By E. B Dynkin, A. A. Yushkevich

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**Sample text**

If A. = O, that is, if µ 1 = µ 2 = · · · = µ'P = O, the noncentral chisquare distribution degenerates into the central chi-squaredistribution. 'D-1µ. Proof: Sin ce D is positive definite, we know that there exista a nonsingular p x p matrix B (not orthogonal) such that B'DB = l. ,I). 'BB'¡i.. But from B'DB = I, we get BB' = D-1 ; so Z'Z = Y'BB'Y = Y'D-1 Y and ).. , and the proof is complete. ral F Since the noncentral chi-square has properties similar to those of the central chi-square, we consider the possibility that the ratio of two noncentral chi-squares might have sorne properties similar to Snedecor's F distribution.

1 Example. Let y1 and Yz be jointly distributed as thc uivariate normal with quadratic form Q = Y~ 2 Y1Y2 - 3y1 - 2y2 +4 Q =(Y - fL)'R(Y - fL} = Y'RY - fL'RY - Y'RfL Y1 -1 + 2yi - Suppose we desire to find the following quantities: E(y¡), E(y 2), cov ·(y 1 ,y 2), var (y 1), and var (y 2), where var and cov denote variance and covariance, respectively. These quantitieswill befound by getting the vector of means and the covariance matrix. Since the quadratic fo~m Q can be written · -6)'( 3-l)f\y2 -6) -2 61 THE l\IULTIVARIA TE NORl\IAL DISTRIBUTION -2 + fL'RfL we sce that Y'RY is the only term that involves only second-degree terms in y 1 and y 2• Thus, from Q we select only the second-degree terms and evaluate the matrix R.

900, = V 13'fL = 5; so_ l - {J(A. A. 905. 900. The probability of rejecting H 0 when l = Il. 905. 3 Example. Suppose that a random variable y is distributed N(µ,l) and that we want to test the hypothesis H 0 : µ =O. 2 Example. U= y2n(n - 1) :E(y. - 'fi) 2 as the test function. , \vhen µ =O. Since the basic variables y 1 , y 2 , ••• , Yn are independent and normal with mean O (assuming H 0 is true) and variance 1, it follows that l. y is distributed normally with mean O and variance l/n. Therefore, yVn is distributed normally with mean O and variance 1, and ny2 has a chi-squarc distribution with 1 degree of freedom.