22
Apr 18

Solution to Question 1 from UoL exam 2016, Zone A




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.

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