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Part BCSIR NET December 2024the-schur-complement-loses-one-degree-of-freedom

The schur complement loses one degree of freedom

Let X, Y and Z be random variables such that S = [[X, Y], [Y, Z]] ~ W2(10,∑)W_{2}(10, \sum), where W2W_{2} denotes the Wishart distribution and ∑=[[1,1/2],[1/2,1]]\sum = [[1, 1/2], [1/2, 1]]. Define T=Z−Y2/XT = Z - Y^{2}/X. Then, Var(T) equals

  1. A.77/6
  2. B.81/8✓ Correct
  3. C.83/9
  4. D.79/7

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: If every marginal is normal then the vector is multivariate normal

More on this topic

The chapter behind this: Multivariate normal and Wishart — free to read

From Linear Models and Multivariate › Multivariate normal distribution

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