22
Apr 18

## Solution to Question 1 from UoL exam 2016, Zone A  (FN3142)

Frankly, this is a crazy exercise. I deserve full 100 marks for this solution but I wouldn't be able to solve this during an exam. In the problem statement, I took the liberty to change some terminology and notation. In particular, I use the definitions of the Value at Risk and expected shortfall that give negative values. You can go ahead and redo everything with definitions that give positive values.

## Problem statement

Assume daily returns that are normally distributed with constant mean (equal to zero) and variance, i.e., they are given by

$R_{t+1}=\sigma_{t+1} z_{t+1},$ where $z_{t+1}\vert\mathcal{F}_t\overset{i.i.d.}{\sim}N(0,1),$

where the time increment $t+1$ is 1-day.

(a) [25 marks] Derive the following formula for the Value-at-Risk at the $\alpha$% critical level and 1-day horizon

(1) $VaR^\alpha_{t+1}=\sigma_{t+1}\Phi^{-1}(\alpha)$

where $\Phi$ is the standard normal cumulative density function.

(b) [25 marks] The expected shortfall $ES^\alpha_{t+1}$ at the critical level $\alpha$% and 1-day horizon can be defined as

$ES^\alpha_{t+1}=E_t(R_{t+1}|R_{t+1}\le VaR^\alpha_{t+1})$.

Using the VaR formula from part (a) derive the following formula for the 1-day expected shortfall

(2) $ES^\alpha_{t+1}=-\frac{\sigma_{t+1}}{\alpha}p_z(\Phi^{-1}(\alpha))$

where $p_z$ is the standard normal probability density function.

(c) [50 marks] Prove that the relative difference between the 1-day expected shortfall and 1-day Value-at-Risk, as a proportion of the 1-day Value-at-Risk, converges to zero when $\alpha$ goes to zero, i.e., show that

(3) $\lim_{\alpha\rightarrow 0}\frac{ES^\alpha_{t+1}-VaR^\alpha_{t+1}}{VaR^\alpha_{t+1}}=0.$

## Solution

(a) The answer is contained in this post.

(b) This part has been solved here.

(c) Plug (1) and (2) in (3):

$\lambda\equiv\frac{ES^\alpha_{t+1}-VaR^\alpha_{t+1}}{VaR^\alpha_{t+1}}=\frac{-\frac{\sigma_{t+1}}{\alpha}p_z(\Phi^{-1}(\alpha))-\sigma_{t+1}\Phi^{-1}(\alpha)}{\sigma_{t+1}\Phi^{-1}(\alpha)}$

(crossing out $\sigma_{t+1}$ and multiplying everything by $\alpha$)

$=\frac{-p_z(\Phi^{-1}(\alpha))-\alpha\Phi^{-1}(\alpha)}{\alpha\Phi^{-1}(\alpha)}.$

At this point it helps to replace $v=\Phi^{-1}(\alpha)$ and note that $\alpha\rightarrow 0$ is equivalent to $v\rightarrow -\infty$. Then we get

$\lambda=\frac{-p_z(v)-\Phi(v)v}{\Phi(v)v}.$

This is indeterminacy of type $0/0$. In such cases people use the L'Hôpital's Rule. The above expression has the same limit as

$\mu=\frac{(-p_z(v)-\Phi(v)v)'}{(\Phi(v)v)'}=\frac{-p'_z(v)-p_z(v)v-\Phi(v)}{p_z(v)v+\Phi(v)}$ (we know that $p'_z(v)=-p_z(v)v$)

$=-\frac{\Phi(v)}{p_z(v)v+\Phi(v)}.$

This is again indeterminacy of type $0/0$ and by the L'Hôpital's Rule this expression has the same limit as

$\nu=-\frac{\Phi'(v)}{(p_z(v)v+\Phi(v))'}=-\frac{p_z(v)}{p'_z(v)v+p_z(v)+p_z(v)}$ (replacing the derivative)

$=-\frac{p_z(v)}{-v^2p_z(v)+2p_z(v)}=-\frac{1}{-v^2+1}\rightarrow 0.$