Skip to content

FRM Part I · FRM Exam Part I · Random Variables

Let X be lognormally distributed so that ln X is normal with mean 0.04 and standard deviation 0.20. Which expression gives E[X], and what is its approximate value? (e^0.06 = 1.0618; e^0.04 = 1.0408; e^0.08 = 1.0833)

The mean of a lognormal variable is exp(mu + sigma squared / 2). With mu = 0.04 and sigma squared = 0.04, this is exp(0.06), about 1.0618. The value exp(0.04) is only the median, which is lower because of positive skewness.

  1. Aexp(0.04 + 0.5 x 0.04) = exp(0.06), about 1.0618Correct
  2. Bexp(0.04) , about 1.0408
  3. Cexp(0.04 + 0.04) = exp(0.08), about 1.0833
  4. Dexp(0.04 - 0.5 x 0.04) = exp(0.02), about 1.0202

Explanation

For a lognormal, E[X] = exp(mu + sigma^2/2). Sigma^2 = 0.04, so half is 0.02, giving exp(0.06) = 1.0618. exp(0.04) is the median, not the mean. exp(0.08) adds the full variance, and exp(0.02) subtracts the half variance.

Did you get it right without looking?

One question tells you little. A timed set on Random Variables shows your real accuracy, how long you take and where you lose marks.

Practise Random VariablesFree diagnostic mock

More Random Variables questions