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Part BCSIR NET June 2023wishart-linear-form

Wishart linear form

Suppose X=(X1,X2,X3,X4)TX = (X_{1}, X_{2}, X_{3}, X_{4})^{T} ~ N4(0,I2N_{4}(0, I_{2} ⊗ ∑)\sum) with ∑=[[2,−1],[−1,2]]\sum = [[2, -1], [-1, 2]]. Define Z=[[X1,X2],[X3,X4]]Z = [[X_{1}, X_{2}], [X_{3}, X_{4}]] and Q = ZᵀZ. If Wm(n,∑)W_{m}(n, \sum) denotes a Wishart distribution of order m with n degrees of freedom, the distribution of Q11+Q12+Q21+Q22Q_{11} + Q_{12} + Q_{21} + Q_{22} is

  1. A.W1(2,2)W_{1}(2, 2)✓ Correct
  2. B.W1(1,2)W_{1}(1, 2)
  3. C.W1(2,1)W_{1}(2, 1)
  4. D.2χ422\chi^{2}_{4}

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

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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