David,
Q 20 in 2009 Annoted Practice exam has some previous discussion in
http://forum.bionicturtle.com/viewthread/1401/ which is not accessible.
Can you please explain the following if possible.
Page 28 of 67 Full Exam II Sample Questions FRM 2009
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The loss from model risk at the 95th percentile corresponds to the 75th percentile for the normal distribution with mean 25 and standard deviation 5, so unexpected loss for model risk at 95% confidence level: < (25MM + 1.645*5MM) - 5MM < 30MM. Reference: Dowd, Chapter 16.
20b [my adds]. What exactly is the maximum expected loss from model risk at the 95th percentile? As the 5% loss tail for the entire distribution corresponds to the 25% of the normal distribution (i.e., 25% of the normal “tail” * 20% of the overall “parent” = 5%), we need the normal inverse at 75%: =NORMSINV(75%) = 0.675 0.675 * $5 MM + $25 MM mean = $28.372 So that max EL = $28.372 - $5 = $23.372 UL
20c. What is the 95% conditional Value at Risk (CVaR; a.k.a., expected shortfall, or expected tail risk) for the rogue trader? ES is the average of the loss *conditional* on the loss exceeding the VaR. In this case, this is the expected loss conditional on the loss exceeeding the $50 MM. As the distribution is uniform: ES [loss | loss > $50 MM] = average [uniform from +$0 to +$50 MM] + $50 VaR = $25 + $50 = $75 MM
20d. In regard to the rogue trader, the question assumes independence between loss severity and loss frequency. Assume instead the correlation between frequency and severity is 0.5. What is the expected loss (EL) given this positive correlation (hint: you need the variance of the a Bernouilla and a uniform distribution). StdDev of frequency (Bernoulli) = SQRT[90%*10%] = about 30% StdDev of severity (uniform) = SQRT[(1.0 - 0)^2/12] = about 30% Covariance = 30%*30%*0.5 = 0.045. You must know this! or could get covariance in dollar terms = 30% * $30 MM * 0.5 = $4.5 million
Thanks.
Turtle2
Q 20 in 2009 Annoted Practice exam has some previous discussion in
http://forum.bionicturtle.com/viewthread/1401/ which is not accessible.
Can you please explain the following if possible.
Page 28 of 67 Full Exam II Sample Questions FRM 2009
=====================================
The loss from model risk at the 95th percentile corresponds to the 75th percentile for the normal distribution with mean 25 and standard deviation 5, so unexpected loss for model risk at 95% confidence level: < (25MM + 1.645*5MM) - 5MM < 30MM. Reference: Dowd, Chapter 16.
20b [my adds]. What exactly is the maximum expected loss from model risk at the 95th percentile? As the 5% loss tail for the entire distribution corresponds to the 25% of the normal distribution (i.e., 25% of the normal “tail” * 20% of the overall “parent” = 5%), we need the normal inverse at 75%: =NORMSINV(75%) = 0.675 0.675 * $5 MM + $25 MM mean = $28.372 So that max EL = $28.372 - $5 = $23.372 UL
20c. What is the 95% conditional Value at Risk (CVaR; a.k.a., expected shortfall, or expected tail risk) for the rogue trader? ES is the average of the loss *conditional* on the loss exceeding the VaR. In this case, this is the expected loss conditional on the loss exceeeding the $50 MM. As the distribution is uniform: ES [loss | loss > $50 MM] = average [uniform from +$0 to +$50 MM] + $50 VaR = $25 + $50 = $75 MM
20d. In regard to the rogue trader, the question assumes independence between loss severity and loss frequency. Assume instead the correlation between frequency and severity is 0.5. What is the expected loss (EL) given this positive correlation (hint: you need the variance of the a Bernouilla and a uniform distribution). StdDev of frequency (Bernoulli) = SQRT[90%*10%] = about 30% StdDev of severity (uniform) = SQRT[(1.0 - 0)^2/12] = about 30% Covariance = 30%*30%*0.5 = 0.045. You must know this! or could get covariance in dollar terms = 30% * $30 MM * 0.5 = $4.5 million
Thanks.
Turtle2
