LO 1.6: Evaluate estimators o f risk measures by estimating their standard errors.

LO 1.6: Evaluate estimators o f risk measures by estimating their standard errors.
Sound risk management practice reminds us that estimators are only as useful as their precision. That is, estimators that are less precise (i.e., have large standard errors and wide confidence intervals) will have limited practical value. Therefore, it is best practice to also compute the standard error for all coherent risk measures.
Professors Note: The process o f estimating standard errors for estimators o f coherent risk measures is quite complex, so your focus should be on interpretation o f this concept.
First, lets start with a sample size of n and arbitrary bin width of h around quantile, q. Bin width is just the width of the intervals, sometimes called bins, in a histogram. Computing standard error is done by realizing that the square root of the variance of the quantile is equal to the standard error of the quantile. After finding the standard error, a confidence interval for a risk measure such as VaR can be constructed as follows:
[q + se(q)X za ] > VaR > [q se(q)X za ] 2018 Kaplan, Inc.
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Topic 1 Cross Reference to GARP Assigned Reading – Dowd, Chapter 3
Example: Estimating standard errors
Construct a 90% confidence interval for 3% VaR (the 93% quantile) drawn from a standard normal distribution. Assume bin width = 0.1 and that the sample size is equal to 500.
Answer:
The quantile value, q, corresponds to the 5% VaR which occurs at 1.65 for the standard normal distribution. The confidence interval takes the following form:
[1.65 + 1.65 x se(q)] > VaR > [1.65 1.65 x se(q)] Professors Note: Recall that a confidence interval is a two-tailed test (unlike VaR), so a 90% confidence level will have 5% in each tail. Given that this is equivalent to the 5% significance level o f VaR, the critical values o f 1.65 will he the same in both cases.
Since bin width is 0.1, q is in the range 1.65 0.1/2 = [1.7, 1.6]. Note that the left tail probability, p , is the area to the left o f1.7 for a standard normal distribution.
Next, calculate the probability mass between [1.7, 1.6], represented 2&f(q). From the standard normal table, the probability of a loss greater than 1.7 is 0.045 (left tail). Similarly, the probability of a loss less than 1.6 (right tail) is 0.945. Collectively, f(q) = 1 -0 .0 4 5 -0 .9 4 5 = 0.01
The standard error of the quantile is derived from the variance approximation of q and is equal to:
yp(l ~ p ) / n
f(q)
Now we are ready to substitute in the variance approximation to calculate the confidence interval for VaR:
1.65 + 1.65V 4 –
0.01
= 3.18 > VaR > 0 .1 2
‘ 500 > V aR >
, 1.65 1.65
, , c V0.045(l 0.045) / 500
0.01
Lets return to the variance approximation and perform some basic comparative statistics. What happens if we increase the sample size holding all other factors constant? Intuitively, the larger the sample size the smaller the standard error and the narrower the confidence interval.
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2018 Kaplan, Inc.
Topic 1 Cross Reference to GARP Assigned Reading – Dowd, Chapter 3
Now suppose we increase the bin size, h, holding all else constant. This will increase the probability mass/fg’) and reducep , the probability in the left tail. The standard error will decrease and the confidence interval will again narrow.
Lastly, suppose that p increases indicating that tail probabilities are more likely. Intuitively, the estimator becomes less precise and standard errors increase, which widens the confidence interval. Note that the expression p(l p) will be maximized at p = 0.3.
The above analysis was based on one quantile of the loss distribution. Just as the previous section generalized the expected shortfall to the coherent risk measure, we can do the same for the standard error computation. Thankfully, this complex process is not the focus of the LO.
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