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

A 99% one-day VaR is estimated from 1,000 independent observations of normally distributed P&L with mean zero and standard deviation USD 1 million. Using the quantile standard error formula sqrt(p(1-p)/n)/f(q) with p = 0.01, the normal quantile 2.326 and standard normal density at 2.326 of 0.02665, which approximate 95% confidence interval for the true VaR, centred on the true VaR of USD 2.326 million, results?

The interval is about USD 2.326 million plus or minus 0.23 million. The standard error is sqrt(0.01×0.99/1000) divided by 0.02665, which is about 0.118 million. Multiplying by 1.96 gives a 95% half-width of roughly 0.23 million.

  1. AUSD 2.326m ± 0.07m
  2. BUSD 2.326m ± 0.14mCorrect
  3. CUSD 2.326m ± 0.23m
  4. DUSD 2.326m ± 0.73m

Explanation

sqrt(0.01*0.99/1000)=sqrt(0.0000099)=0.003146. Divide by f=0.02665/1m scaling: SE = 0.003146/0.02665 = 0.1181 million. 95% half-width = 1.96*0.1181 = 0.2314 million. So the half-width is 0.23m, which makes the third option correct, not 0.14m (that uses one SE rounded up incorrectly).

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