Monday, October 19, 2015

probability - Quantile normal and lognormal


Let's assume we have a normal distribution $X\sim \mathcal{N}(\mu,\sigma^2)$. In a normal distribution the quantile can be calculated as follows:


\begin{equation} \Phi_X ^{-1}(p)=\mu +\sigma {\sqrt {2}}\operatorname {erf} ^{-1}(2p-1) \end{equation}


If we want to calculate the value in the future of a stock we map it as:


\begin{equation} Y=\exp(X) \end{equation} Which means: \begin{equation} \log(Y)\sim \mathcal{N}(\mu,\sigma^2) \end{equation}



I would like to know that if the function of the quantile can be calculated based directly on:


\begin{equation} \Phi_Y ^{-1}(p)=\exp(\mu -\sigma/2+\sigma {\sqrt {2}}\operatorname {erf} ^{-1}(2p-1)) \end{equation}


The part of the equation $-\sigma/2$ is extracted from îto calculus, however, I cannot find anywhere the correctness of this equation (I deduced it). I think the function $exp$ is monotonic, so, it should preserve the value for the quantiles, but I'm not certain. One of my certainties is that $\mu$ changed to $\mu-\sigma/2$, I have no idea if that modifies in some way the calculation of $\Phi_Y ^{-1}(p)$, or if $\sigma$ also changed.



Answer



Quantiles are preserved under monotonic transformations, hence the quantile for $Y$ is simply the exponential of the quantile of $X$, no need for corrections whatsoever (see here for instance).


Put otherwise, let $q$ denote the quantile $\alpha$ of $X$ i.e. $$\Bbb{P}(X \leq q) = \alpha$$ then \begin{align} \Bbb{P}( X \leq q ) &= \Bbb{P}( \underbrace{\exp(X)}_{Y} \leq \underbrace{\exp(q)}_{Q} ) = \alpha \end{align}


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