Inverse CDF of normal variable

Question

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?

Answer 1

1. The author should not have denoted variance as sigma. Sigma traditionally is standard deviation, so that's confusing.

2. The VaR formula is just a scaling. subtracting the mean aligns the mean of the normal distribution at zero, and then multiplying by the stdDev scales the distribution. To answer your question, you can do this because a zero-mean normal random variable times a scalar is another zero-mean normal RV, just with the standard deviations scaled by that scalar. I.e. x*N(mu, sigma^2) ~ N(mu, (x*sigma)^2). This is a theorem, yes. It can be proven via the Central Limit Theorem, since scaling an underlying distribution causes the stdDev to scale the same way, and when you sum over the iid RV's after scaling you get a normal RV back, just with a scaled stdDev.

Sorry if that seemed longwinded and jumbled, it's 6PM on a Friday!