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Identify what makes a sequence of random variables independent and identically distributed (iid).

FRM® Part I · Quantitative Analysis · Multivariate Random Variables · FRM1.QA.4.i

The idea

A sequence of random variables is independent and identically distributed (iid) when two conditions hold. The variables are mutually independent, so the outcome of one carries no information about the others. And each has exactly the same probability distribution, with the same mean, variance and shape. Rolls of a fair die, or repeated draws with replacement from one population, are iid. Observations with volatility clustering or autocorrelation are not iid, because they are dependent. Draws without replacement from a small population are dependent too. Data whose mean or variance shifts across regimes are not identically distributed. Being uncorrelated is not the same as being independent, because correlation only detects linear dependence, and equal means are not enough for identical distribution. Risk models often assume iid returns for convenience. Recognising when this fails is the exam skill.

A practice question

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

What makes a sequence of random variables X1, X2, ..., Xn independent and identically distributed (iid)?

  1. The variables are uncorrelated with each other and have the same mean.Why not: Uncorrelated and equal means are weaker conditions; independence and a common distribution (including the same variance) are required.
  2. The variables are mutually independent and every one of them has the same probability distribution. - correct
  3. The variables have the same distribution and may be dependent on each other, so long as the pairwise correlations are equal.Why not: Drops independence; identical marginals with common correlations are exchangeable but not independent.
  4. The variables are independent and normally distributed with possibly different means and variances.Why not: Requires normality (not part of the definition) and allows different parameters, which violates 'identically distributed'.

iid has two parts: independence (knowing some draws tells you nothing about others) and identical distribution (each draw comes from the same distribution, with the same parameters). Neither alone is enough.

A harder one

Two cards are drawn one after the other without replacement from a standard 52-card deck, and X1 and X2 record whether each is red. Is (X1, X2) an iid pair?

  1. Yes. Each draw has the same marginal probability of red, 1/2, which is all that iid requires.Why not: Checks only identical marginals; iid also needs independence, and here P(X2 red | X1 red) = 25/51, not 1/2.
  2. No. The second draw has a different marginal distribution from the first, though the draws are independent.Why not: The marginal probability of red on the second draw is also 1/2 (by symmetry); the failure is of independence, not of the marginals.
  3. Yes. Without replacement the deck is smaller, which makes the two draws independent.Why not: A smaller deck makes the second draw depend on the first; sampling with replacement would give independence.
  4. No. Each card is red with probability 1/2 marginally, so they are identically distributed, but they are not independent. - correct

By symmetry each draw is red with probability 1/2, so the marginals are identical. But given a red first card, 25 of the remaining 51 cards are red, so P(X2 red | X1 red) = 25/51 differs from 1/2. The draws are dependent, hence not iid.

The hardest tier

A desk's daily returns are serially uncorrelated at every lag, but the squared returns are strongly positively autocorrelated, because calm periods and turbulent periods tend to persist. Which statement is correct?

  1. The returns are iid, because zero autocorrelation at every lag is equivalent to independence.Why not: Treats zero correlation as equivalent to independence; correlation only detects linear dependence, and the squared returns reveal dependence in the variance.
  2. The returns are independent but not identically distributed, because volatility changes over time.Why not: If volatility persists as a function of past returns the draws are dependent; this answer wrongly assigns the failure to the identical-distribution condition.
  3. The returns are uncorrelated but not independent, so they cannot be iid. - correct
  4. The returns are iid with fat tails; autocorrelation of squared returns is just a feature of the tails.Why not: Fat tails alone do not generate autocorrelation in squared returns; that pattern is evidence of dependence (volatility clustering).

Independence implies that any function of the observations, including their squares, is uncorrelated across time. Autocorrelated squared returns therefore show dependence even though the returns themselves are uncorrelated. Uncorrelated does not mean independent.

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