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.


1 Answer 1


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}

  • $\begingroup$ However, I still have a question. Based on îtos lemma, the mean was transformed to $\mu-\sigma/2$ doesn't it affect anything? I mean, if the mean changed, shouldn't the variance also? (btw, nice the demonstration of your last equation, that makes me understand more) $\endgroup$
    – silgon
    Jul 27, 2017 at 14:51
  • $\begingroup$ Mean does change, but variance (and the other higher moments) as well. All in all in does not affect anything. $\endgroup$
    – Quantuple
    Jul 27, 2017 at 15:20
  • 1
    $\begingroup$ I think I didn't explain myself correctly. I mean that in order to keep the mean from the normal to the exponential, your mean becomes $\mu-\sigma/2$, but I do not know if that's the same case for volatility (since it is one of the parameters for the quantile). In any case, I was just trying with a montecarlo empirical quantile in time to see how these analytical quantiles behave (image for reference: i.stack.imgur.com/h0rl5.png). I did the test to verify if the volatility changes as input, and it apparently works. Sorry for not accepting the answer before, I wanted to verify that. =) $\endgroup$
    – silgon
    Jul 28, 2017 at 8:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.