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Apply the Central Limit Theorem (CLT) and the Law of Large Numbers (LLN) to a sample mean, saying what each result implies.

FRM® Part I · Quantitative Analysis · Sample Moments · FRM1.QA.5.g

The idea

The Law of Large Numbers says that for independent identically distributed observations with a finite mean, the sample mean converges to the population mean as n grows. It identifies where the estimator ends up. The Central Limit Theorem needs a finite variance. It says that the standardised sample mean is approximately standard normal for large n, whatever the shape of the data. The standardised mean divides the gap from μ by the standard error. z = (x̄ − μ) ÷ (σ ÷ √n). So the standard error of the mean is SE = σ ÷ √n. Quadrupling the sample size halves it. Confidence intervals and tail probabilities for the average come from the CLT, not the LLN. A usual error is to claim the LLN makes averages normal, or to forget that CLT results are approximate and require finite variance.

Formulas

Standard error of the sample mean

SE = σ ÷ √n

σ
the population standard deviation
n
the sample size

Quadrupling the sample halves the standard error. The normal shape of the sample mean comes from the CLT and needs a finite variance.

Standardised sample mean under the CLT

z = (x̄ − μ) ÷ (σ ÷ √n)

x̄
the sample mean
μ
the population mean
σ
the population standard deviation
n
the sample size

Approximately standard normal for large n, whatever the shape of the data, provided the variance is finite. It is the CLT, not the LLN, that gives the normal shape.

A practice question

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

The Law of Large Numbers, applied to a sample mean of independent identically distributed observations with a finite mean, implies that:

  1. the sample mean becomes normally distributed as the sample size grows.Why not: This is the content of the Central Limit Theorem, not the LLN.
  2. the sample mean equals the population mean once the sample is large enough.Why not: The LLN gives convergence in the limit, not exact equality at any finite n.
  3. the sample mean converges to the population mean as the sample size grows. - correct
  4. the standard deviation of the individual observations falls to zero.Why not: The LLN concerns the sample mean; individual observations keep their variance.

LLN: x-bar tends to mu in probability as n grows. It tells us where the sample mean ends up, not how the estimation error is distributed; that is the CLT's job.

A harder one

Daily returns are independent with mean 0.05% and standard deviation 1.5%. Using the CLT, what is the approximate probability that the average of 225 daily returns is below zero?

  1. 0.31 - correct
  2. 0.49Why not: Standardises with sigma instead of the standard error sigma/sqrt(n).
  3. 0.69Why not: Computes the probability that the average is above zero, the wrong tail.
  4. Less than 0.01Why not: Divides sigma by n instead of sqrt(n), giving a far-too-small standard error.

Standard error = 1.5/sqrt(225) = 0.10%. z = (0 - 0.05)/0.10 = -0.50. P(Z < -0.50) = 0.3085, about 0.31.

The hardest tier

Daily returns have a standard deviation of 2.0%. Using the CLT, how many observations are needed for a 95% confidence interval for the mean with a half-width (margin of error) of 0.25% (take z = 1.96)?

  1. 64Why not: Omits the 1.96 factor, which sets the confidence level.
  2. 16Why not: Forgets to square: n = z sigma / E instead of (z sigma / E)².
  3. 246 - correct
  4. 984Why not: Treats 0.25% as the full width of the interval, halving the half-width.

The margin is z x sigma/sqrt(n) = 0.25, so n = (z sigma / E)² = (1.96 x 2.0 / 0.25)² = 245.9, which rounds up to 246.

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