A Counter-Example Relating to Certain Multivariate Generalizations of $t$ and $F$

Abstract

It is shown that the distributions of certain multivariate analogues of $t$ and $F$ are dependent on the population covariance matrix. Suppose that $\mathbf{z} = (x_1, \cdots, x_p)'$ has the multivariate normal distribution with mean vector $\mathbf{\xi}$ and covariance matrix $\mathbf{\sigma}$. Let $\mathbf{S, S}^\ast$ be independent Wishart matrices also with covariance matrix $\mathbf{sigma}$, based on $n, n^\ast$ degrees of freedom respectively. Then if $\mathbf{T}$ is a $p \times p$ matrix such that \begin{equation*}\tag{1} \mathbf{TT}' = S\end{equation*} natural candidates for the multivariate analogues of $t$ and $F$ are \begin{equation*}\begin{align*}\tag{2}\mathbf{t} = \mathbf{t}^{-1} (\mathbf{z} - \mathbf{\xi}), \\ \tag{3}\mathbf{W} = \mathbf{T} ^{-1}\mathbf{S}^{\ast}\mathbf{T}^{'-1}.\end{align*}.\end{equation*} Olkin and Rubin (1964), Theorems 3.2 and 4.2, have shown that if $\mathbf{T}$ is taken to be upper or lower triangular, then $\mathbf{t}$ and $\mathbf{W}$ do in fact have distributions which independent of $\mathbf{sigma}$. However, if $\mathbf{T}$ is taken to by symmetrical and positive definite, $\mathbf{T} = \mathbf{S}^\frac{1}{2}$, then they remark (Section 3) that the distribution of $\mathbf{W}$ is unknown for general $\mathbf{\sigma}$. It seems worthwhile to present the following example, which shows that for $\mathbf{T = S}^\frac{1}{2}$ the distributions of $t$ and $\mathbf{W}$ depend on $\mathbf{\sigma}$; the contrary is occasionally asserted (see Bennett and Cornish (1964), p. 907. Let $p = 2$, and assume that\begin{equation*}\tag{4} \mathbf{\sigma}^{-1} = \begin{pmatrix}a^2@0\\0@b^2.\end{pmatrix}\end{equation*} Since $\mathbf{S}^\frac{1}{2}$ is positive definite, it may be written in the form\begin{equation*}\tag{5} \mathbf{S}^\frac{1}{2} = \begin{pmatrix}x,@(xy)^\frac{1}{2}q\\(xy)^\frac{1}{2}q,@y (\mathbf{0} < x, y < \infty, q^2 < 1).\end{pmatrix}\end{equation*} Noting the Jacobian\begin{equation*}\tag{6}\partial(\mathbf{S})/\partial(x,y,q) = 4(1 - q^2)(xy)^\frac{1}{2}(x + y),\end{equation*} it is readily found by transforming the Wishart distribution that $\mathbf{S}^\frac{1}{2}$ has the distribution. \begin{equation*}\begin{align*}\tag{7}f(\mathbf{S} ^\frac{1}{2}) d\mathbf{S}^\frac{1}{2} = \{(nab)^n/ \pi\Gamma(n - 1)\} (1 - q^2)^{n-2}(xy)^{n-\frac{3}{2}}(x + y) \\ \cdot\exp \{ -\frac{1}{2}n \lbrack a^2x^2 + (a^2 + b^2)xyq^2 + b^2y^2\rbrack\} dx dy dq, \\ (0 < x, y < \infty, q^2 < 1)\end{align*}.\end{equation*} (See Olkin and Rubin, p. 266.) Now \begin{equation*}\begin{align*}\tag{8} \epsilon(\mathbf{tt}' &= \epsilon\mathbf{W} \\ &= \epsilon (\mathbf{S}-^\frac{1}{2} \mathbf{\sigma S}^{-\frac{1}{2}}) &= \begin{bmatrix}\nu_1/a^2 + \nu_3/b^2, \quad - \nu_4/ a^2 - \nu_5/b^2 \\ - \nu_4/a^2 - \nu_5/b^2, \quad \nu_3/a^2 + \nu_2/b^2\end{bmatrix}\end{align*},\end{equation*} where\begin{equation*}\begin{align*}\tag{9}\nu_1 &= \mathscr{E}\lbrack x^{-2} (1 - q^2)^{-2}\rbrack, \quad \nu_2 = \mathscr{E}\lbrack y^{-2} (1 - q^2)^{-2}\rbrack, \quad \nu_3 = \mathscr{E} \lbrack q^2/xy (1 - q^2)^2\rbrack, \\ \nu_4 = \mathscr{E}\lbrack q/x^{\frac{3}{2}} y ^{\frac{1}{2}}(1 - q^2)^2\rbrack, \quad \nu_5 = \mathscr{E}\rbrack 1/x^{\frac{1}{2}} y^{\frac{3}{2}} (1 - q^2)^2\lbrack\end{align*}. \end{equation*} It will be sufficient to show that the matrices in (8) are not independent of $a$ and $b$. In order to simplify the calculations still further, we shall take the case $n = 4$, and consider only the elements $\epsilon(t^2_1), \epsilon(t^2_2)$ on the main diagonal in (8). The density (7) now takes the form: \begin{equation*}\tag{10}f(\mathbf{S}^{\frac{1}{2}} d\mathbf{S}^{\frac{1}{2}} = 2^{7_\pi^{-1}} (ab)^4(1 - q^2)^2(xy)^{\frac{5}{2}} (x + y) \cdot\exp \{-2\lbrack a^2x^2 + (a^2 + b^2)xyq^2 + b^2y^2\rbrack\} dx dy dq,\end{equation*} and we wish to evaluate $\nu_1, \nu_2, \nu_3$. For $\nu_1$, let us first consider \begin{equation*}\begin{align*}\tag{11}I(q) &= \int_0^\infty \int_0^\infty x^{-2}(1 - q^2)^{-2}f(\mathbf{S} ^{\frac{1}{2}}) dx dy \\ &= 2^{7_\pi^{-1}}(ab)^4 \int_0^\infty \int_0^\infty (x^{\frac{3}{2}}y^{\frac{5}{2}} + x^{\frac{1}{2}} y^{7/2}) e^{-2a2x2-2b2y2} \\ \cdot \sum^\infty_{k=0} K!^{-1}\lbrack -2(a^2 + b^2)xyq^2\rbrack^k dx dy\end{align*}.\end{equation*} Making use of the formulas \begin{equation*}\begin{align*}\tag{12}\Gamma(\omega)\Gamma(\omega + \frac{1}{2}) &= \pi^{\frac{1}{2}}\Gamma)(2\omega)/2^{2\omega-1}, \\ \int_0^\infty \omega^\lambda e^{-a^2\omega^{2}} d\omega &= \Gamma((\lambda + 1)/2)/2a^{\lambda+1},\\ \sum^\infty_{k=0} x^k\Gamma(k + \lambda)/k! = \Gamma(\lambda)(1 - x)^{-\lambda}, \quad (| x | < 1)\end{align*}\end{equation*}, it is found that \begin{equation*} \begin{align*}\tag{13}I(q) &= (2a^3/\pi b^{\frac{1}{2}} \sum^\infty_{k=0} k!^{-1}\lbrack -(a^2 + b^2)q^2/2ab\rbrack^k \cdot\{a\Gamma(k + \frac{3}{2}) + (a + b)\Gamma(k + \frac{5}{2})\} \\ \tag{14} &=(a^3/2b)^{\frac{1}{2}} \{a\phi^{-\frac{3}{2}}(q) + \frac{3}{2}(a + b)\phi^{-\frac{5}{2}}(q)\} \end{align*}\end{equation*} where \begin{equation*}\tag{15} \phi(q) = 1 + (a^2 + b^2)q^2/2ab\end{equation*} The series in (13) is convergent only for\begin{equation*}\tag{16} | q |^2 < 2ab/(a^2 + b^2)\end{equation*} However, since $I(q)$ and the expression (14) are both analytic functions of the complex variable $q = u + iv$ over the region $u^2 - v^2 > -2ab/(a^2 + b^2)$ in the $q$ plane, it follows by analytic continuation that $I(q)$ is certaintly equal to (14) for all real $q$. Noting that \begin{equation*}\begin{align*}\tag{17}\int_0^1\phi ^{-\frac{3}{2}}(q) dq &= (2ab)^{\frac{1}{2}}/(a + b), \\ \int_0^1\phi^{-\frac{5}{2}}(q) dq = \frac{2}{3}(2ab)^{\frac{1}{2}}(a^2 + 3ab + b^2)/(a + b)^3\end{align*},\end{equation*} we have \begin{equation*}\begin{align*}\tag{18}\nu_1 &= 2 \int_0^1 I(q) dq \\ &= 2a^2\lbrack 2 - (b/(a + b))^2\rbrack\end{align*},\end{equation*} $\nu_2$ being obtained by interchanging $a$ and $b$. Similarly, it is found that \begin{equation*}\tag{19}\nu_3 = 2(ab/(a + b))^2.\end{equation*} Hence \begin{equation*}\begin{align*}\tag{20}\mathscr{E}(t^2_1) 7= \nu_1/a^2 + \nu_3/b^2 = 4 + 2(a - b)/(a + b), \\ \mathscr{E}(t^2_2) &= \nu_3/a^2 + \nu_2/b^2 = 4 - 2(a - b)/(a + b)\end{align*},\end{equation*} and the matrices in (8) are not dependent of $a$ and $b$. As a check on the working, we have: \begin{equation*}\tag{21}\mathscr{E}(t^2_1 + t^2_2) = \mathscr{E} (\mathbf{z} -\xi)'\mathbf{S}^{-1}(\mathbf{z} - \xi) = 8,\end{equation*} which is independent of $\Sigma$ and in accordance with the known distribution of Hotelling's $T^2$.