Fin 500Q – Quantitative Risk Management Homework #4 Solutions
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Fin 500Q – Quantitative Risk Management
Homework #4 Solutions
1. The probability density function for an exponential distribution is λe_λx where x is the value of the variable and λ is a parameter. The cumulative probability distribution is 1 _ e_λx . Suppose that two variables V1 and V2 have exponential distributions with λ parameters of 1.0 and 2.0, respectively. Use a Gaussian copula to define the correlation structure between V1 and V2 . You can use the file “bivar.xls” to compute values of the cumulative bivariate normal distribution function.
(a) What is the probability that V1 s 1?
Answer: The probability is 1 _ e_ 1 .0 . 1 = 0.632.
(b) What is the probability that V2 s 1?
Answer: The probability is 1 _ e_2 .0 . 1 = 0.865.
(c) With a copula correlation of 0, what is the probability that V1 s 1 and V2 s 1?
Answer: The probability that V1 s 1 is transformed to the normal value U1 = Φ _ 1 (0.632) = 0.337. This probability can be calculated in Excel with the formula =NORM.INV(0.632, 0, 1). Similarly, the probability that V2 s 1 is transformed to the normal value U2 = Φ _ 1 (0.865) = 1.102. With a copula correlation of 0, we can use the provided Excel file to find that the joint probability is M (0.337, 1.102, 0) = 0.547 (= 0.632 x 0.865).
(d) With a copula correlation of 0.5, what is the probability that V1 s 1 and V2 s 1?
Answer: With a copula correlation of 0.5, the joint probability is M (0.337, 1.102, 0.5) = 0.591. (e) With a copula correlation of _0.2, what is the probability that V1 s 1 and V2 s 1?
Answer: With a copula correlation of _0.2, the joint probability is M (0.337, 1.102, _0.2) = 0.531.
2. Suppose that a bank has made a large number of loans of a certain type. The one-year probability of default on each loan is 1.2%. The bank uses a Gaussian copula for time to default. It is interested in estimating a 99.97% worst case for the percent of loans that default on the portfolio. Show how this worst case percentage varies with the copula correlation, using copula correlations of 0 .2, 0.4, 0.6, and 0.8.
Answer: The WCDR with a 99.97% confidence level is
Φ / 、 ,
where Φ is the CDF of a standard normal distribution. We compute that Φ _ 1 (0.012) = _2.257 (Excel: =NORM.INV(0.012, 0, 1)) and Φ _ 1 (0.9997) = 3.432. For ρ = 0.2, we find
WCDR = Φ ╱ \ = Φ(_0.807) = 0.210,
which is computed in Excel as =NORM.DIST(_0.807, 0, 1, 1).
The table below gives the WCDR for different values of the copula correlation.
ρ |
WCDR(%) |
0.2 |
21.0 |
0.4 |
45.5 |
0.6 |
73.7 |
0.8 |
96.5 |
2022-02-17