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Σ −1 (X − M)Φ −1 )) , (2.7) the place h and ψ make certain one another for distinctive p and n. evidence: I. First, we end up that if X ∼ E p,n (M, Σ ⊗ Φ , ψ ) and E p,n (M, Σ ⊗ Φ , ψ ) is admittedly non-stop, then the p.d.f. of X has the shape (2.7). Step 1. think that M = zero and Σ ⊗ Φ = I pn . Then, X ∼ E p,n (0, I p ⊗ In , ψ ). we wish to express that the p.d.f. of X relies on X merely via tr(X X). permit x = vec(X ). From Theorem 2.1 we all know that x ∼ E pn (0, I pn , ψ ). enable H ∈ O(pn), then, in view of.

. . . . . . . . . . . . . . . . . 2.5 Stochastic illustration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Conditional Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.1 One-Dimensional Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Φ AΣ + M tr(A M ) = c0 Φ AΣ + M tr(AM). 3.2 extra on anticipated Values seventy nine (iv) X ∼ En,p (M , Φ ⊗ Σ , ψ ). E(X tr(AX )) = c0 Φ A Σ + M tr(AM ). (v) p p n n ∑ ∑ ∑ ∑ xi j a jk xkl bli E(tr(XAXB)) = E i=1 j=1 k=1 l=1 = ∑ (c0 σik φ jl + mi j mkl )a jk bli i, j,k,l ∑ = c0 σik a jk φ jl bli + i, j,k,l ∑ mi j a jk mkl bli i, j,k,l = c0tr(Σ A Φ B) + tr(MAMB). (vi) p n n p ∑ ∑ ∑ ∑ xi j a jk xlk bli E(tr(XAX B)) = E i=1 j=1 k=1 l=1 ∑ (c0 σil φ jk + mi j mlk )a jk bli = i, j,k,l ∑.

Have P(log x < zero) = 1. in spite of the fact that, P(log x < zero) = 1, including (5.8), signifies that log x ≈ log y (see Marcinkiewicz, 1938). therefore, we get x ≈ y. Theorem 5.6. enable (W1 , W2 , . . ., Wm ) ∼ G p,m Σ , n21 , n22 , . . ., n2m , ψ , the place ni is confident integer and Wi is p × p, i = 1, 2, . . ., m and Σ > zero. allow v be a p-dimensional consistent nonzero vector. Then, (v W1 v, v W2 v, . . ., v Wm v) ∼ G1,m the place ψ ∗ (z) = ψ (v Σ vz). n1 n2 nm , , . . ., , ψ ∗ , 2 2 2 (5.9) 5.1 Extension of Cochran’s.

proven fact that h1 (tr(X1 X1 )), being marginal p.d.f. of X, may be bought from h(tr(XX )) via integration. enable X = (X1 , X2 ), then h(tr(XX )) = h(tr(X1 X1 + X2 X2 )) = h(tr(X1 X1 ) + tr(X2 X2 )). consequently we have now h1 (tr(X1 X1 )) = IR(p−q)×n h(tr(X1 X1 ) + tr(X2 X2 ))dX2 . (6.19) From (6.18) and (6.19), we get h1 (tr(X1 X1 )) = c IR(p−q)×n h1 (tr(X1 X1 ) + tr(X2 X2 ))dX2 . for that reason h1 (z) = c IR(p−q)×n h1 (z + tr(X2 X2 ))dX2 , z ≥ zero. h1 (z + tr(X2 X2 ))dX2 , z ≥ zero. utilizing (6.17) back, we.