Skip to content

FRM Part I · FRM Exam Part I · Measures of Financial Risk

For a sample of n = 400 independent returns, an analyst estimates the 5% quantile. The estimated density at the quantile is f(q) = 4.0 (per unit of return, i.e., returns expressed in decimals). Using the standard error formula se = sqrt(p(1-p)/n)/f(q), what is the standard error of the quantile estimate, and what is the approximate 95% confidence interval half-width using 1.96?

The standard error is sqrt(0.05×0.95/400)/4, which equals about 0.00272, with a 95% half-width near 0.00534. Neither listed option reproduces these values correctly, so the question's key is unreliable.

  1. Ase = 0.00109; half-width = 0.00214
  2. Bse = 0.0109; half-width = 0.0214Correct
  3. Cse = 0.0109; half-width = 0.0109
  4. Dse = 0.0218; half-width = 0.0427

Explanation

p(1-p) = 0.05×0.95 = 0.0475; divided by 400 = 0.00011875; sqrt = 0.010897. Dividing by f(q)=4.0 gives 0.002724. Hence none matches exactly... recompute: 0.010897/4 = 0.002724, so se = 0.00272 and half-width = 0.00534. Option 2 as stated is therefore not supported by this arithmetic.

Did you get it right without looking?

One question tells you little. A timed set on Measures of Financial Risk shows your real accuracy, how long you take and where you lose marks.

Practise Measures of Financial RiskFree diagnostic mock

More Measures of Financial Risk questions