The quantile function for log-normal distribution is given by
$$F^{-1}(p)=\exp(\mu+\sigma\Phi^{-1}(p)),$$
where $0<p<1$ and $\Phi(p)$ is the CDF of a normal distribution.
I am trying to derive $F^{-1}(p)$ and arrive at the solution above.
Let $X$ be log-normally distributed and $Z\sim N(\mu,\sigma^2)$. Now
$$F(x)=\Phi\left(\frac{\ln x-\mu}{\sigma}\right)=\frac{1}{\sqrt{2\pi}}\int^{\frac{\ln x-\mu}{\sigma}}_{0}\exp \left(-\frac{1}{2} \left(\frac{z-\mu}{\sigma}\right)^2\right)dz.$$
Now let's solve it in terms of $x$.
$$x=\frac{1}{\sqrt{2\pi}}\int^{\frac{\ln x-\mu}{\sigma}}_{0}\exp \left(-\frac{1}{2} \left(\frac{z-\mu}{\sigma}\right)^2\right)dz \iff \\
x\sqrt{2\pi}=\int^{\frac{\ln x-\mu}{\sigma}}_{0}\exp \left(-\frac{1}{2} \left(\frac{z-\mu}{\sigma}\right)^2\right)dz \iff \\
x\sqrt{2\pi}=\exp \left(-\frac{1}{2} \left(\frac{\frac{\ln x-\mu}{\sigma}-\mu}{\sigma}\right)^2\right)-\exp \left(-\frac{1}{2} \left(\frac{0-\mu}{\sigma}\right)^2\right) \iff \\
x\sqrt{2\pi}=\exp \left(-\frac{1}{2} \left(\ln x -\mu -\sigma\mu\right)^2\right)-\exp \left(\frac{1}{2} \left(\frac{\mu}{\sigma}\right)^2\right) \iff \\
x\sqrt{2\pi}+\exp \left(\frac{1}{2} \left(\frac{\mu}{\sigma}\right)^2\right)=\exp \left(-\frac{1}{2} \left(\ln x -\mu -\sigma\mu\right)^2\right) \iff \\
\log\left(x\sqrt{2\pi}+\exp \left(\frac{1}{2} \left(\frac{\mu}{\sigma}\right)^2\right)\right)=-\frac{1}{2} \left(\ln x -\mu -\sigma\mu\right)^2$$
I am not sure how to go on from here.
Best Answer
Let $\log (X) \sim \mathcal{N}(\mu,\sigma^2)$, than indeed $$ F(x) = \Phi\left( \frac{\log(X) - \mu}{\sigma} \right). $$ We know that $F(x) \in [0,1]$, so let $F(x) = p$, than we see that $$ p = \Phi\left( \frac{\log(F^{-1}(p)) - \mu}{\sigma} \right). $$ Solving for the quantile function $F^{-1}(p)$ gives us $$ \Phi^{-1}(p) = \frac{\log(F^{-1}(p)) - \mu}{\sigma} \iff\\ \sigma \Phi^{-1}(p) = \log(F^{-1}(p)) - mu \iff\\ \mu + \sigma\Phi^{-1}(p) = \log(F^{-1}(p)) \iff\\ F^{-1}(p) = \exp(\mu + \sigma\Phi^{-1}(p)). $$