1. Show that E(Z)
= 0 (the zero vector in Rn).Virtual Laboratories > Special Distributions > 1 2 3 4 5 6 7 [8] 9 10 11 12 13 14 15
The multivariate normal distribution is a natural generalization of the bivariate normal distribution. The exposition is very compact and elegant using expected value and covariance matrices, and would be horribly complex without these matrices. Thus, this section requires prerequisite knowledge of linear algebra at the undergraduate level.
Suppose that Z1, Z2, ..., Zn are independent each with the standard normal distribution. The random vector Z = (Z1, Z2, ..., Zn) is said to have the n-dimensional standard normal distribution.
1. Show that E(Z)
= 0 (the zero vector in Rn).
2. Show that VC(Z)
= I (the n × n identity matrix).
3. Show that Z
has density function
g(z) = [1 / (2
)n/2] exp(-zTz
/ 2) for z in Rn.
4. Show that Z
has moment generating function given by
E[exp(tTZ)] = exp(tTt / 2) for t in Rn.
Now suppose that Z has the n-dimensional standard normal distribution. Suppose that µ is a vector in Rn and that A is an invertible n × n matrix. The random vector X = µ + AZ. is said to have an n-dimension normal distribution.
5. Show that E(X)
= µ.
6. Show that VC(X)
= AAT and that this matrix is invertible and hence
positive definite.
7. Let V
= VC(X) = AAT. Use the multivariate change of variables theorem to show that X
has density function
f(x) = {1 / [(2
8. Show that X
has moment generating function given by
E[exp(tTX)] = exp(tTµ + tTVt / 2) for t in Rn.
Note that the matrix A that occurs in the transformation is not unique, but of course the variance-covariance matrix V is unique. In general, for a given positive definite matrix V, there are many invertible matrices A such that AAT = V. A theorem in matrix theory states that there is a unique lower triangular matrix L with this property.
9. Identify the
lower triangular matrix L for the bivariate normal distribution.
The multivariate normal distribution is invariant under two basic types of transformations: affine transformation with an invertible matrix, and the formation of subsequences.
10. Suppose that X
has an n-dimensional normal distribution. Suppose also that a
is in Rn and that B is an
invertible n × n matrix. Show that Y = a
+ BX has a multivariate normal distribution. Identify the mean
vector and the variance-covariance matrix of Y.
11. Suppose that X
has an n-dimensional normal distribution. Show that any permutation of the
coordinates of X also has an n-dimensional normal
distribution. Identify the mean vector and the variance-covariance matrix. Hint:
Permuting the coordinates of X corresponds to multiplication of X
by a permutation matrix--a matrix of 0's and 1's in which each row and column has
a single 1.
12. Suppose that X
= (X1, X2, ..., Xn) has an n-dimensional
normal distribution. Show that if k < n, W
= (X1, X2, ..., Xk) has a k-dimensional
normal distribution. Identify the mean vector and the variance-covariance matrix.
13. Use the
results of Exercises 11 and 12 to show that if X = (X1,
X2, ..., Xn) has an n-dimensional normal
distribution and if i1, i2, ..., ik
are distinct indices, then W = (Xi1,
Xi2, ..., Xik) has a k-dimensional
normal distribution.
14. Suppose that X
has an n-dimensional normal distribution, a is in
Rn,
and B is an m × n matrix with linearly
independent rows (thus, m 
n). Show
that Y = a + BX
has an m-dimensional normal distribution. Hint: There exists an
invertible n × n matrix C for which the first
m rows are the rows of B. Now use Exercises 10 and 12.
Note that the results in Exercises 10, 11, 12, and 13 are special cases of the result in Exercise 14.
15. Suppose that X
has an n-dimensional normal distribution, Y has an m-dimensional
normal distribution, and that X and Y
are independent. Show that (X, Y) has
an n + m-dimensional normal distribution. Identify the mean vector and
the variance-covariance matrix.
16. Suppose that X
is a random vector in Rn, Y
is a random vector in Rm, and that (X,
Y) has an n + m-dimensional normal
distribution. Show that X and Y are
independent if and only if cov(X, Y)
= 0.