The Multivariate Normal Distribution
Now let the components of X = (x,…, xn)T be independent, standard normally distributed random variables. Then, E(X) = 0(e Kn) and Var(X) = In. Moreover, the joint density f (x) = f (x,.xn )of X in this case is the product of the standard normal marginal densities:
A exp(x2
f(x) = f(X1, …,xn) = П ——————
j=1 jln
_ exp (2 ЕП=1xj) exp (1XTx)
(fbn )n (fbn )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 = д + AX, where д = (д1,…, дп )T is a vector of constants and A is a nonsingular n x n matrix with nonrandom elements. Because A is nonsingular and therefore invertible, this transformation is a onetoone mapping with inverse X = A1 (Y – д). Then the density function g(y) of Yis equal to
f (x )det(9 x/дy ) f (A1 y – AV)det(d(A1 y
f (A1 y – AV)det(A1) =
exp [2(y – д)T(A 1)TA 1(y – д)]
(V2n )n det( A)
exp [2(y – д)T(AAT)1(y – д)] (V2n )V det( AAT) .
Observe that д is the expectation vector of Y : E(Y) = д + A (E(X)) = д. But what is AAT? We know from (5.2) that Var(Y) = E [iYT] – дд’1′. Therefore, substituting Y = д + AX yields
Var(Y) = E[(д + AX)(^T + XTAT) – ддТ]
= д( E (XT)) AT + A(E ^д^ + A( E (XXT)) AT = AAT

because E(X) = 0 and E[XXT] = In. Thus, AAT is the variance matrix of Y. This argument gives rise to the following definition of the nvariate normal distribution:
Definition 5.1: Let Y be an n x 1 random vector satisfying E(Y) = д and Var(Y) = £, where £ is nonsingular. Then Y is distributed Nn (д, £) if the density g(y) ofY is of the form
In the same way as before we can show that a nonsingular (hence onetoone) linear transformation of a normal distribution is normal itself:
Theorem 5.1: Let Z = a + BY, where Y is distributed Nn(д, £) and B is a nonsingular matrix of constants. Then Z is distributed Nn (a + B д, B£ BT).
Proof: First, observe that Z = a + BY implies Y = B1(Z – a). Let h(z) be the density of Z and g(y) the density of Y. Then
h(z) = g(y)det(9y/d z)
= g(B1 z – B1a)det(9(B1 z – B1a)/9z)
_ g(B1 z – B1a) g(B1(z – a))
det(B) det(BBT)
exp [2(B1(z – a) – д)^1(B1(z – a) – д)]
(V2n)V det(£Vdet(BBT)
exp [2(z – a – B ifT(B£BT)1(z – a – B д)]
= (V2n )V det( B £ B T) ‘
Q. E.D.
I will now relax the assumption in Theorem 5.1 that the matrix B is a nonsingular n x n matrix. This more general version of Theorem 5.1 can be proved using the momentgenerating function or the characteristic function of the multivariate normal distribution.
Theorem 5.2: Let Y be distributed Nn(д, E). Then the momentgenerating function of Y is m(t) = exp(tTд + tTEt/2), and the characteristic of Y is tp(t) = exp(i ■ tтд — tTEt/2).
Proof: We have
Because the last integral is equal to 1, the result for the momentgenerating function follows. The result for the characteristic function follows from p(t) = m(i ■ t). Q. E.D.
Theorem 5.3: Theorem 5.1 holds for any linear transformation Z = a + BY.
Proof: Let Z = a + BY, where B is m x n. It is easy to verify that the characteristic functionof Z is vz(t) = E[exp(i ■ tTZ)] = E[exp(i ■ tT(a + BY))] =
exp(i ■ tTa)E[exp(i ■ tTBY)] = exp(i ■ (a + Bд)тt – 1 tTBEBTt). Theorem
Leave a reply