24
Apr 19

## Solution to Question 3b) from UoL exam 2018, Zone A

I thought that after all the work we've done with my students the answer to this question would be obvious. It was not, so I am sharing it.

Question. Consider a position consisting of a $20,000 investment in asset $X$ and a$20,000 investment in asset $Y$. Assume that returns on these two assets are i.i.d. normal with mean zero, that the daily volatilities of both assets are 3%, and that the correlation coefficient between their returns is 0.4. What is the 10-day VaR at the $\alpha =1\%$ critical level for the portfolio?

Solution. First we have to work with returns and then translate the result into dollars.

Let $R_X,$ $R_Y$ be the daily returns on the two assets. We are given that $ER_X=ER_Y=0,$ $\sigma (R_X)=\sigma(R_Y)=0.03$, $\rho(R_X,R_Y)=0.4.$

Since the total investment is $40,000, the shares of the investment are $s_X=s_Y=20,000/40,000=0.5.$ Therefore the daily return on the portfolio is $R=0.5R_X+0.5R_Y,$ see Exercise 2. It follows that $ER=0.5ER_X+0.5ER_Y=0,$ $Var(R)=0.5^2\sigma^2(R_X)+2\cdot 0.5\cdot 0.5\rho (R_X,R_Y)\sigma (R_X)\sigma(R_Y)+0.5^2\sigma^2(R_Y)=0.5^2\cdot 0.03^2(1+0.8+1)=0.015^2\cdot 2.8.$ These figures are for daily returns. We need to make sure that $R$ is normally distributed. The sufficient condition for this is that the returns $R_X,$ $R_Y$ are jointly normally distributed. It is not mentioned in the problem statement, and we have to assume that it is satisfied. Let $R_i$ denote the return on day $i.$ Under continuous compounding the daily returns are summed: if we invest $M_0$ initially, after the first day we have $M_1=M_0e^{R_1},$ after the second day we have $M_2=M_1e^{R_2}=M_0e^{R_1+R_2}$ and so on. So the 10-day return is $r=R_1+...+R_{10}.$ Since the daily returns are independent and identically distributed, by additivity of variance we have $Var(r)=\sum Var(R_i)=10\cdot 0.015^2\cdot 2.8,$ $\sigma (r)=0.015\sqrt{28}=0.079,$ $Er=0.$ $r$ is normally distributed because the daily returns are independent. It remains to apply the VaR formula $VaR^\alpha=\mu+\sigma\Phi^{-1}(\alpha)$. for normal distributions. From the table of the distribution function of the standard normal $\Phi ^{-1}(0.01)=-2.33.$ Thus, $VaR^{\alpha }=0.079\cdot (-2.33)=-0.184.$ This translates to the minimum loss of $0.184\times 40,000=7362.$ Thus, with probability 1% the loss can be$7362 or more.