Suppose you have a sample $X_1, \ldots, X_n$, $n$ large, from a multivariate normal distribution $N_p(\mu, \Sigma)$. It is easy to show that $D_i := k(X_i - \bar{X}_n)'S^{-1}_n(X_i-\bar{X}_n) \sim T^2_{p, n-1}$, for some constant $k$ and $\bar{X}_n = \frac{1}{n}\sum_{i=1}^n X_i$ and $S_n = \frac{1}{n-1}\sum_{i=1}^n (X_i - \bar{X}_n)(X_i - \bar{X}_n)'$.

What is the distribution of $D_i$ when $n\rightarrow \infty$?

At a first sight I think $D_i \rightarrow_d \chi^2_{p}$ because $S^{-1} \rightarrow_p \Sigma$.


1 Answer 1


Using the relation between Hotelling's T-squared and F distributions, we know that $$T^2\sim T^2_{p,n}\implies\frac{T^2}{n}\cdot\frac{n-p+1}{p}\sim F_{p,n-p+1}$$

Or, $$\frac{T^2}{n}(n-p+1)\sim pF_{p,n-p+1}$$

Now if $F\sim F_{n_1,n_2}$, one can write $F=\frac{U/n_1}{V/n_2}$ where $U\sim \chi^2_{n_1}$ and $V\sim \chi^2_{n_2}$ are independent.

We can also write $V=\sum_{j=1}^{n_2} Z_i^2$ where $Z_j$'s are i.i.d $N(0,1)$.

By law of large numbers, as $n_2\to \infty$, $$\frac1{n_2}V \stackrel{P}\longrightarrow 1$$

Then by Slutsky's theorem, as $n_2\to \infty$,

$$n_1F \stackrel{d}\longrightarrow \chi^2_{n_1}$$

So if $p$ is fixed and $n\to \infty$, we must have $$\frac{T^2}{n}(n-p+1)\stackrel{d}\longrightarrow \chi^2_p$$

In other words, for fixed $p$,

$$T^2\stackrel{d}\longrightarrow \chi^2_p \quad\text{ as }\quad n\to \infty$$


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