Part CCSIR NET December 2025scaling-a-wishart-matrix-by-c-scales-sigma-by-c-not-by-1-over-c

Scaling a wishart matrix by c scales sigma by c not by 1 over c

Let $X_{1}, X_{2}, \dots, X_{20}$ be a random sample from $N_{12}(\mu,\sum)$ , where $\mu\in\mathbb{R}^{12}$ and $\sum$ is positive definite. Suppose ${\bar{X}} = (1/20)\sum_{i}$ ₌ $_{1}^{20}X_{i}$ and $S = (1/19)\sum_{i}$ ₌ $_{1}^{20}(X_{i}-{\bar{X}})(X_{i}-{\bar{X}})^{T}$ . Then which of the following statements are true?

A.19(X̄ᵀSX $)({\bar{X}}^{T}\sum{\bar{X}})$ ⁻ $^{1}$ ~ $\chi^{2}_{19}$ ✓ Correct
B.E(S⁻ $^{1}) = (19/6)\sum$ ⁻ $^{1}$ ✓ Correct
C.S ~ $W_{12}(19, 19\sum)$
D.trace $(19\sum$ ⁻ $^{1}S)$ ~ $\chi^{2}_{228}$ ✓ Correct

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

Scaling a wishart matrix by c scales sigma by c not by 1 over c

More on this topic