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def. def. using visual .net todeploy code-128 for asp.net web,windows application iPhone cov(x1 , xn ) visual .net Code 128 Code Set C cov(x2 , xn ) . .

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(5.1). cov(xn , x2 ) .net framework barcode 128 cov(xn , xn ). Recall that the diagonal elements of the matrix (5.1) are variances: cov(x j , x j ) = var(x j ). Obviously, a variance matrix is symmetric and positive (semi)de nite.

Moreover, note that (5.1) can be written as Var(X ) = E[X X T ] (E[X ])(E[X ])T . (5.

2). Similarly, the c Code 128 Code Set C for .NET ovariance matrix of a pair of random vectors X and Y is the matrix of covariances of their components:2. To distinguish t he variance of a random variable from the variance matrix of a random vector, the latter will be denoted by Var with capital V. The capital C in Cov indicates that this is a covariance matrix rather than a covariance of two random variables..

The Multivariate Normal Distribution Cov(X, Y ) = E (X E(X ))(Y E(Y ))T . def. (5.3). Note that Cov(Y, .net framework Code 128 Code Set A X ) = Cov(X, Y )T . Thus, for each pair X , Y there are two covariance matrices, one being the transpose of the other.

5.2. The Multivariate Normal Distribution Now let the components of X = (x1 , .

. . , xn )T be independent, standard normally distributed random variables.

Then, E(X ) = 0 ( Rn ) and Var(X ) = In . Moreover, the joint density f (x) = f (x1 , . .

. , xn ) of X in this case is the product of the standard normal marginal densities: f (x) = f (x1 , . .

. , xn ) = exp x 2 /2 j 2 j=1. exp 1 n x 2 ex p 1 x T x j=1 j 2 = = . 2 ( 2 )n ( 2 )n The shape of this density for the case n = 2 is displayed in Figure 5.1.

Next, consider the following linear transformations of X : Y = + AX , where = ( 1 , . . .

, n )T is a vector of constants and A is a nonsingular n n matrix with nonrandom elements. Because A is nonsingular and therefore invertible, this transformation is a one-to-one mapping with inverse X = A 1 (Y ). Then the density function g(y) of Y is equal to g(y) = f (x).

det( x/ y). = f (A 1 y A 1 ). det( (A 1 y A 1 )/ y). = f (A 1 y A 1 ). det(A 1 ). = = f (A 1 y A 1 ) . det(A). exp 1 (y )T (A 1 )T A 1 (y ) 2 ( 2 )n det(A). exp 1 (y )T (A AT ) 1 (y ) 2 = . ( 2 )n det(A AT ). Observe that is the expectation vector of Y : E(Y ) = + A (E(X )) = . But what is AAT We know from (5.2) that Var(Y ) = E[YYT ] T .

Therefore, substituting Y = + AX yields Var(Y ) = E ( + AX)( T + X T AT ) T = (E(X T ))AT + A(E(X )) T + A(E(XXT ))AT = AAT. The Mathematical and Statistical Foundations of Econometrics Figure 5.1. The bivariate standard normal density on [ 3, 3] [ 3, 3].

. because E(X ) = .net vs 2010 Code 128 Code Set B 0 and E[XXT ] = In . Thus, AAT is the variance matrix of Y .

This argument gives rise to the following de nition of the n-variate normal distribution: Definition 5.1: Let Y be an n 1 random vector satisfying E(Y ) = and Var(Y ) = , where is nonsingular. Then Y is distributed Nn ( , ) if the density g(y) of Y is of the form g(y) = exp 1 (y )T 1 (y ) 2 .

( 2 )n det( ) (5.4). In the same way as before we can show that a nonsingular (hence one-to-one) linear transformation of a normal distribution is normal itself: Theorem 5.1: Let Z = a + BY, where Y is distributed Nn ( , ) and B is a nonsingular matrix of constants. Then Z is distributed Nn (a + B , B B T ).

Proof: First, observe that Z = a + BY implies Y = B 1 (Z a). Let h(z) be the density of Z and g(y) the density of Y . Then h(z) = g(y).

det( y/ z). = g(B 1 z B 1 a). det( (B 1 z B 1 a)/ z). = = = Q.E.D.

g( B 1 z B 1 a) g(B 1 (z a)) = . det(B). det(BBT ) exp 1 (B 1 (z a) )T 1 (B 1 (z a) ) 2 ( 2 )n det( ) det(BBT ) exp 1 (z a B )T (B B T ) 1 (z a B ) 2 . ( 2 )n det(B B T ).
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