# Value at Risk for normal r.v. with shock (regimes)

I am struggling to understand how was this simple Value-at-Risk calculated. It's Example 1 in Daníelsson, Jón, et al. "Fat tails, VaR and subadditivity." Journal of econometrics 172.2 (2013): 283-291 (a draft of which can be found in this link).

The authors consider a random variable $X_i$ defined as the sum of a standard normal random variable and a discrete random variable (which is independent from the normal r.v.).

I have attempted to obtain the distribution of $X_i$ by considering that it can be seen a mixture of gaussian distributions, with each component of the mixture representing a "regime", but I don't arrive at the same value as the authors (3.1).

I am conscious that this is a very basic exercise, but I must be missing something....

Thank you in advance for any suggestions.

You can approximate the pdf of the mixture as: $$f_X(x) = 0.009 g(x+10) + 0.991 g(x)$$ for $g(x)$ the pdf of $\mathcal{N}(0,1)$

You will observe that since the standard deviation is 1 almost all of the density attributed to the component $g(x+10)$ is about the point x=-10 and certainly for about 7 standard deviations away about the point x=-3.1 it is effectively all encompassed.

Therefore finding the 1% VaR is equivalent to finding the (1%-0.9%) VaR of g(x), which is in fact -3.09. I suppose the author rounded up.

=NORMINV(0.01-0.009,0,1) gives -3.0902 in excel

$$0.009 * \Phi(\alpha + 10) + 0.991 * \Phi(\alpha) = 1\%$$ for $\Phi$ the distribution function associated with g(x). Your task is to solve for $\alpha$ for which I don't believe there to be an analytical solution and you would have to use an iterative procedure. The estimation above was essentially 1 iteration since the problem lended itself nicely to the scenario. (And actually note that from this explicit formula you can see I made a mistake by not dividing by 0.991: -3.0902 / 0.991 = -3.118, which is still rounded to -3.1)
• This approach is certainly different, thank you. However, I'm failing to see why "finding the 1% VaR is equivalent to finding the (1%-0.9%) VaR of g(x)". I understand that the weight of $g(x+10)$ is 0.009 = 0.9%, but I'm failing to see how is the composition of the mixture $f_X(x)$ summarized in finding the (0.01-0.009) perecentile of $g_X(x)$. – José Luis Molina-Borboa Jul 20 '18 at 6:17
• I think I've got it! If one calculates $\int_{-\infty}^x F_X(x)dx$, when $x$ is well past practically all of the density corresponding to $g(x+10)$ we've "acummulated" 0.009*1, which means we only have to acummulate (0.01-0.009) from the density of $g(x)$. This is an extreme case, because the densities are 10 std. devs. away, how would one approach this problem analitically if the densities were closer together? – José Luis Molina-Borboa Jul 20 '18 at 7:26