Chapter 5

The Multivariate Normal Distribution

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Problem 5.1

In this chapter we have (so far) met three equivalent definitions of a multivariate normal distribution. Here is a fourth one: X\mathbf{X} is normal if and only if there exists an orthogonal transformation C\mathbf{C} such that the random vector CX\mathbf{CX} has independent, normal components. Show that this definition is indeed equivalent to the usual ones (e.g., by showing that it is equivalent to the first one).

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Problem 5.2

Suppose that XX and YY have a two-dimensional normal distribution with means 00, variances 11, and correlation coefficient ρ\rho, ∣ρ∣<1\left|\rho \right| < 1. Let (R,Θ)(R, \Theta ) be the polar coordinates. Determine the distribution of Θ\Theta.

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Problem 5.3

The random variables X1X_{1} and X2X_{2} are independent and N(0,1)N(0,1)-distributed. Set

Y1=X12−X22X12+X22andY2=2X1⋅X2X12+X22 . Y_{1} = \frac{X_{1}^{2}-X_{2}^{2}}{\sqrt{X_{1}^{2}+X_{2}^{2}}} \quad \text{and} \quad Y_{2} = \frac{2 X_{1} \cdot X_{2}}{\sqrt{X_{1}^{2}+X_{2}^{2}}} \, .

Show that Y1Y_{1} and Y2Y_{2} are independent, N(0,1)N(0,1)-distributed random variables.

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Problem 5.4

The random vector (X,Y)′(X, Y)' has a two-dimensional normal distribution with Var⁡[X]=Var⁡[Y]\operatorname {Var}\left[X\right] = \operatorname {Var}\left[Y\right]. Show that X+YX+Y and X−YX-Y are independent random variables.

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Problem 5.5

Suppose that XX and YY have a joint normal distribution with E[X]=E[Y]=0\mathbb {E}\left[X\right] = \mathbb {E}\left[Y\right] = 0, Var⁡[X]=σx2\operatorname {Var}\left[X\right] = \sigma_{x}^{2}, Var⁡[Y]=σy2\operatorname {Var}\left[Y\right] = \sigma_{y}^{2}, and correlation coefficient ρ\rho. Compute E[XY]\mathbb {E}\left[XY\right] and Var⁡[XY]\operatorname {Var}\left[XY\right].

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Remark.
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One may use the fact that XX and a suitable linear combination of XX and YY are independent.

Problem 5.6

The random variables XX and YY are independent and N(0,1)N(0,1)-distributed. Determine

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(a)

E[X∣X>Y]\mathbb {E}\left[X \mid X > Y\right],

(b)

E[X+Y∣X>Y]\mathbb {E}\left[X+Y \mid X > Y\right].

Problem 5.7

We know from Section 7 that if XX and YY are jointly normally distributed then they are independent iff they are uncorrelated. Now, let X∈N(0,1)X \in N(0,1) and c≥0c \geq 0. Define YY as follows:

Y={X,for ∣X∣≤c,−X,for ∣X∣>c. Y = \begin{cases} X, & \text{for } \left|X\right| \leq c, \\ -X, & \text{for } \left|X\right| > c. \end{cases}
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(a)

Show that Y∈N(0,1)Y \in N(0,1).

(b)

Show that XX and YY are not jointly normal.

Next, let g(c)=Cov⁡[X,Y]g(c) = \operatorname {Cov}\left[ X, Y \right].

(c)

Show that g(0)=−1g(0) = -1 and that g(c)→1g(c) \rightarrow 1 as c→∞c \rightarrow \infty. Show that there exists c0c_{0} such that g(c0)=0g(c_{0}) = 0 (i.e., such that XX and YY are uncorrelated).

(d)

Show that XX and YY are not independent (when c=c0c = c_{0}).

Problem 5.8

In Section 6 we found that conditional distributions of normal vectors are normal. The converse is, however, not true. Namely, consider the bivariate density

fX,Y(x,y)=C⋅exp⁡{−(1+x2)(1+y2)},−∞<x,y<∞, f_{X,Y}(x,y) = C \cdot \exp \left\{ -(1+x^{2})(1+y^{2})\right\} , \quad -\infty < x, y < \infty ,

where CC is a normalizing constant. This is not a bivariate normal density. Show that in spite of this the conditional distributions are normal, that is, compute the conditional densities fY∣X=x(y)f_{Y \mid X=x}(y) and fX∣Y=y(x)f_{X \mid Y=y}(x) and show that they are normal densities.

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Problem 5.9

Suppose that the random variables XX and YY are independent and N(0,σ2)N(0,\sigma^{2})-distributed.

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(a)

Show that X/Y∈C(0,1)X/Y \in C(0,1).

(b)

Show that X+YX+Y and X−YX-Y are independent.

(c)

Determine the distribution of (X−Y)/(X+Y)(X-Y)/(X+Y) (see also Problem 1.43(b)).

Problem 5.10

Suppose that the moment generating function of (X,Y)′(X,Y)' is

ψX,Y(t,u)=exp⁡{2t+3u+t2+atu+2u2}. \psi _{X,Y}(t,u) = \exp \left\{ 2t+3u+t^{2}+atu+2u^{2}\right\} .

Determine aa so that X+2YX+2Y and 2X−Y2X-Y become independent.

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Problem 5.11

Let X\mathbf{X} have a three-dimensional normal distribution. Show that if X1X_{1} and X2+X3X_{2}+X_{3} are independent, X2X_{2} and X1+X3X_{1}+X_{3} are independent, and X3X_{3} and X1+X2X_{1}+X_{2} are independent, then X1X_{1}, X2X_{2}, and X3X_{3} are independent.

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Problem 5.12

Let X1X_{1} and X2X_{2} be independent, N(0,1)N(0,1)-distributed random variables. Set Y1=X1−3X2+2Y_{1} = X_{1}-3X_{2}+2 and Y2=2X1−X2−1Y_{2} = 2X_{1}-X_{2}-1. Determine the distribution of

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(a)

Y\mathbf{Y}, and

(b)

Y1∣Y2=yY_{1} \mid Y_{2} = y.

Problem 5.13

Let X1X_{1}, X2X_{2}, and X3X_{3} be independent, N(1,1)N(1,1)-distributed random variables. Set U=2X1−X2+X3U = 2X_{1}-X_{2}+X_{3} and V=X1+2X2+3X3V = X_{1}+2X_{2}+3X_{3}. Determine the conditional distribution of VV given that U=3U = 3.

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Problem 5.14

Let X1,X2,X3X_{1}, X_{2}, X_{3} be independent N(2,1)N(2,1)-distributed random variables. Determine the distribution of X1+3X2−2X3X_{1}+3X_{2}-2X_{3} given that 2X1−X2=12X_{1}-X_{2} = 1.

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Problem 5.15

Let Y1,Y2Y_{1}, Y_{2}, and Y3Y_{3} be independent, N(0,1)N(0,1)-distributed random variables, and set

X1=Y1−Y3,X2=2Y1+Y2−2Y3,X3=−2Y1+3Y3. X_{1} = Y_{1}-Y_{3}, \quad X_{2} = 2Y_{1}+Y_{2}-2Y_{3}, \quad X_{3} = -2Y_{1}+3Y_{3}.

Determine the conditional distribution of X2X_{2} given that X1+X3=xX_{1}+X_{3} = x.

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Problem 5.16

The random variables X1X_{1}, X2X_{2}, and X3X_{3} are independent and N(0,1)N(0,1)-distributed. Consider the random variables

Y1=X2+X3,Y2=X1+X3,Y3=X1+X2. Y_{1} = X_{2}+X_{3}, \quad Y_{2} = X_{1}+X_{3}, \quad Y_{3} = X_{1}+X_{2}.

Determine the conditional density of Y1Y_{1} given that Y2=Y3=0Y_{2} = Y_{3} = 0.

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Problem 5.17

The random vector X\mathbf{X} has a three-dimensional normal distribution with mean vector 0\mathbf{0} and covariance matrix Λ\mathbf{\Lambda } given by

Λ=(20−1031−115). \mathbf{\Lambda } = \begin{pmatrix} 2 & 0 & -1 \\ 0 & 3 & 1 \\ -1 & 1 & 5 \end{pmatrix} .

Find the distribution of X2X_{2} given that X1−X3=1X_{1}-X_{3} = 1 and that X2+X3=0X_{2}+X_{3} = 0.

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Problem 5.18

The random vector X\mathbf{X} has a three-dimensional normal distribution with expectation 0\mathbf{0} and covariance matrix Λ\mathbf{\Lambda } given by

Λ=(12−1240−107). \mathbf{\Lambda } = \begin{pmatrix} 1 & 2 & -1 \\ 2 & 4 & 0 \\ -1 & 0 & 7 \end{pmatrix} .

Find the distribution of X3X_{3} given that X1=1X_{1} = 1.

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Problem 5.19

The random vector X\mathbf{X} has a three-dimensional normal distribution with expectation 0\mathbf{0} and covariance matrix Λ\mathbf{\Lambda } given by

Λ=(21−1130−105). \mathbf{\Lambda } = \begin{pmatrix} 2 & 1 & -1 \\ 1 & 3 & 0 \\ -1 & 0 & 5 \end{pmatrix} .

Find the distribution of X2X_{2} given that X1+X3=1X_{1}+X_{3} = 1.

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Problem 5.20

The random vector X\mathbf{X} has a three-dimensional normal distribution with mean vector μ=0\boldsymbol {\mu } = \mathbf{0} and covariance matrix

Λ=(3−21−220101). \mathbf{\Lambda } = \begin{pmatrix} 3 & -2 & 1 \\ -2 & 2 & 0 \\ 1 & 0 & 1 \end{pmatrix} .

Find the distribution of X1+X3X_{1}+X_{3} given that

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(a)

X2=0X_{2} = 0,

(b)

X2=2X_{2} = 2.

Problem 5.21

Let X∈N(μ,Λ)\mathbf{X} \in N(\boldsymbol {\mu }, \mathbf{\Lambda }), where

μ=(201)andΛ=(3−21−220102). \boldsymbol {\mu } = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} \quad \text{and} \quad \mathbf{\Lambda } = \begin{pmatrix} 3 & -2 & 1 \\ -2 & 2 & 0 \\ 1 & 0 & 2 \end{pmatrix} .

Determine the conditional distribution of X1−X3X_{1}-X_{3} given that X2=−1X_{2} = -1.

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Problem 5.22

Let X∈N(μ,Λ)\mathbf{X} \in N(\boldsymbol {\mu }, \mathbf{\Lambda }), where

μ=(201)andΛ=(3−21−220103). \boldsymbol {\mu } = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} \quad \text{and} \quad \mathbf{\Lambda } = \begin{pmatrix} 3 & -2 & 1 \\ -2 & 2 & 0 \\ 1 & 0 & 3 \end{pmatrix} .

Determine the conditional distribution of X1+X2X_{1}+X_{2} given that X3=1X_{3} = 1.

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Problem 5.23

The random vector X\mathbf{X} has a three-dimensional normal distribution with expectation μ\boldsymbol {\mu } and covariance matrix Λ\mathbf{\Lambda } given by

μ=(110)andΛ=(4−21−230101). \boldsymbol {\mu } = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \quad \text{and} \quad \mathbf{\Lambda } = \begin{pmatrix} 4 & -2 & 1 \\ -2 & 3 & 0 \\ 1 & 0 & 1 \end{pmatrix} .

Find the conditional distribution of X1+2X2X_{1}+2X_{2} given that

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(a)

X2−X3=1X_{2}-X_{3} = 1.

(b)

X2+X3=1X_{2}+X_{3} = 1.

Problem 5.24

The random vector X\mathbf{X} has a three-dimensional normal distribution with mean vector μ\boldsymbol {\mu } and covariance matrix Λ\mathbf{\Lambda } given by

μ=(10−2)andΛ=(3−21−24−11−12). \boldsymbol {\mu } = \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix} \quad \text{and} \quad \mathbf{\Lambda } = \begin{pmatrix} 3 & -2 & 1 \\ -2 & 4 & -1 \\ 1 & -1 & 2 \end{pmatrix} .

Find the conditional distribution of X1X_{1} given that X1=−X2X_{1} = -X_{2}.

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Problem 5.25

Let X\mathbf{X} have a three-dimensional normal distribution with mean vector and covariance matrix

μ=(111)andΛ=(21113−11−12), \boldsymbol {\mu } = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \quad \text{and} \quad \mathbf{\Lambda } = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 3 & -1 \\ 1 & -1 & 2 \end{pmatrix} ,

respectively. Set Y1=X1+X2+X3Y_{1} = X_{1}+X_{2}+X_{3} and Y2=X1+X3Y_{2} = X_{1}+X_{3}. Determine the conditional distribution of Y1Y_{1} given that Y2=0Y_{2} = 0.

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Problem 5.26

Let X∈N(0,Λ)\mathbf{X} \in N(\mathbf{0}, \mathbf{\Lambda }), where

Λ=(21−1130−105). \mathbf{\Lambda } = \begin{pmatrix} 2 & 1 & -1 \\ 1 & 3 & 0 \\ -1 & 0 & 5 \end{pmatrix} .

Find the conditional distribution of X1X_{1} given that X1=X2X_{1} = X_{2} and X1+X2+X3=0X_{1}+X_{2}+X_{3} = 0.

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Problem 5.27

The random vector X\mathbf{X} has a three-dimensional normal distribution with expectation 0\mathbf{0} and covariance matrix Λ\mathbf{\Lambda } given by

Λ=(210121012). \mathbf{\Lambda } = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix} .

Find the distribution of X2X_{2} given that X1=X2=X3X_{1} = X_{2} = X_{3}.

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Problem 5.28

Let X∈N(0,Λ)\mathbf{X} \in N(\mathbf{0}, \mathbf{\Lambda }), where

Λ=(1−1232−122−132−14). \mathbf{\Lambda } = \begin{pmatrix} 1 & -\frac{1}{2} & \frac{3}{2} \\ -\frac{1}{2} & 2 & -1 \\ \frac{3}{2} & -1 & 4 \end{pmatrix} .

Determine the conditional distribution of (X1,X1+X2)′(X_{1}, X_{1}+X_{2})' given that X1+X2+X3=0X_{1}+X_{2}+X_{3} = 0.

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Problem 5.29

Suppose that the characteristic function of (X,Y,Z)′(X,Y,Z)' is

φ(s,t,u)=exp⁡{2is−s2−2t2−4u2−2st+2su}. \varphi (s,t,u) = \exp \left\{ 2is-s^{2}-2t^{2}-4u^{2}-2st+2su\right\} .

Compute the conditional distribution of X+ZX+Z given that X+Y=0X+Y = 0.

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Problem 5.30

Let X1X_{1}, X2X_{2}, and X3X_{3} have a joint moment generating function as follows:

ψ(t1,t2,t3)=exp⁡{2t1−t3+t12+2t22+3t32+2t1t2−2t1t3}. \psi (t_{1},t_{2},t_{3}) = \exp \left\{ 2t_{1}-t_{3}+t_{1}^{2}+2t_{2}^{2}+3t_{3}^{2}+2t_{1}t_{2}-2t_{1}t_{3}\right\} .

Determine the conditional distribution of X1+X3X_{1}+X_{3} given that X1+X2=1X_{1}+X_{2} = 1.

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Problem 5.31

The moment generating function of (X,Y,Z)′(X,Y,Z)' is

ψ(s,t,u)=exp⁡{s22+t2+2u2−st2+3su2−tu2}. \psi (s,t,u) = \exp \left\{ \frac{s^{2}}{2}+t^{2}+2u^{2}-\frac{st}{2}+\frac{3su}{2}-\frac{tu}{2}\right\} .

Determine the conditional distribution of XX given that X+Z=0X+Z = 0 and Y+Z=1Y+Z = 1.

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Problem 5.32

Suppose (X,Y,Z)′(X,Y,Z)' is normal with density

C⋅exp⁡{−12(4x2+3y2+5z2+2xy+6xz+4zy)}, C \cdot \exp \left\{ -\frac{1}{2}(4x^{2}+3y^{2}+5z^{2}+2xy+6xz+4zy)\right\} ,

where CC is a normalizing constant. Determine the conditional distribution of XX given that X+Z=1X+Z = 1 and Y+Z=0Y+Z = 0.

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Problem 5.33

Let XX and YY be random variables, such that

Y∣X=x∈N(x,τ2)withX∈N(μ,σ2). Y \mid X = x \in N(x, \tau ^{2}) \quad \text{with} \quad X \in N(\mu , \sigma ^{2}).
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(a)

Compute E[Y]\mathbb {E}\left[Y\right], Var⁡[Y]\operatorname {Var}\left[Y\right] and Cov⁡[X,Y]\operatorname {Cov}\left[ X, Y \right].

(b)

Determine the distribution of the vector (X,Y)′(X,Y)'.

(c)

Determine the (posterior) distribution of X∣Y=yX \mid Y = y.

Problem 5.34

Let XX and YY be jointly normal with means 00, variances 11, and correlation coefficient ρ\rho. Compute the moment generating function of X⋅YX \cdot Y for

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(a)

ρ=0\rho = 0, and

(b)

general ρ\rho.

Problem 5.35

Suppose X1X_{1}, X2X_{2}, and X3X_{3} are independent and N(0,1)N(0,1)-distributed. Compute the moment generating function of Y=X1X2+X1X3+X2X3Y = X_{1}X_{2}+X_{1}X_{3}+X_{2}X_{3}.

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Problem 5.36

If XX and YY are independent, N(0,1)N(0,1)-distributed random variables, then X2+Y2∈χ2(2)X^{2}+Y^{2} \in \chi^{2}(2) (recall Exercise 3.3.6). Now, let XX and YY be jointly normal with means 00, variances 11, and correlation coefficient ρ\rho. In this case X2+Y2X^{2}+Y^{2} has a noncentral χ2(2)\chi^{2}(2)-distribution. Determine the moment generating function of that distribution.

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Problem 5.37

Let (X,Y)′(X,Y)' have a two-dimensional normal distribution with means 00, variances 11, and correlation coefficient ρ\rho, ∣ρ∣<1\left|\rho \right| < 1. Determine the distribution of (X2−2ρXY+Y2)/(1−ρ2)(X^{2}-2\rho XY+Y^{2})/(1-\rho^{2}) by computing its moment generating function.

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Remark.
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Recall Example 8.6 and Remark 9.1.

Problem 5.38

Let X1,X2,…,XnX_{1}, X_{2}, \ldots , X_{n} be independent, N(0,1)N(0,1)-distributed random variables, and set X‾k=1k−1∑i=1k−1Xi\overline{X}_{k} = \dfrac {1}{k-1}\sum_{i=1}^{k-1} X_{i}, 2≤k≤n2 \leq k \leq n. Show that

Q=∑k=2nk−1k(Xk−X‾k)2 Q = \sum _{k=2}^{n} \frac{k-1}{k}(X_{k}-\overline{X}_{k})^{2}

is χ2\chi^{2}-distributed. What is the number of degrees of freedom?

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Problem 5.39

Let X1X_{1}, X2X_{2}, and X3X_{3} be independent, N(1,1)N(1,1)-distributed random variables. Set U=X1+X2+X3U = X_{1}+X_{2}+X_{3} and V=X1+2X2+3X3V = X_{1}+2X_{2}+3X_{3}. Determine the constants aa and bb so that E[(U−a−bV)2]\mathbb {E}\left[(U-a-bV)^{2}\right] is minimized.

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Problem 5.40

Let XX and YY be independent, N(0,1)N(0,1)-distributed random variables. Then X+YX+Y and X−YX-Y are independent; see Example 7.1. The purpose of this problem is to point out a (partial) converse. Suppose that XX and YY are independent random variables with common distribution function FF. Suppose, further, that FF is symmetric and that σ2=E[X2]<∞\sigma^{2} = \mathbb {E}\left[X^{2}\right] < \infty. Let φ\varphi be the characteristic function of XX (and YY). Show that if X+YX+Y and X−YX-Y are independent then we have

φ(t)=(φ(t/2))4. \varphi (t) = \left(\varphi (t/2)\right)^{4} .

Use this relation to show that φ(t)=e−σ2t2/2\varphi (t) = e^{-\sigma^{2}t^{2}/2}. Finally, conclude that FF is the distribution function of a normal distribution (N(0,σ2))(N(0,\sigma^{2})).

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Remark.
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Remark 1. The assumptions that the distribution is symmetric and the variance is finite are not necessary. However, without them the problem becomes much more difficult.

Remark 2. Results of this kind are called characterization theorems. Another characterization of the normal distribution is provided by the following famous theorem due to the Swedish probabilist and statistician Harald Cramer (1893--1985): If XX and YY are independent random variables such that X+YX+Y has a normal distribution, then XX and YY are both normal.