24
Apr 19

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




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.

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