# Chi-Squared Distribution – Understanding the Distribution of X’??¹X for X Following a Multivariate t Distribution

According to Golam Kibria & Joarder (2006, p.7) available here and Kotz & Nadarajah (2004, p. 19) visible in google, the distribution of $$X'\Sigma^{-1}X /p$$, for a known correlation matrix $$\Sigma$$ when $$X$$ is a $$p$$-long vector of $$\{X_i\}$$ with (assuming uncorrelated items)

$$X_i \sim t_{v, \mu, \sigma^2}$$

a student-t distribution with location $$\mu$$ and scale $$\sigma^2$$, follows a $$F$$ distribution with degrees of freedom $$p$$ and $$v$$ and noncentrality parameter $$\lambda$$ = $$\mu'\Sigma^{-1}\mu /p$$.

Here is a cut-and-paste of the whole section 3.5 of Golam Kibrian & Joarder (result presented with no proof):

I can't replicate this result with simulations so looked at the maths, and found something different (see below). Which is correct? Edit Actually, both may be incorrect as neither fit the results from simulations… Where did I make an error?

Herein, the demonstration is restricted to the case where correlation is null and assuming a unique variance $$\sigma^2$$ for all items in $$X$$. Setting the correlation to zero in $$\Sigma$$, we find that $$X'\Sigma^{-1}X/p$$ equals $$\frac{1}{p}\sum_{i=1}^p X_i^2/\sigma^2$$. As

$$X_i \sim t_{v,\mu_i,\sigma^2}$$
implies
$$X_i/\sigma \sim t_{v,\mu_i/\sigma,1} = \frac{\mathcal{N}(\mu_i/\sigma,1)}{\sqrt{\chi_v^2/v}}$$
Thus,
$$\frac{X_i^2}{\sigma^2} \sim \frac{\chi^2_1(\mu_i^2/\sigma^2)}{\chi^2_v/v}$$
and consequently,
$$\begin{split} \frac{1}{p}\sum_{i=1}^p \frac{X_i^2}{\sigma^2} &\sim \frac{\chi^2_p\left(\sum \mu_i^2/\sigma^2\right)/p}{\chi^2_v/v} \\ &= F_{p,v}\left(\sum \mu_i^2/\sigma^2\right)=F_{p,v}(\mu'\Sigma^{-1}\mu) \end{split}$$

The line before the last comes from the fact that the sum of independent $$\chi^2$$ is a $$\chi^2$$ with the sum of the df and the sum of the noncentrality parameters. The last line is from the definition of a $$F$$ distribution.

This derivation differs from Kibria & Joarder's result as the noncentrality parameter is not divided by $$p$$ when the data are. Can we expand this demonstration to the case where correlation is non-null?

Suppose the random variable $$\textbf{X}$$ has a multivariate t-distribution with mean vector $$\mu$$, covariance matrix $$\boldsymbol{\Sigma}$$ and degrees of freedom $$\nu$$, then $$\textbf{X}^\prime \boldsymbol{\Sigma}^{-1} \textbf{X}/p$$ has the $$F$$ distribution with $$\nu$$ and $$p$$ degrees of freedoms and non-centrality parameter $$\mu^\prime \boldsymbol{\Sigma}^{-1} \mu/p$$

You can make this work when you define the multivariate t-distribution as

$$\begin{array}{c} \mathbf{X} = \frac{\mathbf{Y}+\boldsymbol{\mu}}{\sqrt{Z/\nu}}\\ \\ \text{where} \qquad \textbf{Y} \sim N(0,\boldsymbol{\Sigma}) \qquad \text{and} \qquad Z \sim \chi^2_\nu \end{array}$$

Note that this differs from

$$\mathbf{X} = \frac{\mathbf{Y}}{\sqrt{Z/\nu}} +\boldsymbol{\mu}\\$$

which is, I believe, the more standard from of the multivariate t-disribution.

The distribution of the product of only the numerator is $$\chi^2$$ disributed $$(\mathbf{Y}+\boldsymbol{\mu})^\prime \boldsymbol{\Sigma}^{-1} (\mathbf{Y}+\boldsymbol{\mu}) \sim \chi^2_{p}(\text{ncp} = \boldsymbol{\mu}^\prime \boldsymbol{\Sigma}^{-1} \boldsymbol{\mu})$$

The distribution of product with the entire fraction $$\mathbf{X}$$ is distributed as

$$\begin{array}{c}\mathbf{X}^\prime \boldsymbol{\Sigma}^{-1} \mathbf{X}/p = \frac{(\mathbf{Y}+\boldsymbol{\mu})^\prime \boldsymbol{\Sigma}^{-1} (\mathbf{Y}+\boldsymbol{\mu})/p}{Z/\nu} \sim \frac{Z_1/p}{Z_2/\nu}\\ \\ \text{where} \qquad Z_1 \sim \chi^2_{p}(\text{ncp} = \boldsymbol{\mu}^\prime \boldsymbol{\Sigma}^{-1} \boldsymbol{\mu}) \qquad \text{and} \qquad Z_2 \sim \chi^2_\nu \end{array}$$

This ratio of two $$\chi^2$$ distributed variables, $$Z_1$$ and $$Z_2$$ (where $$Z_1$$ may be a non-central $$\chi^2$$-distribution), divided by their degrees of freedom, $$\frac{Z_1/p}{Z_2/\nu}$$, is equivalent to a F-distributed variable.