Hi @Ali Ehsan Abbas I love seemingly straightforward questions that are thought-provoking, like this! My view is that if we only know the alternative distribution's 95% VaR is 1.56, then we know very little and actually we cannot say is has thinner tails. To demonstrate, consider the exponential distribution https://en.wikipedia.org/wiki/Exponential_distribution which has the remarkably convenient feature that its kurtosis is 6.0; i.e., the exponential is very heavy tailed. Given it is also parsimonious, we can solve for the lambda, λ, param that gives us a 95% quantile of 1.56, which is about 1.92. So that, because exponential CDF = 1 - exp(-λ*X), we can observe that 95.0% = 1 - exp(-1.92*1.56) and we've illustrated a heavy-distribution with a 95% quantile of 1.56.
I think the shows the "problem:" we don't know the mean and variance of the alternative distribution. Please keep in mind the 1.65 refers to a standard normal distribution; i.e., mean, µ, of zero and variance, σ^2, of one. My illustrative exponential distribution has mean of 0.5 and variance of 0.27. Changes in location and scale make the bigger difference here. Which leads to me to an even simpler example: assume a normal distribution with µ=-0.09 and variance of one. It's 95% VaR is 1.56 because (1.56-0.09)/1 = 1.65; yet is has normal tails by definition. In this way, we could manufacture all sorts of non-standard normal distributions with a 95% quantile of 1.56.
Now, if the alternative distribution has mean of zero and variance of one, then I think we can infer the alternative distribution has so-called light tails (kurtosis < 3 or excess kurtosis < 0) ... I hope that helps!
Helpful, but a bit of sherlock holmes re. your last statement!
Wouldn't a distribution with mean 0 and variance of 1 by default be a standard normal distribution, hence 1.65 deviate to 95% VaR quantile?!!
I think the key is for the distribution to be elleyptical! In short, an ellyptical distribution with 1.56 deviate would exhibit thinnner loss tail viz. standard normal distribution! Please correct me if i'm wrong.
Hi @Ali Ehsan AbbasRe: Wouldn't a distribution with mean 0 and variance of 1 by default be a standard normal distribution. Nope a distribution with non-zero skew and/or non-zero excess kurtosis is non-normal. For example, a normal mixture distribution can mix two normals, each with µ=0 and variance of 1.0, such that the normal mixture distribution that mixes them equally has mean of zero, variance of 1.0, but heavy-tails (as do all variance mixture distributions). But that's just to illustrate that the definition of normal depends on different third (skew) and fourth (kurtosis) moments.
Re: elliptical, I like your thinking! I'm not sure, I have my doubts only because I *think* elliptical includes fat-tailed distributions (https://en.wikipedia.org/wiki/Elliptical_distribution) so I don't mean to be a stick in the mud, but couldn't you just "shift left" a heavy-tailed distribution ? ... The "problem" with 1.56 deviate in isolation is that you don't know where the mean is, so you don't know if that's in the tail or in the body, or the extreme tail etc! However, as usual, I could be wrong Thanks,
append: I gave this a bit more thought, and I'm going to double-down on my previous response. I think the condition should be that the 1.56 is a standardized deviate. Any non-normal deviate, say given by (D), can be standardized with (D - µ)/σ, but as mentioned, to standardize does not imply the distribution normal or somehow "made normal." That is, a standardized 1.56 ought to have lighter-than-normal tails (aka, platykurtosis). To be real strict, since I can still imagine exceptions even in this case, I'd say the condition is actually: if the standardized deviate is 1.56 and the distribution is unimodal. Thanks!