UDC 519.251.8

MATHEMATICS

Academician Yu. V. LINNIK, V. A. PLISS, O. V. SHALAEVSKII

ON THE THEORY OF HOTELLING’S TEST

Let \(X_1,\ldots,X_N\) be independent normal vectors of a \(p\)-dimensional Euclidean space, having common mean \(\xi=EX_i\) (a \(p\)-dimensional column vector) and common correlation matrix
\[ \Sigma=E'[(X_i-\xi)(X_i-\xi)^\tau], \]
\(i=1,\ldots,N\); here \(\tau\) denotes transposition. Put
\[ \overline X=\frac1N\sum_{i=1}^N X_i; \]
\[ S=\sum_{i=1}^N (X_i-\overline X)(X_i-\overline X)^\tau . \]

Consider the problem of testing the statistical hypothesis \(H_0:\xi=0\) against the (composite) alternative \(H_\delta:N\xi^\tau\Sigma^{-1}=\delta\), where \(\delta\) is an arbitrary fixed positive number. Obviously, this problem may be interpreted, in a certain sense, as testing for the presence, in a given Gaussian noise with zero mean, of a signal of given energy. For its solution one usually uses Hotelling’s \(T^2\)-test (see \((^1)\)), rejecting the hypothesis \(H_0\) when
\[ T^2=N(N-1)\overline X^\tau S^{-1}\overline X>T_0^2, \]
where \(T_0^2\) is a constant determined by the chosen level \(\alpha\). We note that for a constant value of \(\delta\) the power of the \(T^2\)-test is also constant (see \((^1)\)).

The statistical properties of the \(T^2\)-test were studied by J. Simaika \((^2)\), C. Stein \((^3,^4)\), N. Giri, J. Kiefer, and C. Stein \((^5,^6)\). The test has a simple structure, but its properties are mysterious. Up to now not a single nontrivial case is known for which the admissibility of the \(T^2\)-test has been proved. A weaker property here, minimaxity, was justified in \((^6)\) only for \(p=2,\ N=3\). In this very interesting work the authors, relying on the Hunt–Stein theorem (see \((^7)\)), reduce the problem of minimaxity of the \(T^2\)-test to the investigation of a Fredholm integral equation of the first kind, which for \(p=2,\ N=3\) is transformed into an “overdetermined” linear differential equation of the first order; the explicit form of the solution of the latter makes it possible to prove the minimaxity of the \(T^2\)-test in this special case. In the general case the Fredholm integral equation has the form
\[ \int_{\Gamma_1} \exp\left\{\sum_{j=1}^p t_j\sum_{i>j}\frac{\beta_i}{2}\right\} \cdot \prod_{i=1}^p \Phi\left(\frac{N-i+1}{2},\frac12,\frac{t_i\beta_i}{2}\right) \,d\lambda^*(\beta_1,\ldots,\beta_p) = \Phi\left(\frac N2,\frac p2,\frac\gamma2\right), \]
where \(\Gamma_1\) denotes the \((p-1)\)-dimensional simplex
\[ \{(\beta_1,\ldots,\beta_p):\ \beta_i\ge 0,\ \sum_{i=1}^p \beta_i=1\} \]
and \(\Phi\) is the confluent hypergeometric function. The equation must hold identically in \(t_1,\ldots,t_p\), satisfying the condition
\[ \sum_{i=1}^p t_i=\gamma, \]
where \(\gamma\) is any fixed positive number, and the unknown function \(\lambda^*\) must be a probability measure on \(\Gamma_1\). The minimaxity of the \(T^2\)-test is equivalent to the existence of such a \(\lambda^*\).

In the present note we continue the investigation of N. Giri, J. Kiefer, and C. Stein and prove the minimaxity of the \(T^2\)-test under the alternative

$H_\delta$ for $p=2$, $N=4$. A slight complication of our arguments also gives the construction of an overdetermined boundary-value problem with a linear differential operator in the general case $(p=2)$.

For $p=2$, $N=4$ the problem consists in showing that on the interval $0<\beta<1$ there exists a probability density $\lambda(\beta)$ for which

\[ \int_0^1 e^{t(1-\beta)/2}\Phi\left(2,\frac12,\frac{t\beta}{2}\right) \Phi\left(\frac32,\frac12,\frac{(\gamma-t)(1-\beta)}{2}\right)\lambda(\beta)\,d\beta = \Phi\left(2,1,\frac{\gamma}{2}\right). \tag{1} \]

With a view to solving this problem, let us consider the equation

\[ \int_0^1 e^{t(1-\beta)/2}\Phi\left(2,\frac12,\frac{t\beta}{2}\right) \Phi\left(\frac32,\frac12,\frac{(\gamma-t)(1-\beta)}{2}\right)\lambda(\beta)\,d\beta = \]

\[ = \int_0^1 \Phi\left(\frac32,\frac12,\frac{\gamma(1-\beta)}{2}\right)\lambda(\beta)\,d\beta. \tag{2} \]

Using the fact that

\[ \Phi\left(\frac32,\frac12,\frac{x}{2}\right)=(1+x)e^{x/2}, \]

we rewrite it in the form

\[ \int_0^1 e^{-\gamma\beta/2}\Phi\left(2,\frac12,\frac{t\beta}{2}\right) [1+\gamma-\gamma\beta-t(1-\beta)]\lambda(\beta)\,d\beta = \]

\[ = \int_0^1 e^{-\gamma\beta/2}(1+\gamma-\gamma\beta)\lambda(\beta)\,d\beta. \]

The left-hand side of the last equality can be expanded in a power series in $t$ and the system obtained

\[ \int_0^1 e^{-\gamma\beta/2} \bigl[(r-2r^2)\beta^{r-1}+(1+\gamma+\gamma r+2r^2)\beta^r-(\gamma+\gamma r)\beta^{r+1}\bigr] \lambda(\beta)\,d\beta=0, \]

\[ r=1,2,\ldots \tag{3} \]

Let

\[ e^{\gamma\beta/2}f(\beta)=\sum_{r=1}^{\infty} a_{r-1}\beta^{r-1} \]

be an arbitrary function representable by a series whose radius of convergence is greater than one. Multiply the $r$-th equation (3) by $a_{r-1}$ and sum over $r$ from $1$ to $\infty$. After some calculations we find

\[ \int_0^1 \left[ (2\beta^3-2\beta^2)f'' +(-5\beta+6\beta^2-\gamma\beta^2+\gamma\beta^3)f' +\left(-1+3\beta-\frac{\gamma\beta}{2}+\gamma\beta^2\right)f \right]\lambda(\beta)\,d\beta = \int_0^1 L(f)\lambda(\beta)\,d\beta=0. \tag{4} \]

Choose a small $\varepsilon>0$ and apply Green’s formula to the differential form $L(f)$:

\[ \int_{\varepsilon}^{1-\varepsilon} L(f)\lambda(\beta)\,d\beta = \int_{\varepsilon}^{1-\varepsilon} L^*(\lambda)f(\beta)\,d\beta + \mathscr{L}[f,\lambda]\biggr|_{\varepsilon}^{1-\varepsilon}, \]

where the adjoint form is

\[ L^*(\lambda) = 2\beta^2(\beta-1)\lambda'' +\beta(-3+6\beta+\gamma\beta-\gamma\beta^2)\lambda' +\left(3\beta+\frac{3\gamma\beta}{2}-2\gamma\beta^2\right)\lambda \]

and the bilinear form is

\[ \mathscr{L}[f,\lambda] = 2\beta^2(1-\beta)(f\lambda'-f'\lambda) - \bigl[\beta+\gamma\beta^2(1-\beta)\bigr]f\lambda. \]

From Frobenius theory (8) it follows that the equation \(L^*(\lambda)=0\) has two fundamental solutions in the interval \((0,1)\),

\[ \lambda_{01}=\sum_{k=0}^{\infty} a_k\beta^k,\qquad \lambda_{02}=\beta^{-1/2}\sum_{k=0}^{\infty} c_k\beta^k \tag{5} \]

and two fundamental solutions

\[ \lambda_{11}=\sum_{k=0}^{\infty} g_k(1-\beta)^k,\qquad \lambda_{12}=(1-\beta)^{-1/2}\sum_{k=0}^{\infty} h_k(1-\beta)^k, \tag{6} \]

where \(a_0,c_0,g_0,h_0\ne 0\). Hence any solution of the equation \(L^*(\lambda)=0\) is integrable on \([0,1]\). We assert, moreover, that there exists a solution of this equation for which

\[ \lim_{\varepsilon\to 0}\mathcal L[f,\lambda]\big|_{\varepsilon}^{1-\varepsilon}=0 . \tag{7} \]

Indeed, considering linear combinations of the solutions (5), and then (6), one can easily find that (7) holds if and only if \(\lambda=\lambda_{12}\).

The preceding arguments allow us to conclude that \(\lambda=\lambda_{12}\) is a solution of equation (4) and hence of equation (2). We now prove that \(\lambda=\lambda_{12}\) does not vanish in the interval \((0,1)\).

The equation \(L^*(\lambda)=0\), after the substitution \(\beta=1-x\), \(v(x)=\dfrac{1}{h_0}\sqrt{x}\lambda(1-x)\), becomes the equation

\[ 2x(x-1)v''+(-1+4x-\gamma x+\gamma x^2)v'+(3/2+3\gamma x/2)v=0, \tag{8} \]

and the subsequent substitution \(u=-v'/v\) gives us the Riccati equation

\[ u'=u^2-\frac{1-4x+\gamma x-\gamma x^2}{2x(1-x)}u-\frac{3+3\gamma x}{4x(1-x)} . \tag{9} \]

Recall that

\[ v(x)=\sum_{k=0}^{\infty}\frac{h_k}{h_0}x^k,\quad |x|<1. \]

Suppose that the function \(v(x)\) vanishes somewhere in the interval \((0,1)\). Since \(v(0)=1\), among the roots of the equation \(v(x)=0\) there will be a smallest one. Denote it by \(x_0\). Then the function \(u\) is analytic in the disk \(|x|<x_0\). From (8), \(v'(0)=3/2\), i.e. \(u(0)=-3/2\). Therefore \(u(x)\) cannot vanish in the interval \((0,x_0)\), since at the point where this occurs for the first time one would have to have \(u'(x)\ge 0\), but equation (9) shows that at this point \(u'(x)<0\). The negativity of the function \(u(x)\) leads to unbounded decrease as one approaches \(x_0\) from the left \((v'(x_0)\ne 0\), since \(v\not\equiv 0)\), which again contradicts (9). All this means that \(\lambda=\lambda_{12}\) has no zeros in the interval \((0,1)\).

Let us return to equation (2), which, as we have seen, is satisfied by the function \(\lambda=\lambda_{12}\ne 0\). Using the method of Laplace transformations, it is not difficult to obtain the relation

\[ \int_{0}^{1}x^{b-1}(1-x)^{c-b-1}e^{qx}\Phi(a,b,px)\Phi(a-b,c-b,q(1-x))\,dx \]

\[ =\frac{\Gamma(b)\Gamma(c-b)}{\Gamma(c)}\Phi(a,c,p+q),\qquad 0<b<c. \tag{10} \]

First putting \(t=\gamma\tau\) in (2), we multiply both sides of this equality by \(\tau^{-1/2}(1-\tau)^{-1/2}\) and integrate with respect to \(\tau\) from 0 to 1. We find the integral with respect to \(\tau\), taking in (10) \(a=2\), \(b=1/2\), \(c=1\), \(p=\gamma\beta/2\), \(q=\gamma(1-\beta)/2\). As a result it turns out that

\[ \int_{0}^{1}\Phi\left(2,1,\frac{\gamma}{2}\right)\lambda(\beta)\,d\beta = \int_{0}^{1}\Phi\left(\frac{3}{2},\frac{1}{2},\frac{\gamma(1-\beta)}{2}\right)\lambda(\beta)\,d\beta . \]

To obtain (1), it suffices to use the homogeneity of the written relation and to normalize the function \(\lambda=\lambda_{12}\) accordingly.

In conclusion we make a small remark. It can be shown that from the minimaxity of the \(T^2\)-test under any alternative \(H_\delta\), \(\delta>0\), there follows the minimaxity of the \(T^2\)-test under the more natural alternative \(H_\delta' : N\xi'\Sigma^{-1}\xi \geq \delta\). The proof is by contradiction, if one takes into account that, as we have already mentioned, for fixed \(\delta\) the power of the \(T^2\)-test is constant. This remark is also substantiated by I. V. Romanovskii.

REFERENCES

1. T. W. Anderson, Introduction to Multivariate Statistical Analysis, N. Y., 1958.
2. J. Simaika, Biometrika, 32, 70 (1941).
3. C. Stein, Ann. Math. Stat., 26, 769 (1955).
4. C. Stein, Ann. Math. Stat., 27, 616 (1956).
5. N. Giri, J. Kiefer, Ann. Math. Stat., 33, 1490 (1962).
6. N. Giri, J. Kiefer, C. Stein, Ann. Math. Stat., 34, 1524 (1963).
7. E. L. Lehmann, Testing Statistical Hypotheses, “Nauka,” 1964.
8. E. L. Ince, Ordinary Differential Equations, Kharkov, 1939.