Judge how reliable risk measure estimators are by working out their standard errors.

FRM® Part II · Market Risk Measurement and Management · Estimating Market Risk Measures: An Introduction and Overview · FRM2.MR.1.e

The idea

A risk measure estimated from a sample is itself a random variable, so we judge its reliability with a standard error. For a quantile at tail probability p, estimated from n independent observations, first work out the term V, where V = p × (1 − p) ÷ n. The standard error of the quantile estimate, written SE, divides the square root of V by the density at the quantile, so SE = √V ÷ f(q). Here f(q) is the probability density of the distribution at the quantile. A rough 95 percent confidence interval is the estimate plus or minus 1.96 standard errors. The error shrinks with the square root of n, so quadrupling the sample halves it, and cutting it to a third needs nine times the data. It grows as the quantile moves further into the tail, because fewer observations lie there and the density is lower. Common errors are forgetting the square root, omitting the density, and ignoring the volatility scale. Exam point: tail estimates such as 99 percent VaR are far less precise than central ones.

Formulas

Variance term for the standard error of a quantile

V = p × (1 − p) ÷ n

p: the tail probability
n: the number of independent observations

The first step only. It must be square-rooted and divided by the density at the quantile; stopping here understates the standard error badly.

Standard error of a quantile estimate

SE = √V ÷ f(q)

V: p × (1 − p) ÷ n
f(q): the probability density at the quantile, in the same units as the estimate

Independent observations. The error falls with √n, so quadrupling the sample halves it; it grows deeper in the tail where the density is lower.

A practice question

From the question bank, with the reason each wrong option is wrong.

With the same sample size and a normally distributed P&L, which estimate of VaR has the larger standard error?

The 95 percent VaR, because more observations lie beyond the quantile and so add dispersion.Why not: Reverses the effect: a larger tail count makes the quantile estimate more precise, not less.
They are equal, because the standard error of a quantile depends only on the sample size.Why not: Ignores the dependence on the tail probability and on the density at the quantile.
Neither can be compared, because the standard error depends on the sample mean of P&L.Why not: The quantile standard error depends on the density and tail probability, not on the sample mean alone.
The 99 percent VaR, because fewer observations lie in the tail and the density there is lower. - correct

The standard error is sqrt(p(1-p)/n) divided by the density at the quantile. Moving deeper into the tail lowers the density faster than it shrinks p(1-p), so extreme quantiles are estimated less precisely.

A harder one

Daily P&L is assumed to be normal with mean zero and standard deviation USD 2 million. A 99 percent VaR is estimated from 500 independent observations. The standard error of the VaR estimate is closest to USD (million):

0.167Why not: Computes the standard error for a standard normal and does not multiply by sigma.
0.334 - correct
0.083Why not: Divides by sigma instead of multiplying, so the density scale is applied the wrong way round.
0.668Why not: Multiplies by the variance (4) instead of the standard deviation (2).

se = sqrt(p(1-p)/n) / f(q). In standard units: sqrt(0.01 x 0.99/500) = 0.00445, divided by phi(2.326) = 0.02665, gives 0.1670. Times sigma of 2: 0.334.

The hardest tier

Daily P&L is normally distributed. A risk manager compares the standard error of a 95 percent VaR estimated from 500 observations with that of a 99 percent VaR estimated from 2,000 observations. The ratio of the 99 percent standard error to the 95 percent standard error is closest to:

1.77Why not: Compares the two quantiles at the same sample size and ignores that the 99 percent estimate uses four times the data.
0.44Why not: Divides by 4 rather than by the square root of 4 when adjusting for the larger sample.
0.23Why not: Includes the tail probability term sqrt(p(1-p)) but leaves out the division by the density at each quantile.
0.88 - correct

se is proportional to sqrt(p(1-p))/phi(z). At 99 percent: 0.0995/0.0267 = 3.733; at 95 percent: 0.2179/0.1031 = 2.113. Same-n ratio 1.767; four times the data halves it, giving 0.88.

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