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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.

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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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  • $\begingroup$ 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 '17 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 '17 at 15:20
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    $\begingroup$ 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 '17 at 8:08

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