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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?

zsljulius
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    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
  • @whuber That is the part where I got stuck. How should I formulate this? I don't know how to express the inverse CDF of a normal distribution – zsljulius Jan 22 '14 at 20:29
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    It's just a change of units of measurement. I have explained this in various answers (http://stats.stackexchange.com/a/49794, http://stats.stackexchange.com/a/5876, and http://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 http://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
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    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
  • The edit clarifies things well--good work! However, this is not an issue of linearity of an inverse CDF; in fact, it is impossible for any inverse CDF to be linear. It would help to be more explicit in your notation, using something like "$F_{\mu,\sigma}$" to denote the CDF associated with the parameters $\mu$ and $\sigma$, for otherwise formulas like your last one become nonsensical. Take some care, too, in considering what kinds of quantities you are dealing with: the argument to an inverse cdf like $F^{-1}$ is a *probability*: but how can $X=\mu+\sigma Z$ be a probability? – whuber Jan 24 '14 at 16:33

1 Answers1

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  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!

TheBigAmbiguous
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    The CLT is not needed to demonstrate any of this. In fact, the relationships among Normal distributions of differing means and variances follow immediately from their definitions (or are actually used as definitions). – whuber Jan 25 '14 at 01:34
  • It depends how exactly you define the Normal R.V. Personally I've always considered it to be defined via the CLT, since a sum/mean of IID RV's is the only real situation where normal distributions arise. That's kinda-sorta what I meant when I said the theorem can be proven via CLT. – TheBigAmbiguous Jan 27 '14 at 17:07
  • The CLT does not provide a *definition* for a normal distribution. Historically, normal distributions arose because maximum likelihood equations are linear--and therefore easy to analyze--when the log of the PDF is quadratic: that defines normal distributions. There are [plenty of other characterizations](http://stats.stackexchange.com/questions/4364), many of which could also provide definitions. – whuber Jan 27 '14 at 18:04
  • That may have been how they "arose" on paper (citation needed, though), but the way they "arise" in real life is when a bunch of iid RV's are summed. When you shoot an arrow at a target, the distribution of the final location of the arrow is the sum of a series of tiny motions by your hand and the arrow, each of which is (somewhat) iid. Therefore the final distribution is normal. It's the same reason Plinko discs fall in a normal distribution and stock returns have a (mostly) normal distribution. To understand any RV distribution you have to know where it comes from. – TheBigAmbiguous Jan 27 '14 at 19:30
  • The citation (to Gauss's work) has been offered in many places on this site; see Stigler's books on statistical history for instance. You appear to confuse theory with application. Theory, not applications, is used for proofs and manipulations like those requested in the current question. Whether that theory is applicable in any particular case is a separate issue. – whuber Jan 27 '14 at 20:14