Section 4.4

Suppose X and Y are two random variables such that

(1)

(2)

(3) Y | X = x has a normal distribution for any real number x.

E(Y | X = x) is a linear function of x,

Var(Y | X = x) is a constant for each x.

Then we know that

(1)

(2)

E(Y | X = x) =Y

Y + — (x – X)X

(because when a conditional mean is a linear function, that linear function must be the least squares line).

Var(Y | X = x) =

–

2Y|x f1(x) dx =

Var(Y | X = x) =

–

2Y|x f1(x) dx =

–

y – h(y | x) dy f1(x) dx =

–

Y

Y + — (x – X)X

2

–

f(x,y) dy dx =

–

Y

(y – Y) – — (x – X) X

2

E Y

(Y – Y) – — (X – X) = X

2

E Y

(Y – Y) – — (X – X) = X

2

E Y 2

Y

(Y – Y)2 – 2 — (X – X)(Y – Y) + 2 — (X – X)2 = X 2

X

Y 2Y

E[(Y – Y)2] – 2 — E[(X – X)(Y – Y)] + 2 — E[(X – X)2] = X 2

X

2Y –

Y2 — Cov(X,Y) + X

22Y = 2

Y – 222Y + 22

Y = 2Y (1 – 2)

Suppose X and Y are two random variables such that

(1)

(2)

(3)

Then we know that

(1)

(2)

E(Y | X = x) =Y

Y + — (x – X)X

Var(Y | X = x) = 2Y (1 – 2)

h(y | x) =

Y | X = x has a normal distribution for any real number x,

E(Y | X = x) is a linear function of x,

Var(Y | X = x) is a constant for each x.

exp

Y

– y – Y + — (x – X)X

2

22Y (1 – 2)

(2)1/2Y (1 – 2)1/2

– < y <

(3) For all x,

Now, suppose that X has a normal distribution. Then, the joint p.d.f. of (X,Y) is f(x,y) = f1(x) h(y | x) =

exp– (x – X)2

22X

(2)1/2X

exp

Y

– y – Y + — (x – X)X

2

22Y (1 – 2)

(2)1/2Y (1 – 2)1/2

exp

Y 2Y

(y – Y)2 – 2 — (x – X)(y – Y) + 2 — (x – X)2

X 2X

22Y (1 – 2)

1 x – X– — ——— – 2 X

2

2X Y (1 – 2)1/2

=

=

exp

Y 2Y

(y – Y)2 – 2 — (x – X)(y – Y) + 2 — (x – X)2

X 2X

22Y (1 – 2)

1 x – X– — ——— – 2 X

2

2X Y (1 – 2)1/2=

exp

y – Y x – X y – Y x – X ——— – 2 ——— ——— + 2 ——— Y X Y X

2 (1 – 2)

1 x – X– — ——— – 2 X

2

2X Y (1 – 2)1/2=

2 2

exp

y – Y x – X y – Y x – X ——— – 2 ——— ——— + 2 ——— Y X Y X

2 (1 – 2)

1 x – X– — ——— – 2 X

2

2X Y (1 – 2)1/2=

2 2

exp

2X Y (1 – 2)1/2

2 (1 – 2)

2 x – X——— – X

x – X y – Y y – Y 2 ——— ——— + ——— X Y Y

2

This is called a bivariate normal p.d.f., and (X, Y) are said to have a bivariate normal distribution with correlation coefficient .

(Note: Completing this derivation is Text Exercise 4.4-2.)

– < x <

– < y <

exp

2X Y (1 – 2)1/2

2 (1 – 2)

2 x – X——— – X

x – X y – Y y – Y 2 ——— ——— + ——— X Y Y

2

– < x <

– < y <

From our derivation of the bivariate normal p.d.f., we know that

(1) X has a N( X , 2X ) distribution,

(2) Y | X = x has a N( , ) distribution.Y

Y + — (x – X)X

2Y (1 – 2)

From the symmetry of the bivariate normal p.d.f., we find that

(1) Y has a distribution,

(2) X | Y = y has a N( , ) distribution.X

X + — (y – Y)Y

2X (1 – 2)

N( Y , 2Y )

Important Theorem in the Text:

If (X, Y) have a bivariate normal distribution with correlation coefficient , then X and Y are independent if and only if = 0.

Theorem 4.4-1

1.

(a)

Random variables X and Y are respectively the height (inches) and weight (lbs.) of adult males in a particular population. It is known that X and Y have a bivariate normal distribution with

X = 70, Y = 150, X2 = 9, Y

2 = 225, and = +0.4 .

Find each of the following:

P(140 < Y < 170)

Y has a distribution.N( , )150 225

P(140 < Y < 170) = 140 – 150 Y – 150 170 – 150P( ———— < ——— < ———— ) = 15 15 15

P(– 0.67 < Z < 1.33) = (1.33) – (– 0.67) =

(1.33) – (1 – (0.67)) = 0. 9082 – (1 – 0.7486) = 0.6568

1.-continued

(b) P(X > 65)

X has a distribution.N( , ) 70 9

P(X > 65) = X – 70 65 – 70P( ——— > ———) = 3 3

P(Z > –1.67) =

1 – (– 1.67) = 1 – (1 – (1.67)) = 1 – (1 – 0.9525) = 0.9525

(c)

(d)

E(Y | X = x)

Var(Y | X = x)

=Y

Y + — (x – X) =X

15150 + 0.4 — (x – 70) =

310 + 2x

= 2Y (1 – 2) = 225(1 – 0.42) = 189

1.-continued

(e) P(140 < Y < 170 | X = 72)

Y | X = 72 has a distribution.N( , )154 189

P(140 < Y < 170 | X = 72 ) = 140 – 154 Y – 154 170 – 154P( ———— < ——— < ———— ) = 189 189 189

P(– 1.02 < Z < 1.16) = (1.16) – (– 1.02) =

(1.16) – (1 – (1.02)) = 0. 8770 – (1 – 0.8461) = 0.7231

Y | X = 68 has a distribution.N( , )146 189

P(140 < Y < 170 | X = 68 ) = 140 – 146 Y – 146 170 – 146P( ———— < ——— < ———— ) = 189 189 189

P(– 0.44 < Z < 1.75) = (1.75) – (– 0.44) =

(1.75) – (1 – (0.44)) = 0. 9599 – (1 – 0.6700) = 0.6299

(f) P(140 < Y < 170 | X = 68)

1.-continued

(g) P(140 < Y < 170 | X = 70)

Y | X = 70 has a distribution.N( , )150 189

P(140 < Y < 170 | X = 70 ) = 140 – 150 Y – 150 170 – 150P( ———— < ——— < ———— ) = 189 189 189

P(– 0.73 < Z < 1.45) = (1.45) – (– 0.73) =

(1.45) – (1 – (0.73)) = 0.9265 – (1 – 0.7673) = 0.6938

(h)

(i)

E(X | Y = y)

Var(X | Y = y)

=X

X + — (y – Y) =Y

370 + 0.4 — (y – 150) =

1558 + 0.08y

= 2X (1 – 2) = 9(1 – 0.42) = 7.56

1.-continued

(j) P(X > 65 | Y = 140)

X | Y = 140 has a distribution.N( , )69.2 7.56

P(X > 65 | Y = 140 ) = X – 69.2 65 – 69.2P( ———— > ———— ) = 7.56 7.56

P(Z > – 1.53) = 1 – (– 1.53) = 1 – (1 – (1.53)) =

1 – (1 – 0.9370) = 0.9370

2.

(a)

(b)

(c)

(d)

(e)

(f)

The random variables X and Y have a bivariate normal distribution with X = 10, Y = 5, X

2 = 16, Y2 = 36, and = –0.4 .

Find each of the following:

E(X + Y) = 10 + 5 = 15

Var(X + Y) = 16 + 36 + 2(–0.4)(16)1/2(36)1/2 = 32.8

E(X – Y) = 10 – 5 = 5

Var(X – Y) = 16 + 36 – 2(–0.4)(16)1/2(36)1/2 = 71.2

E(X – 3Y) = 10 – 3(5) = – 5

Var(X – 3Y) =16 + (–3)2(36) + (2)(–3)(–0.4)(16)1/2(36)1/2 = 397.6

Complete #2 for homework!

3.

(a)

(b)

Suppose the random variables X and Y have joint p.d.f.

2e + xy e

– (x2 + y2)/2 1 – (x2 + y2)/2

f(x,y) = – < x < if – < y <

Are X and Y independent? Why or why not?

Since f(x,y) cannot be factored into the product of a function of x alone and a function of y alone, then X and Y cannot possibly be independent.

f1(x) =

–

dy =2

e + xy e– (x2 + y2)/2 1 – (x2 + y2)/2

Find the marginal p.d.f. of X and the marginal p.d.f. of Y.

–

2dy +

e– (x2 + y2)/2

–

2dy

xy e1 – (x2 + y2)/2

This is a bivariate normal p.d.f. with

X = , 2X = , Y = , 2

Y = , and = .This is an odd function, which implies that the integral of this function is .Consequently, f1(x) is a N(0,1) p.d.f., and

similarly we see that f2(y) is a N(0,1) p.d.f.

0 1 0 1 0

Decide whether each of the following statements is true or false: (c)

If (X, Y) have a bivariate normal distribution, then the marginal distribution for each of X and Y must be a normal distribution.

If the marginal distribution for each of X and Y is a normal distribution, then (X, Y) must have a bivariate normal distribution.

False

True

0