by the above set of equalities that this $\frac12$ is also the value of $F_Y(y) = F_Y(e^m)$.

So if you can find the median of the Gaussian, just transform it and you're done.