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 $p$,

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.