Free PRMIA 8008 Practice Exam with Questions & Answers | Set: 5

Loss provisioning is intended to cover expected losses. Economic capital is expected to cover unexpected losses. No capital or provisions are set aside for losses in excess of unexpected losses, which will ultimately be borne by equity.

Choice 'd' is the correct answer.

When we do a Monte Carlo simulation, the statistic we obtain (eg, the expected price) is an estimate of the real variable. The difference between the real value (which would be what we would get if we had access to the entire population) and that estimated by the Monte Carlo simulation is measured by the 'standard error', which is the standard deviation of the difference between the 'real' value and the simulated value (ie, the 'error').

As we increase the number of draws in a Monte Carlo simulation, the closer our estimate will be to the true value of the variable we are trying to estimate. But increasing the sample size does not reduce the error in a linear way, ie doubling the sample size does not halve the error, but reduces it by the inverse of the square root of the sample size. So if we have a sample size of 1000, going up to a sample size of 100,000 will reduce the standard error by a factor of 10 (and not 100), ie, SQRT(1/100) = 1/10. In other words, standard error is proportional to 1/√N, where N is the sample size.

Therefore Choice 'c' is correct and the others are incorrect.

If annual variance is 0.0256, then annual volatility (ie standard deviation) is √0.0256. Therefore the daily volatility will be √0.0256/√250 = 1.01%. The other choices are not correct.

The recovery rate, the default hazard rate (also called the average default intensity) and the spread on debt are linked by the equation Hazard Rate = Spread/(1 - Recovery Rate). Therefore, the recovery rate implicit in the given data is = 1 - 8%/10% = 20%.

The economic capital required to avoid insolvency is just the asset VaR, ie $100. This means that if the worst case losses are realized, the institution would need to have a buffer equivalent to those losses which in this case will be $100, and this buffer is the economic capital.

The actual value of liabilities is not relevant as they are considered 'riskless' from the institution's point of view, ie they will be taken at full value. In this particular case, the institution has $200 in capital which is more than the economic capital required.

Therefore Choice 'c' is the correct answer.

Choice 'a' is the correct answer. Exposures such as those to a line of credit of which only a part (or none) may be drawn at the time of assessment present a difficulty when attempting to quantify credit risk. It is not correct to take the entire amount of the line as the exposure at default, and likewise the current exposure is likely to be too aggressively low a number to consider.

While the borrower has an information advantage in that he would be aware of the deterioration in credit standing before the bank and would probably draw cash prior to default, it is unlikely that the entire amount of the line of credit would be drawn in all cases. In some cases, none may be drawn. In other cases, the bank would become aware of the situation and curtail or cancel access to the credit line in a timely fashion.

Therefore a fixed proportion of existing credit lines is considered a reasonable approximation of the exposure at default against credit lines.

A recap would be in order here: calculating operational risk capital is a multi-step process.

First, we fit curves to estimate the parameters to our chosen distribution types for frequency (eg, Poisson), and severity (eg, lognormal). Note that these curves are fitted at the UoM level - which is the lowest level of granularity at which modeling is carried out. Since there are many UoMs, there are are many frequency and severity distributions. However what we are interested in is the loss distribution for the entire bank from which the 99.9th percentile loss can be calculated. From the multiple frequency and severity distributions we have calculated, this becomes a two step process:

- Step 1: Calculate the aggregate loss distribution for each UoM. Each loss distribution is based upon and underlying frequency and severity distribution.

- Step 2: Combine the multiple loss distributions after considering the dependence between the different UoMs. The 'dependence' recognizes that the various UoMs are not completely independent, ie the loss distributions are not additive, and that there is a sort of diversification benefit in the sense that not all types of losses can occur at once and the joint probabilities of the different losses make the sum less than the sum of the parts.

Step 1 requires simulating a number, say n, of the number of losses that occur in a given year from a frequency distribution. Then n losses are picked from the severity distribution, and the total loss for the year is a summation of these losses. This becomes one data point. This process of simulating the number of losses and then identifying that number of losses is carried out a large number of times to get the aggregate loss distribution for a UoM.

Step 2 requires taking the different loss distributions from Step 1 and combining them considering the dependence between the events. The correlations between the losses are described by a 'copula', and combined together mathematically to get a single loss distribution for the entire bank. This allows the 99.9th percentile loss to be calculated.

The correct answer is Choice 'c'. The other choices do not refer to 'regulatory arbitrage'.

If the confidence levels for a VaR estimate are set too high, there may never be any exceedences, ie actual losses will never exceed VaR estimates. For limit setting, we want actual losses to exceed the VaR estimates enough number of times as during the year so that the limits are considered seriously. If the VaR estimate is exceeded too many times, or never, then it is unlikely to be considered seriously. Therefore Choice 'd' is the correct answer.

The other answers are incorrect as they either require the VaR to be too high (ie zero or rare excess loss situations) or too low (ie there will be too many cases of excess loss situations to be taken seriously).

For a dataset with 250 observations, the top 2% of the losses will be the top 5 observations. Expected shortfall is the average of the losses beyond the VaR threshold. Therefore the correct answer is (20 + 19 + 19 + 17 + 16)/5 = 18.2m .

Note that Expected Shortfall is also called conditional VaR (cVaR), Expected Tail Loss and Tail average.