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May 18

Solution to Question 2 from UoL exam 2016, zone A

Solution to Question 2 from UoL exam 2016, Zone A (FN3142)

This is another difficult question, and I don't think it will appear again in its entirety. However, the ideas applied here and here are worth repeating and will surely be on future UoL exams in some form.

Problem statement

You hold two different corporate bonds (bond A and bond B), each with a face value of $1 mln. The issuing firms have a 2% probability of defaulting on the bonds, and both the default events and the recovery values are independent of each other. Without default, the notional value is repaid, while in case of default, the recovery value is uniformly distributed between 0 and the notional value.
(a1) (20 points) Find the 1% VaR for bond A or bond B and report your calculations. (HINT: you will need the definition and properties of a uniform distribution U[a,b] on [a,b].)
(a2) (60 points) Taking into account the formula for the distribution function of the sum of two independent uniformly distributed random variables on [a,b] explain how you would find the 1% VaR for a portfolio combining the two bonds (A + B). Report your calculations without finding the actual value.
(b) (20 points) The 1% expected shortfall is defi ned as the expected loss given that the loss exceeds the 1% VaR. Find the 1% expected shortfall for bond A and report your calculations.

I made the statement shorter by referencing my post on the uniform distribution. The formulas for the triangular distribution given in the exam and examiner's comments are both wrong.

Solution

As with previous solutions related to VaR, I emphasize general methods that reduce the amount of guessing to a minimum. Besides, I work with the definition that gives a negative VaR in percents (that is, I work with returns) and then translate that value to positive dollar amounts. Translating the problem statement to returns: without default, the return on each bond is zero, while in case of default the loss is uniformly distributed between -1 and 0 (-1 corresponds to -100%). The recovery value is the notional value minus the dollar loss.

The method consists of two steps. Step 1. Derive the distribution function. Step 2. Use the usual or generalized inverse to find the VaR.

(a1) R_A denotes the return on A, D_A is the event that A defaults and \bar{D}_A is the complement event of no default. Similar notation is employed for B. By the law of total probability for any real x

(1) F_{R_A}(x)=P(R_A\le x)=P(R_A\le x|D_A)P(D_A)+P(R_A\le x|\bar{D}_A)P(\bar{D}_A).

We want to show that for our purposes the second term can be made zero. We know that without default, return is zero. This means that conditionally we deal with a random variable that takes value 0 with probability 1. Hence, conditional on no default, by additivity of conditional probability for x\ge 0

(2) P(R_A\le x|\bar{D}_A)=P(R_A<0|\bar{D}_A)+P(R_A=0|\bar{D}_A)+P(0<R_A\le x|\bar{D}_A)=1

because the middle term on the right is 1 and the other two are zero. It follows that for x\ge 0 the second term in (1) is

(3) P(R_A\le x|\bar{D}_A)P(\bar{D}_A)=1\times 0.98>0.01.

To find the VaR, we need the whole expression in (1) to be 0.01, so we have to consider only x<0, in which case the second term on the right of (1) is zero.

In the first term on the right of (1) P(R_A\le x|D_A) is uniformly distributed on [-1,0], so from its distribution function it's clear that we need to consider only -1<x<0:

F_{R_A}(x)=P(R_A\le x|D_A)P(D_A)=(x+1)0.02.

As this should be equal to 0.01, we obtain x=-0.5. The loss is $1 mln times 50% and the recovery value is half a million.

(a2) Denote R=R_A+R_B. This allows us to use the formula for the convolution of two independent uniform distributions. We should remember that this is not a return on the total portfolio because it takes values between -2 and 0. By the  law of total probability for any real x

(4) F_R(x)=P(R\le x|D_A\cap D_B)P(D_A\cap D_B)+P(R\le x|D_A\cap \bar{D}_B)P(D_A\cap \bar{D}_B)

+P(R\le x|\bar{D}_A\cap D_B)P(\bar{D}_A\cap D_B)+P(R\le x|\bar{D}_A\cap \bar{D}_B)P(\bar{D}_A\cap \bar{D}_B).

As above, we can show that the last term in (4) is large if x\ge 0. Indeed, if no company defaults, the return is zero and, as in (2), for such x

P(R\le x|\bar{D}_A\cap \bar{D}_B)=1.

Therefore, as in (3)

P(R\le x|\bar{D}_A\cap \bar{D}_B)P(\bar{D}_A\cap \bar{D}_B)=1\times 0.98^2=0.9604>0.01.

It follows that we can restrict our attention to x<0, in which case the last term on the right of (4) is zero, the second and third terms are the same and we get

(5) F_R(x)=P(R\le x|D_A\cap D_B)0.02^2+2P(R\le x|D_A\cap \bar{D}_B)0.02\cdot 0.98.

In case of two defaults, R_A,\ R_B are independent and distributed as U[-1,0]. Hence, R has a triangular distribution. Denoting its density f_T, we have

(6) P(R\le x|D_A\cap D_B)0.02^2=0.0004\int_{-2}^xf_T(x)dx,\ -2<x<0.

The integral here is

(7) \int_{-2}^xf_T(x)dx=\left\{\begin{array}{ll}x^2/2+2x+2, &-2<x<-1\\1-x^2/2, &-1\le x<0\end{array}\right.

In case of one default, we use the uniform distribution to find

(8) 2P(R\le x|D_A\cap \bar{D}_B)0.02\cdot 0.98=\left\{\begin{array}{ll}0.0392(x+1),&-1\le x<0\\0,&x<-1\end{array}\right..

Plugging (6), (7), (8) in (5) and equating the result to 0.01, we get

(9) 0.0004(x^2/2+2x+2)=0.01, -2<x<-1

and

(10) 0.0004(1-x^2/2)+0.0392(x+1)=0.01, -1\le x<0.

Mathematica gives two solutions for (9), none of which is in the interval (-2,-1). Of the two solutions of (10), the one which belongs to (-1,0) is -0.75. This is 37.5% of -2 and the loss is $750,000 out of $2 mln. The exam question doesn't require you to give numerical values.

(b) From the general definition of expected shortfall

ES^{0.01}=\frac{ER_A1_{R_A\le -0.5}}{P(R_A\le -0.5)}=\frac{\int_{-1}^{-0.5}tdt}{\int_{-1}^{-0.5}dt}=\frac{-3/8}{1/2}=-0.75

which means an expected loss of $750,000.

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