# Inverse CDF of normal variable

The following paragraph was an excerpt from R PerformanceAnalytics documentation on VaR.

The most common estimate is a normal (or Gaussian) distribution $R\sim \mathcal{N}(\mu,\sigma)$ for the return series. In this case, estimation of VaR requires the mean return $\bar{R}$, the return distribution and the variance of the returns $\sigma$. In the most common case, parametric VaR is thus calculated by

$$\sigma=var(R)$$

$$VaR= -mean(R) - \sqrt{\sigma}*qnorm(c)$$

I am curious why is this the case. VaR is just simply the inverse cdf evaluated at c%.

Edit

After reading some articles on standardization suggested by @whuber, I come to the following observations.

Let $Z \sim \mathcal{N}(0, 1)$ and $X \sim \mathcal{N}(\mu, \sigma^2)$, the relationship between the two random variables can be expressed as $$X = \mu + \sigma*Z$$ This can be deduced from the linearity property of normal random variables. The only question left was to show that $$F^{-1}(X) = \mu + \sigma * F^{-1}(Z)$$ That is to show the inverse CDF is a linear function. This is how far I get to. Is there any theorem that says inverse CDF of a normal R.V. is linear?

• Hint: precisely which normal inverse CDF is evaluated by qnorm? – whuber Jan 22 '14 at 20:11
• @whuber I understand that it is standard normal N(0,1) – zsljulius Jan 22 '14 at 20:20
• Right: so what exactly is the relationship between the inverse CDF of a Normal$(\mu,\sigma)$ distribution and the values returned by qnorm? – whuber Jan 22 '14 at 20:22
• It's just a change of units of measurement. I have explained this in various answers (stats.stackexchange.com/a/49794, stats.stackexchange.com/a/5876, and stats.stackexchange.com/a/55613, inter alia) and likely other explanations can be found elsewhere on this site. I see a complete mathematical answer appears in a somewhat disguised form (in terms of error functions) at stats.stackexchange.com/a/22923. For a concrete example, consider what change you would have to make to the formula to convert from dollars to Euros. – whuber Jan 22 '14 at 21:14
• It's a rather odd of theirs (the writers of that documentation) to use $\sigma$ to represent variance rather than using the near universal $\sigma^2$ for that ... odd to the point of being actively misleading. If one wishes to communicate, one doesn't break such a strong convention lightly. – Glen_b Jan 22 '14 at 21:56