# Limit of Hotelling $T^2$ distribution

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 $$S^{-1} \rightarrow_p \Sigma$$.

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 $$p$$,

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