MANG 6020 Financial Risk Management
Seminar 6
Question 1
Suppose that a one-day 97.5% VaR is estimated as $13 million from 2,000 observations. The
one-day changes are approximately normal with mean zero and standard deviation $6 million.
Estimate a 99% confidence interval for the VaR estimate.
Solution:
The standard error is
where f(q) is an estimate of the loss probability density at the VaR point. In this case the 0.975
point on the approximating normal distribution is NORMINV(0.975,0,6) = 11.76. f(q) is
estimated as NORMDIST(11.76,0,6,FALSE) = 0.0097. The standard error is therefore
A 99% confidence interval for the VaR is 13 − 2.576 × 0.358 to 13 + 2.576 × 0.358, or 12.077 to
13.923.
Question 2
Based on a 90% confidence level, how many exceptions in backtesting a VAR would be
expected over a 250-day trading year?
Solution:
This is p × T = 10% × 250 = 25.
Question 3
A large, international bank has a trading book whose size depends on the opportunities perceived
by its traders. The market risk manager estimates the one-day VAR, at the 95% confidence level,
to be USD 50 million. You are asked to evaluate how good a job the manager is doing in
estimating the one-day VAR. Which of the following would be the most convincing evidence
that the manager is doing a poor job, assuming that losses are identical and independently
distributed (i.i.d.)?
a. Over the past 250 days, there are eight exceptions.
b. Over the past 250 days, the largest loss is USD 500 million.
c. Over the past 250 days, the mean loss is USD 60 million.
d. Over the past 250 days, there is no exception.
Seminar 6
Question 1
Suppose that a one-day 97.5% VaR is estimated as $13 million from 2,000 observations. The
one-day changes are approximately normal with mean zero and standard deviation $6 million.
Estimate a 99% confidence interval for the VaR estimate.
Solution:
The standard error is
where f(q) is an estimate of the loss probability density at the VaR point. In this case the 0.975
point on the approximating normal distribution is NORMINV(0.975,0,6) = 11.76. f(q) is
estimated as NORMDIST(11.76,0,6,FALSE) = 0.0097. The standard error is therefore
A 99% confidence interval for the VaR is 13 − 2.576 × 0.358 to 13 + 2.576 × 0.358, or 12.077 to
13.923.
Question 2
Based on a 90% confidence level, how many exceptions in backtesting a VAR would be
expected over a 250-day trading year?
Solution:
This is p × T = 10% × 250 = 25.
Question 3
A large, international bank has a trading book whose size depends on the opportunities perceived
by its traders. The market risk manager estimates the one-day VAR, at the 95% confidence level,
to be USD 50 million. You are asked to evaluate how good a job the manager is doing in
estimating the one-day VAR. Which of the following would be the most convincing evidence
that the manager is doing a poor job, assuming that losses are identical and independently
distributed (i.i.d.)?
a. Over the past 250 days, there are eight exceptions.
b. Over the past 250 days, the largest loss is USD 500 million.
c. Over the past 250 days, the mean loss is USD 60 million.
d. Over the past 250 days, there is no exception.
