Prove $T^{2}$ Hotelling central statistic

Question

I've been given the following definition of the Hotelling distribution:

Let $\underline{d}$~$N_{p}$($\underline{0},I_{p}$) Multivariate normal distribution

and $\textbf{W}$~(n,$I_{p}$) Whishart distribution independent between them,

then $T^2 \,= n \,\underline{d}^{t}\textbf{W}^{-1}\underline{d}$ ~ $T^2$ (Hotelling distribution)

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I need to prove the Hotelling central statistic:

Lemma: Let $\underline{X}$~$N_{p}$($\underline{\mu},\sum$) and $\textbf{W}$~(n,$\sum$) independent then:

\begin{equation} T^2 = n(\underline{X}-\underline{\mu})^t\textbf{W}^{-1} (\underline{X}-\underline{\mu}) \sim T^2_{p,n} \end{equation}

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I've tried to standardize the Normal distribustion as it follows: $\sum^{-1}(\underline{X}-\underline{\mu})$ ~ $N_{p}(\underline{0},I_{p})$ and use the Wishart property that says that $CWC^{t} \sim W_{p}(n,CWC^{t})$ but I can't figure out the conclusion.

Thanks in advance to any help.

Answer 1

Everything settled, bad standardization.

$\sqrt{\sum}^{-1}(\underline{X}-\underline{\mu})$ ~ $N_{p}(\underline{0},I_{p})$

Now you can proceed without mistakes, just using the definition you will get to the conclusion.