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Explain how a smoothed, non-parametric estimate of the density can underpin a historical simulation.

FRM® Part II · Market Risk Measurement and Management · Non-Parametric Approaches · FRM2.MR.2.b

The idea

Basic historical simulation treats the sample as a set of discrete points. VaR can then only be read at observed losses, and it jumps as single observations enter or leave the window. A smoothed estimate fits a continuous density through the data without assuming a parametric form. One route treats each observation as the midpoint of a bin and interpolates linearly. Let i be the rank of a loss counting from the smallest, and n the number of losses. The i-th smallest loss is then taken to sit at a quantile level that is a little below its plain rank share, given by Quantile = (i − 0.5) ÷ n. Another route places a kernel, often a normal curve with a chosen bandwidth, on every observation and averages them. Quantiles then come from the smoothed cumulative distribution at any confidence level. The bandwidth trades noise against bias. Too small leaves a spiky density and too large blurs the tail. Smoothing spreads the sample's information. It does not add any, so tail estimates still rest on the same few extreme observations.

Formulas

Quantile assigned to an observation under the midpoint convention

quantile = (i − 0.5) ÷ n

i
the rank of the loss, 1 for the smallest
n
the number of losses

Quantiles between neighbours come from linear interpolation. The largest loss sits at (n − 0.5) ÷ n, beyond which there is no information.

A practice question

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

What is the main purpose of fitting a smoothed, non-parametric density to historical returns before reading off VaR?

  1. To replace the historical data with a normal distribution having the same mean and variance as the sampleWhy not: That is the parametric (variance-covariance) approach; a smoothed non-parametric density keeps the shape of the data.
  2. To supply information on losses beyond the worst observation by fitting an extreme value tail modelWhy not: Smoothing only redistributes the sample's own information; extending the tail with a fitted model is extreme value theory, a different technique.
  3. To obtain a continuous density from which quantiles can be read at any confidence level, without assuming a parametric distribution - correct
  4. To give recent observations a larger weight than older onesWhy not: That is age-weighting; smoothing concerns the shape of the density between observations, not their weights.

Smoothing, whether by interpolating bin midpoints or by kernel estimation, turns discrete observations into a continuous density. VaR and ES can then be read at any confidence level without assuming normality or any other form.

A harder one

A risk manager has 20 daily losses and smooths the empirical distribution by treating the i-th smallest loss as the (i - 0.5)/20 quantile, interpolating linearly between neighbours. The largest loss is USD 4.6 million and the second largest is USD 3.1 million. The 97% VaR is closest to:

  1. USD 4.60 millionWhy not: Takes the largest loss as VaR, the unsmoothed historical simulation reading for a tail of only 0.6 observations.
  2. USD 3.85 millionWhy not: Takes the simple midpoint of the two losses instead of interpolating at the 97% position.
  3. USD 3.25 millionWhy not: Interpolates in the wrong direction, giving the larger loss a weight of 0.1 instead of 0.9.
  4. USD 4.45 million - correct

The 19th smallest loss is the 0.925 quantile and the 20th the 0.975 quantile. 97% is (0.97 - 0.925)/0.05 = 0.9 of the way between them: 3.1 + 0.9 x (4.6 - 3.1) = 4.45.

The hardest tier

A risk manager smooths a small sample of four losses (USD million), 2, 3, 4 and 9, by placing a normal kernel with standard deviation (bandwidth) 2 on each observation and averaging the four densities. Under the smoothed density, the probability that a loss exceeds USD 6 million is closest to:

  1. 29.5% - correct
  2. 25.0%Why not: Counts only the raw observation above 6 (1 of 4), ignoring that the kernels spread probability across the threshold.
  3. 27.0%Why not: Treats the bandwidth of 2 as a variance, so the kernel standard deviation is 1.41 instead of 2.
  4. 70.5%Why not: Computes the probability that the loss is below 6 (using the cumulative probability) rather than above it.

P(loss > 6) = average of 1 - Phi((6 - x)/2): for x = 2, 3, 4, 9 this is 0.0228, 0.0668, 0.1587, 0.9332, which averages to 0.2954. Smoothing moves mass across the threshold, so the answer differs from the raw 25%.

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