Multivariate Normal Distribution and Generalized Distance Analysis

Running head: Assignment 1
Assignment 1
Name of the student
Name of the University
Author Note

Assignment 1
Question 1:
a) The pdf of the multivariate normal distribution is given by,
f x ( x p∗1 )=c∗e
( − ( x−μ ) T
∑
( −1 )
( x−μ )
2 }
Here, μ = E(Xp*1) and ∑ = var(Xp*1) > 0 and c is a constant which is given by
c= 1
( 2∗π )
p
2 ¿ ¿
Here, ¿ ∑∨¿ is the value of the determinant of ∑.
Now, by the method of singular value decomposition ∑ = UD U T = U D (1
2 ) D ( 1
2 )U T
Here, U = eigenvector matrix of and
D=diagonal of eigenvalues λ 1 , λ 2 , λ 3 ... . λp
Now, in order to generate the random vectors from multivariate n∑ormal distribution
Np(μ,∑) the following methods must be followed.
Step 1: At first the eigenvalues λ 1 , λ 2 , λ 3 ,... λp and the corresponding eigenvectors
Uj of ∑ and then D^(0.5) = diag(λ 10.5 , λ 20.5 , λ 30.5 , ... λ p0.5 ¿ is needed to be
calculated.
Then the matrix A = U D
1
2 ( or A = UD^(1/2)U^(T)) is formed.
Step 2: Now, the z values z1,z2...zp from N(0,1) is obtained. Now, Let z =
( z 1 , ... , zp ) T
Step 3: Now, one random vector yp*1 = Az + μ from the multivariate normal
distribution Np(μ,∑) can be obtained.
b) Now, the multivariate normal distribution is chosen as 4-variate normal distribution.
The R code to generate a random vector from 4-variate normal distribution in SVD
method is given below.

Assignment 1
R-code:
m = c(2,3,5,1) #defining the means of quadravariate normal distribution
s = cbind(c(2,-1.2,-1,1),c(-1.2,4,1.5,1),c(-1,1.5,1,-0.5),c(1,1,-0.5,4)) # defining the
variance matrix S
lam = eigen(s)$values #calculating eigen values of S
U = eigen(s)$vectors #calculating eigen vectors of S
A = U%*%diag(sqrt(lam)) # creating the matrix A
z = rnorm(4) # generating 4 independent random numbers from standard normal
distribution.
A%*%z+m # generating one random vector from the quadravariate normal
distribution
Output:
[,1]
[1,] 2.9869710
[2,] 2.2991517
[3,] 5.2024936
[4,] -0.7154101
So, the generated random vector is (2.987, 2.299, 5.202, -0.715)
c) Now, according to question 2 the 3-variate normal distribution is used to generate 100
random vector of size 3*1 from mean μ = (1,-2,2)^T and
Variance = ∑ = (1 1 1
1 3 2
1 2 2 ). Here the target distribution is normal distribution and as
multivariate normal distribution is used for random vector generation, hence each
random vector must be a sample from a normal distribution.

Assignment 1
R code:
m = c(1,-2,2) #defining the means of quadravariate normal distribution
s = cbind(c(1,1,1),c(1,3,2),c(1,2,2)) # defining the variance matrix S
lam = eigen(s)$values #calculating eigen values of S
U = eigen(s)$vectors #calculating eigen vectors of S
A = U%*%diag(sqrt(lam)) # creating the matrix A
z = rnorm(3) # generating 3 independent random numbers from standard normal
distribution.
for (x in c(1:100)) print(A%*%z+m)
Question 2:
a) The linear transformation of the multivariate normal distribution x= (x1,x2,x3)^T
with mean μ = ( 1
−2
2 ) and ∑ = (1 1 1
1 3 2
1 2 2 ) will also follow a multivariate normal
distribution.
Now, μ(x1,x2) = ( 1
−2 ), ∑(x1,x2) = (1 1
1 3 )
Now, here the linear combination is z= 2x1 – 3x2.
Or, z = BX
Where, B = (2 -3)
So, the mean of z, E(z) = Bμ = (2 -3) ( 1
−2 ) = 2 + 6 = 8
Variance of z Var(z) = B ∑ ( x 1 , x 2 ) BT = (2 -3)(1 1
1 3 )( 2
−3 ) = (2 -3)(2−3
2−9 )
= (2 -3)(−1
−7 ) = -2 +21 = 19.
So, the distribution of 2x1 -3x2 is a normal distribution with mean 8 and variance 19.