Sections 4.1, 4.2, 4.3
Important Definitions in the Text:
The definition of joint probability mass function (joint p.m.f.)
Definition 4.1-1
The definitions of marginal probability mass function (marginal
p.m.f.) and the independence of random variables Definition 4.1-2
If the joint p.m.f. of (X, Y) is f(x,y), and S is the corresponding outcome
space, then the mathematical expectation, or expected value, of u(X,Y)
is
If the marginal p.m.f. of X is f1(x), and S1 is the corresponding outcome
space, then E[v(X)] can be calculated from either
An analogous statement can be made about E[v(Y)] .
1. Twelve bags each contain two pieces of candy, one red and one
green. In two of the bags each piece of candy weighs 1 gram; in
three of the bags the red candy weighs 2 grams and the green candy
weighs 1 gram; in three of the bags the red candy weighs 1 gram and
the green candy weighs 2 grams; in the remaining four bags each
piece of candy weighs 2 grams. One bag is selected at random and
the following random variables are defined:
1/4
1/3
2
X = weight of the red candy ,
y
Y = weight of the green candy .
1/6
1/4
1
The space of (X, Y) is {(1,1) (1,2) (2,1) (2,2)}.
The joint p.m.f. of (X, Y) is f(x, y) =
1
—
6
1
—
4
1
—
3
1
x
2
if (x, y) = (1, 1)
if (x, y) = (1, 2) , (2, 1)
if (x, y) = (2, 2)
The marginal p.m.f. of X is f1(x) =
The marginal p.m.f. of Y is f2(y) =
5 / 12 if x = 1
7 / 12 if x = 2
5 / 12 if y = 1
7 / 12 if y = 2
A formula for the joint p.m.f. of (X,Y) is f(x, y) =
x+y
—— if (x, y) = (1, 1) , (1, 2) , (2, 1) , (2, 2)
12
A formula for the marginal p.m.f. of X is f1(x) =
x+1 x+2
2x + 3
f(x, 1) + f(x, 2) = —— + —— = ———
12
12
12
A formula for the marginal p.m.f. of Y is f2(y) =
1+y 2+y
2y + 3
f(1, y) + f(2, y) = —— + —— = ———
12
12
12
if x = 1, 2
if y = 1, 2
Sections 4.1, 4.2, 4.3
Important Definitions in the Text:
The definition of joint probability mass function (joint p.m.f.)
Definition 4.1-1
The definitions of marginal probability mass function (marginal
p.m.f.) and the independence of random variables Definition 4.1-2
If the joint p.m.f. of (X, Y) is f(x,y), and S is the corresponding outcome
space, then the mathematical expectation, or expected value, of u(X,Y)
is E[u(X,Y)] =  u(x,y)f(x,y)
(x,y)  S
If the marginal p.m.f. of X is f1(x), and S1 is the corresponding outcome
space, then E[v(X)] can be calculated from either
or
 v(x)f1(x)
 v(x)f(x,y)
x  S1
(x,y)  S
An analogous statement can be made about E[v(Y)] .
1. - continued
E(X) = (1)(5/12) + (2)(7/12) = 19/12
E(X2) = (1)2(5/12) + (2)2(7/12) = 11/4
Var(X) = 11/4 – (19/12)2 = 35/144
E(Y) = (1)(5/12) + (2)(7/12) = 19/12
E(Y2) = (1)2(5/12) + (2)2(7/12) = 11/4
Var(Y) = 11/4 – (19/12)2 = 35/144
f(x, y)  f1(x)f2(y)
Since _________________________,
then the random variables X
are not
and Y _______________
independent.
Using the joint p.m.f., E(X + Y) =
(1+1)(1/6) + (1+2)(1/4) + (2+1)(1/4) + (2+2)(1/3) = 19 / 6
Alternatively, E(X + Y) = E(X) + E(Y) = 19/12 + 19/12 = 19 / 6
Using the joint p.m.f., E(X – Y) =
(1–1)(1/6) + (1–2)(1/4) + (2–1)(1/4) + (2–2)(1/3) = 0
Alternatively, E(X – Y) = 19/12 – 19/12 = 0
E(X + Y) can be interpreted as the mean of
the total weight of candy in the bag.
E(X – Y) can be interpreted as the mean of
how much more the red candy in the bag weighs than the
green candy.
E(XY) = (1)(1)(1/6) + (1)(2)(1/4) + (2)(1)(1/4) + (2)(2)(1/3) = 5/2
Cov(X,Y) =
1. - continued
=
The least squares lines for predicting Y from X is
The least squares lines for predicting X from Y is
The conditional p.m.f. of
Y | X = 1 is
Y | X = 2 is
For x = 1, 2, a formula for the conditional p.m.f. of Y | X = x is
1. - continued
The conditional p.m.f. of
X | Y = 1 is
X | Y = 2 is
For y = 1, 2, a formula for the conditional p.m.f. of X | Y = y is
E(Y | X = 1) =
E(Y2 | X = 1) =
Var(Y | X = 1) =
E(Y | X = 2) =
E(Y2 | X = 2) =
Var(Y | X = 2) =
Is the conditional mean of Y given X = x a linear function of the
given value, that is, can we write E(Y | X = x) = a + bx ?
1. - continued
E(X | Y = 1) =
E(X2 | Y = 1) =
Var(X | Y = 1) =
E(X | Y = 2) =
E(X2 | Y = 2) =
Var(X | Y = 2) =
Is the conditional mean of X given Y = y a linear function of the
given value, that is, can we write E(X | Y = y) = c + dy ?
2. An urn contains six chips, one $1 chip, two $2 chips, and three $3
chips. Two chips are selected at random and without replacement.
The following random variables are defined:
X = dollar value of the first chip selected ,
Y = dollar value of the second chip selected .
The space of (X, Y) is {(1,2) (1,3) (2,1) (2,2) (2,3) (3,1) (3,2) (3,3)}.
3
1/10
1/5
1/5
y 2
1/15
1/15
1/5
1/15
1/10
1
1
2
x
3
The joint p.m.f. of (X, Y) is f(x, y) =
1
— if (x, y) = (2, 3) , (3, 2) , (3, 3)
5
1
— if (x, y) = (1, 3) , (3, 1)
10
1
— if (x, y) = (1, 2) , (2, 1) , (2, 2)
15
2. - continued
The marginal p.m.f. of X is f1(x) =
1/6
1/3
1/2
if x = 1
if x = 2
if x = 3
The marginal p.m.f. of Y is f2(y) =
1/6
1/3
1/2
if y = 1
if y = 2
if y = 3
A formula for the joint p.m.f. of (X,Y) is f(x, y) =
(There seems to be no easy formula.)
A formula for the marginal p.m.f. of X is f1(x) = x / 6 if x = 1, 2, 3
A formula for the marginal p.m.f. of Y is f2(y) = y / 6 if y = 1, 2, 3
E(X) = 7 / 3
E(X2) = 6
Var(X) = 6 – (7 / 3)2 = 5 / 9
E(Y) = 7 / 3
E(Y2) = 6
Var(Y) = 6 – (7 / 3)2 = 5 / 9
f(x, y)  f1(x)f2(y)
Since _________________________,
then the random variables X
are not
and Y _______________
independent.
P(X + Y < 4) = P[(X,Y) = (1,2)] + P[(X,Y) = (2,1)] = 1 / 15 + 1 / 15 =
2 / 15
Using the joint p.m.f., E(XY) =
(1)(2)(2/30) + (1)(3)(3/30) + (2)(1)(2/30) + (2)(2)(2/30) +
(2)(3)(6/30) + (3)(1)(3/30) + (3)(2)(6/30) + (3)(3)(6/30) = 16 / 3
2. - continued
Cov(X,Y) =
=
The least squares lines for predicting Y from X is
The least squares lines for predicting X from Y is
The conditional p.m.f. of
Y | X = 1 is
Y | X = 2 is
Y | X = 3 is
For x = 1, 2, 3, a formula for the conditional p.m.f. of Y | X = x is
2. - continued
The conditional p.m.f. of
X | Y = 1 is
X | Y = 2 is
X | Y = 3 is
For y = 1, 2, 3, a formula for the conditional p.m.f. of X | Y = y is
E(Y | X = 1) =
E(X | Y = 1) =
E(Y2 | X = 1) =
E(X2 | Y = 1) =
Var(Y | X = 1) =
Var(X | Y = 1) =
E(Y | X = 2) =
E(X | Y = 2) =
E(Y2 | X = 2) =
E(X2 | Y = 2) =
Var(Y | X = 2) =
Var(X | Y = 2) =
E(Y | X = 3) =
E(X | Y = 3) =
E(Y2 | X = 3) =
E(X2 | Y = 3) =
Var(Y | X = 3) =
Var(X | Y = 3) =
2. - continued
Is the conditional mean of Y given X = x a linear function of the
given value, that is, can we write E(Y | X = x) = a + bx ?
Is the conditional mean of X given Y = y a linear function of the
given value, that is, can we write E(X | Y = y) = c + dy ?
3. An urn contains six chips, one $1 chip, two $2 chips, and three $3
chips. Two chips are selected at random and with replacement. The
following random variables are defined:
X = dollar value of the first chip selected ,
Y = dollar value of the second chip selected .
The space of (X, Y) is
{(1,1) (1,2) (1,3) (2,1) (2,2) (2,3) (3,1) (3,2) (3,3)}.
3
1/12
1/6
1/4
y 2
1/18
1/9
1/6
1
1/36
1/18
1/12
1
2
x
3
xy
x = 1, 2, 3
The joint p.m.f. of (X, Y) is f(x, y) = — if
y = 1, 2, 3
36
3. - continued
The marginal p.m.f. of X is f1(x) =
1/6
1/3
1/2
if x = 1
if x = 2
if x = 3
The marginal p.m.f. of Y is f2(y) =
1/6
1/3
1/2
if y = 1
if y = 2
if y = 3
A formula for the joint p.m.f. of (X,Y) is f(x, y) =
(The formula was found previously)
A formula for the marginal p.m.f. of X is f1(x) = x / 6 if x = 1, 2, 3
A formula for the marginal p.m.f. of Y is f2(y) = y / 6 if y = 1, 2, 3
E(X) = 7 / 3
E(X2) = 6
Var(X) = 6 – (7 / 3)2 = 5 / 9
E(Y) = 7 / 3
E(Y2) = 6
Var(Y) = 6 – (7 / 3)2 = 5 / 9
f(x, y) = f1(x)f2(y)
Since _________________________,
then the random variables X
are
and Y _______________
independent.
P(X + Y < 4) = P[(X,Y) = (1,1)] + P[(X,Y) = (1,2)] + P[(X,Y) = (2,1)] =
1 / 36 + 1 / 18 + 1 / 18 = 5 / 36
3. - continued
3
E(XY) = 
3
3
 (xy) (xy / 36) = 
x=1 y=1
3
3
x=1
y=1
3
 (xy) (x / 6) (y / 6) =
x=1 y=1
 (x) (x / 6)  (y) (y / 6) = E(X) E(Y) = (7/3)(7/3) = 49 / 9
Cov(X,Y) =
=
The least squares lines for predicting Y from X is
The least squares lines for predicting X from Y is
For x = 1, 2, 3,
the conditional p.m.f. of Y | X = x is
E(Y | X = x) =
Var(Y | X = x) =
For y = 1, 2, 3,
the conditional p.m.f. of X | Y = y is
E(X | Y = y) =
Var(X | Y = y) =
Is the conditional mean of Y given X = x a linear function of the
given value, that is, can we write E(Y | X = x) = a + bx ?
Is the conditional mean of X given Y = y a linear function of the
given value, that is, can we write E(X | Y = y) = c + dy ?
For continuous type random variables (X, Y), the definitions of joint
probability density function (joint p.d.f.), independence of X and Y, and
mathematical expectation are each analogous to those for discrete type
random variables, with summation signs replaced by integral signs.
The covariance between random variables X and Y is
The correlation between random variables X and Y is
y
Consider the equation of a line y = a + bx
which comes “closest” to predicting the values
of the random variable Y from the random
variable X in the sense that E{[Y – (a + bX)]2}
is minimized.
x
We let k(a,b) = E{[Y – (a + bX)]2} =
To minimize k(a,b) , we set the partial derivatives with respect to a and
b equal to zero. (Note: This is textbook exercise 4.2-5.)
k
— =
a
k
— =
b
(Multiply the first equation by X , subtract the resulting equation from
the second equation, and solve for b. Then substitute in place of b in the
first equation to solve for a.)
b=
The least squares line for predicting Y from X is
a=
The least squares line for predicting Y from X can be written
The least squares line for predicting X from Y can be written
The conditional p.m.f./p.d.f. of Y given X = x is defined to be
The conditional p.m.f./p.d.f. of X given Y = y is defined to be
The conditional mean of Y given X = x is defined to be
The conditional variance of Y given X = x is defined to be
The conditional mean of X given Y = y and the conditional variance of
X given Y = y are each defined similarly.
For continuous type random variables (X, Y), the definitions of
conditional mean and variance are each analogous to those for discrete
type random variables, with summation signs replaced by integral signs.
Suppose X and Y are two discrete type random variables, and
E(Y | X = x) = a + bx. Then, for each possible value of x,
Multiplying each side by f1(x),
Summing each side over all x,
Now, multiplying each side of
Summing each side over all x,
by x f1(x),
The two equations
and
are essentially the same as those in the derivation of the least squares line
for predicting Y from X. This derivation is analogous for continuous type
random variables with summation signs replaced by integral signs.
Consequently, if E(Y | X = x) = a + bx (i.e., if E(Y | X = x) is a linear
function of x), then a and b must be respectively the intercept and slope
in the least squares line for predicting Y from X.
Similarly, if E(X | Y = y) = c + dy (i.e., if E(X | Y = y) is a linear function
of y), then c and d must be respectively the intercept and slope in the
least squares line for predicting X from Y.
Suppose a set contains N = N1 + N2 + N3 items, where N1 items are of
one type, N2 items are of a second type, and N3 items are of a third type;
n items are selected from the N items at random and without
replacement. If the random variable X1 is defined to be the number of
selected n items that are of the first type, the random variable X2 is
defined to be the number of selected n items that are of the second type,
and the random variable X3 is defined to be the number of selected n
items that are of the third type, then the joint distribution of
(X1 , X2 , X3) is called a trivariate hypergeometric distribution. Since X3
= n – X1 – X2 , X3 is totally determined by X1 and X2 .
The joint p.m.f. of (X1 , X2) is
Each Xi has a
distribution.
If the number of types of items is any integer k > 1 with (X1 , X2 , … , Xk)
defined in the natural way, then the joint p.d.f. is called a multivariate
hypergeometric distribution.
Suppose each in a sequence of independent trials must result in one of
outcome 1, outcome 2, or outcome 3. The probability of outcome 1 on
each trial is p1 , the probability of outcome 2 on each trial is p2 , and the
probability of outcome 3 on each trial is p3 = 1 – p1 – p2 . If the random
variable X1 is defined to be the number of the n trials resulting in
outcome 1, the random variable X2 is defined to be the number of the n
trials resulting in outcome 2, and the random variable X3 is defined to
be the number of the n trials resulting in outcome 3, then the joint
distribution of (X1 , X2 , X3) is called a trinomial distribution. Since X3
= n – X1 – X2 , X3 is totally determined by X1 and X2 .
The joint p.m.f. of (X1 , X2) is
Each Xi has a
distribution.
If the number of outcomes is any integer k > 1 with (X1 , X2 , … , Xk)
defined in the natural way, then the joint p.d.f. is called a multinomial
distribution.
4. An urn contains 15 red chips, 10 blue chips, and 5 white chips.
Eight chips are selected at random and without replacement. The
following random variables are defined:
X1 = number of red chips selected ,
X2 = number of blue chips selected ,
X3 = number of white chips selected .
(a) Find the joint p.m.f. of (X1 , X2 , X3) .
(X1 , X2 , X3) have a trivariate hypergeometric distribution, and X3 =
8 – X1 – X2 is totally determined by X1 and X2 . The joint p.m.f. of
(X1 , X2) is
15
10
5
x1 = 0, 1, …, 8
x1
x2
8–x –x
1
2
if
f(x1, x2) =
30
8
x2 = 0, 1, …, 8
3  x1 + x2  8
(b) Find the marginal p.m.f. for each of X1 , X2 , and X3 .
Each of X1 , X2 , and X3 has a hypergeometric distribution.
15
15
x1
8 – x1
f1(x1) =
if x1 = 0, 1, …, 8
30
8
10
20
x2
8 – x2
f2(x2) =
if x2 = 0, 1, …, 8
30
8
4. - continued
5
25
x3
8 – x3
f3(x3) =
if x3 = 0, 1, …, 5
30
8
(c) Are X1 , X2 , and X3 independent? Why or why not?
X1, X2, X3 cannot possibly be independent, because any one of
these random variables is totally determined by the other two.
(d) Find the probability that at least two of the selected chips are blue
or at least two chips are white.
P({X2  2}  {X3  2}) = 1 – P({X2  1}  {X3  1}) =
1 – [P(X2 = 0 , X3 = 0) + P(X2 = 1 , X3 = 0) +
P(X2 = 0 , X3 = 1) + P(X2 = 1 , X3 = 1)] =
15
15
10
15
5
15
10
5
8
7
1
7
1
6
1
1
1–
+
+
+
30
30
30
30
8
8
8
8
4. - continued
(e) Find the conditional p.m.f. of X1 | x2 .
X1 | x2 can be treated as “the number of red chips selected when
For x2 =
distribution with p.m.f.
X1 | x2 has a
(f) E(X1 | x2) can be written as a linear function of x2 , since
E(X1 | x2) =
Therefore, the least squares
line for predicting X1 from X2 must be
(g) E(X2 | x1) can be written as a linear function of x2 , since
E(X2 | x1) =
Therefore, the least squares
line for predicting X2 from X1 must be
(h) Find the covariance and correlation between X1 and X2 by making
use of the following facts (instead of using direct formulas):
The slope in the least squares line for predicting X1 from X2 is
The slope in the least squares line for predicting X2 from X1 is
The product of the slope in the least squares line for predicting X1
from X2 and the slope in the least squares line for predicting X2
from X1 is equal to .
Suppose a set contains N = N1 + N2 + N3 items, where N1 items are of
one type, N2 items are of a second type, and N3 items are of a third type;
n items are selected from the N items at random and without
replacement. If the random variable X1 is defined to be the number of
selected n items that are of the first type, the random variable X2 is
defined to be the number of selected n items that are of the second type,
and the random variable X3 is defined to be the number of selected n
items that are of the third type, then the joint distribution of
(X1 , X2 , X3) is called a trivariate hypergeometric distribution. Since X3
= n – X1 – X2 , X3 is totally determined by X1 and X2 .
N1 N2 N – N1 – N2
The joint p.m.f. of (X1 , X2) is
x1 x 2 n – x1 – x2
if x1 and x2 are
N
“appropriate”
n
Each Xi has a hypergeometric distribution.
integers
If the number of types of items is any integer k > 1 with (X1 , X2 , … , Xk)
defined in the natural way, then the joint p.d.f. is called a multivariate
hypergeometric distribution.
5. An urn contains 15 red chips, 10 blue chips, and 5 white chips.
Eight chips are selected at random and with replacement. The
following random variables are defined:
X1 = number of red chips selected ,
X2 = number of blue chips selected ,
X3 = number of white chips selected .
(a) Find the joint p.m.f. of (X1 , X2 , X3) .
(X1 , X2 , X3) have a trinomial distribution, and X3 = 8 – X1 – X2 is
totally determined by X1 and X2 . The joint p.m.f. of (X1 , X2) is
x1
x2
8 – x1 – x2
1
1
1
8!
f(x1, x2) =
—
—
—
x1! x2! (8 – x1 – x2)! 2
3
6
if
x1 = 0, 1, …, 8
x2 = 0, 1, …, 8
x 1 + x2  8
(b) Find the marginal p.m.f. for each of X1 , X2 , and X3 .
Each of X1 , X2 , and X3 has a binomial distribution.
8
8!
f1(x1) =
x1! (8 – x1)!
8!
f2(x2) =
x2! (8 – x2)!
1
—
2
1
—
3
if
x2
2
—
3
x1 = 0, 1, …, 8
8 – x2
if
x2 = 0, 1, …, 8
5. - continued
8!
f3(x3) =
x3! (8 – x3)!
1
—
6
x3
5
—
6
8 – x3
if
x3 = 0, 1, …, 8
(c) Are X1 , X2 , and X3 independent? Why or why not?
X1, X2, X3 cannot possibly be independent, because any one of
these random variables is totally determined by the other two.
(d) Find the probability that at least two of the selected chips are blue
or at least two chips are white.
P({X2  2}  {X3  2}) = 1 – P({X2  1}  {X3  1}) =
1 – [P(X2 = 0 , X3 = 0) + P(X2 = 1 , X3 = 0) +
P(X2 = 0 , X3 = 1) + P(X2 = 1 , X3 = 1)] =
8
1–
1
—
2
7
+
8!
7! 1!
1
—
2
1
—
3
+
7
8!
7! 1!
1
—
2
6
1
—
6
8!
+
6! 1! 1!
1
—
2
1
—
3
1
—
6
5. - continued
(e) Find the conditional p.m.f. of X1 | x2 .
X1 | x2 can be treated as “the number of red chips selected when
For x2 =
distribution with p.m.f.
X1 | x2 has a
(f) E(X1 | x2) can be written as a linear function of x2 , since
E(X1 | x2) =
Therefore, the least squares
line for predicting X1 from X2 must be
(g) E(X2 | x1) can be written as a linear function of x2 , since
E(X2 | x1) =
Therefore, the least squares
line for predicting X2 from X1 must be
(h) Find the covariance and correlation between X1 and X2 by making
use of the following facts (instead of using direct formulas):
The slope in the least squares line for predicting X1 from X2 is
The slope in the least squares line for predicting X2 from X1 is
The product of the slope in the least squares line for predicting X1
from X2 and the slope in the least squares line for predicting X2
from X1 is equal to .
Suppose each in a sequence of independent trials must result in one of
outcome 1, outcome 2, or outcome 3. The probability of outcome 1 on
each trial is p1 , the probability of outcome 2 on each trial is p2 , and the
probability of outcome 3 on each trial is p3 = 1 – p1 – p2 . If the random
variable X1 is defined to be the number of the n trials resulting in
outcome 1, the random variable X2 is defined to be the number of the n
trials resulting in outcome 2, and the random variable X3 is defined to
be the number of the n trials resulting in outcome 3, then the joint
distribution of (X1 , X2 , X3) is called a trinomial distribution. Since X3
= n – X1 – X2 , X3 is totally determined by X1 and X2 .
The joint p.m.f. of (X1 , X2) is
x1
x2
n – x1 – x2
n!
p1 p2 (1 – p1 – p2)
x1! x2! (n – x1 – x2)!
if x1 and x2 are non-negative
Each Xi has a b( n , pi ) distribution.
integers such that x1 + x2  n
If the number of outcomes is any integer k > 1 with (X1 , X2 , … , Xk)
defined in the natural way, then the joint p.d.f. is called a multinomial
distribution.
6. One chip is selected from each of two urns, one containing three
chips labeled distinctively with the integers 1 through 3 and the
other containing two chips labeled distinctively with the integers 1
and 2. The following random variables are defined:
X = largest integer among the labels on the selected chips ,
Y = smallest integer among the labels on the selected chips .
The space of (X, Y) is {(1,1) (2,1) (3,1) (2,2) (3,2)}.
(Note: We immediately see that X and Y
cannot be independent, since the joint
space is not “rectangular”.)
2
1/6
1/6
1/3
1/6
y
1/6
1
1
2
x
3
The joint p.m.f. of (X, Y) is f(x, y) =
1
— if (x, y) = (1, 1) , (3, 1) , (2, 2) , (3, 2)
6
1
— if (x, y) = (2, 1)
3
The marginal p.m.f. of X is f1(x) =
1/6
1/x
if x = 1
if x = 2, 3
The marginal p.m.f. of Y is f2(y) = (3 – y) / 3 if y = 1, 2
E(X) = 13 / 6
E(X2) = 31 / 6
Var(X) = 31 / 6 – (13 / 6)2 = 17 / 36
E(Y) = 4 / 3
E(Y2) = 2
Var(Y) = 2 – (4 / 3)2 = 2 / 9
6. - continued
f(x, y)  f1(x)f2(y)
Since _________________________,
then the random variables X
are not
and Y _______________
independent (as we previously noted).
Using the joint p.m.f., E(XY) =
(1)(1)(1/6) + (3)(1)(1/6) + (2)(2)(1/6) + (3)(2)(1/6) + (2)(1)(1/3) = 3
Cov(X,Y) =
=
The least squares lines for predicting Y from X is
The least squares lines for predicting X from Y is
The conditional p.m.f. of
Y | X = 1 is
Y | X = 2 is
Y | X = 3 is
6. - continued
The conditional p.m.f. of
X | Y = 1 is
X | Y = 2 is
E(Y | X = 1) =
E(Y2 | X = 1) =
Var(Y | X = 1) =
E(Y | X = 2) =
E(Y2 | X = 2) =
Var(Y | X = 2) =
E(Y | X = 3) =
E(Y2 | X = 3) =
Var(Y | X = 3) =
6. - continued
E(X | Y = 1) =
E(X2 | Y = 1) =
Var(X | Y = 1) =
E(X | Y = 2) =
E(X2 | Y = 2) =
Var(X | Y = 2) =
Is the conditional mean of Y given X = x a linear function of the
given value, that is, can we write E(Y | X = x) = a + bx ?
Is the conditional mean of X given Y = y a linear function of the
given value, that is, can we write E(X | Y = y) = c + dy ?
For continuous type random variables (X, Y), the definitions of joint
probability density function (joint p.d.f.), independence of X and Y, and
mathematical expectation are each analogous to those for discrete type
random variables, with summation signs replaced by integral signs.
The covariance between random variables X and Y is
The correlation between random variables X and Y is
y
Consider the equation of a line y = a + bx
which comes “closest” to predicting the values
of the random variable Y from the random
variable X in the sense that E{[Y – (a + bX)]2}
is minimized.
x
9. Random variables X and Y have joint p.d.f.
f(x,y) = 5xy2 / 2 if 0 < x/2 < y < 1 .
Skip to #9
The space of (X, Y) displayed graphically is as follows:
y
(Note: We immediately see that X and Y
cannot be independent, since the joint
(0,2)
space is not “rectangular”.)
(0,0)
y=x/2
(2,1)
x
Event A = {(x,y) | 1/2 < x < 1 , 1/2 < y < 3/2} displayed graphically
is as follows:
y
(1/2, 3/2)
(0,2)
(1/2, 1/2)
P(A) =
(1, 3/2)
(2,1)
(1, 1/2)
x
(0,0)
f(x, y) dx dy =
A
1
1/2
1
1
1
1
1
35
5xy2
5x2y2
15y2
5y3
—— dx dy =
—— dy = —— dy = — = –—
128
2
4
16
16
x = 1/2 1/2
y = 1/2
1/2
1/2

9. - continued
The marginal p.d.f. of X is f1(x) =
1
5xy2
—— dy =
2
f(x, y) dy =
–
x/2
1
40x – 5x4
= ———— if 0 < x < 2
48
5xy3
——
6
y=x/2
2
2
40x – 5x4
E(X) = x ———— dx =
48
0
0
40x2 – 5x5
———— dx =
48
2
5x3
5x6
10
— – —— = —
9
18
288
x=0
2
2
2
5x4
5x7
10
40x – 5x4
40x3 – 5x6
2
2
E(X ) = x ———— dx = ———— dx = — – —— = —
7
24
336
48
48
x=0
0
0
110
Var(X) = —–
567

9. - continued
2y
The marginal p.d.f. of Y is f2(y) = f(x, y) dx =
–
5xy2
—— dx =
2
0
2y
5x2y2
——
4
x=0
5
E(Y) = —
6
= 5y4 if 0 < y < 1
E(Y 2)
5
= —
7
5
Var(Y) = —–
252
9. - continued
f(x, y)  f1(x)f2(y)
Since _________________________,
then the random variables X
are not
and Y _______________
independent (as we previously noted).
1
2y
 
5xy2
E(XY) =
xy f(x, y) dx dy =
xy —— dx dy =
2
– –
0
0
1
2y
1
1
1
2y
5x2y3
5x3y3
20y6
20y7
20
—— dx dy =
—— dy = —— dy = —–
= —
21
2
6
3
21
x=0
y=0
0
0
0
0
Cov(X,Y) =
=
The least squares lines for predicting Y from X is
The least squares lines for predicting X from Y is
For 0 < x < 2, the conditional p.d.f. of Y | X = x is
9. - continued
For 0 < x < 2, E(Y | X = x) =
For 0 < x < 2, E(Y2 | X = x) =
For 0 < x < 2, Var(Y | X = x) =
For 0 < y < 1, the conditional p.d.f. of X | Y = y is
For 0 < y < 1, E(X | Y = y) =
For 0 < y < 1, E(X2 | Y = y) =
9. - continued
For 0 < y < 1, Var(X | Y = y) =
Is the conditional mean of Y given X = x a linear function of the
given value, that is, can we write E(Y | X = x) = a + bx ?
Is the conditional mean of X given Y = y a linear function of the
given value, that is, can we write E(X | Y = y) = c + dy ?
7. Random variables X and Y have joint p.d.f.
f(x,y) = (x + y) / 8 if 0 < x < 2 , 0 < y < 2 .
The space of (X, Y) displayed graphically is as follows:
y
(0,2)
(0,0)
(2,2)
(2,0)
x
Events A = {(x,y) | 1/2 < x < 1 , 1/2 < y < 3/2} and B = {(x,y) | x > y}
displayed graphically are as follows:
The set A =
{(x, y) | 1/2 < x < 1 , 1/2 < y < 3/2}
is graphically displayed as follows:
A
y
(0,2)
(1/2, 3/2)
(2,2)
(1, 3/2)
(1/2, 1/2)
(1, 1/2)
(0,0)
(2,0)
The set B = {(x, y) | x > y} is
graphically displayed as follows:
y B
(0,2)
x
(0,0)
(2,2)
(2,2)
(2,0)
x
7. - continued
P(A) = P(1/2 < X < 1 , 1/2 < Y < 3/2) =
f(x, y) dx dy
=
A
3/2
1
3/2
x+y
—— dx
8
1/2
3/2
1/2
dy
=
1/2
3/2
1 + 2y
1/4 + y
3 + 4y
——— – ——— dy =
——— dy =
16
16
64
1/2
1/2
1
x2 + 2xy
——— dy
16
x = 1/2
=
3/2
3y + 2y2
7
———
= —
64
64
y = 1/2
P(B) = P(X > Y) =
f(x, y) dx dy
=
x > y
2
2
x+y
—— dx
8
2
0
dy
y
0
x2
2
x+y
—— dy
8
or
0
x2
2xy + y2
——— dx
16
y=0
2
0
2
2x3 + x4
5x4 + 2x5
——— dx = ———— =
16
160
=
0
dx
x=0
9
—
10
7. - continued
The marginal p.d.f. of X is f1(x) =
E(X) = 7/6
E(X2) = 5/3
Var(X) = 11/36
2
f(x, y) dy =
x+y
—— dy =
8
–
2
2xy + y2
———
16
y=0

=
x+1
——
4
0
if
0<x<2

2
f(x, y) dx =
x+y
—— dx =
8
The marginal p.d.f. of Y is f2(y) =
–
y+1
——
4
0
if
0<y<2
E(Y) = 7/6
E(Y2) = 5/3
Var(Y) = 11/36
f(x, y)  f1(x)f2(y)
Since _________________________,
then the random variables X
are not
and Y _______________
independent
7. - continued
 
E(XY) =
2
0
=
0
2
2x3y + 3x2y2
————— dy =
48
x=0
0
Cov(X,Y) =
x2y + xy2
——— dx
8
xy f(x, y) dx dy =
– –
2
2
2
dy
=
0
2
4y + 3y2
2y2 + y3
4
———— dy = ——— = —
12
12
3
y=0
The least squares lines for predicting Y from X is
The least squares lines for predicting X from Y is
For 0 < x < 2, the conditional p.d.f. of Y | X = x is
7. - continued
For 0 < x < 2, E(Y | X = x) =
For 0 < x < 2, E(Y2 | X = x) =
For 0 < x < 2, Var(Y | X = x) =
For 0 < y < 2, the conditional p.d.f. of X | Y = y is
For 0 < y < 2, E(X | Y = y) =
For 0 < y < 2, E(X2 | Y = y) =
7. - continued
For 0 < y < 2, Var(X | Y = y) =
Is the conditional mean of Y given X = x a linear function of the
given value, that is, can we write E(Y | X = x) = a + bx ?
Is the conditional mean of X given Y = y a linear function of the
given value, that is, can we write E(X | Y = y) = c + dy ?
8. Random variables X and Y have joint p.d.f.
f(x,y) = (y – 1) / (2x2) if 1 < x , 1 < y < 3 .
The space of (X, Y) displayed graphically is as follows:
y
(1,3)
(1,1)
(0,0)
x
8. - continued
Event A = {(x,y) | 1 < x < 3 , 1 < y < (x+1)/2} displayed graphically
is as follows:
y
y = (x + 1) / 2
x =3
(1,3)
(3,2)
(1,1)
x
(0,0)
3
f(x, y) dx dy
P(A) =
A
(x+1)/2
y–1
—— dy
2x2
=
1
1
dx
=
3
(x+1)/2
y–1
—— dy
2x2
1
3
dx =
1
1
3
y2 – 2y
——— dx
4x2
y=1
=
3
(x + 1)2 – 4(x + 1)
1
——————— + —— dx =
16x2
4x2
1
x2 – 2x + 1
———— dx
16x2
1
3
3
1
1
1
— – — + —— dx
16
8x
16x2
1
(x+1)/2
x
ln x
1
=
— – —— – —— =
16
8
16x
x=1
1
ln 3
3
ln 3
1
— – —— – — – 0 = — – ——
6
8
16
8
48
=
8. - continued
Events B = {(x,y) | x > y} displayed graphically is as follows:
y
y=x
(1,3)
(3,3)
Note that describing B as
{1 < x < 3 , 1 < y < x} 
{3 < x <  , 1 < y < 3}
makes the integration
more work than
describing B as
{1 < y < 3 , y < x < }
(1,1)
x
(0,0)
3 
P(B) =
x>y
f(x, y) dx dy =
1 y
3
y–1
—— dx dy =
2x2
1

1–y
—— dy =
2x
x=y
3
1

1–y
—— dy =
2x
x=y
1
3
3
y–1
y
lny
—— dy = — – —–
2y
2
2
ln 3
= 1 – ——
2
y=1
8. - continued

3
y–1
—— dy
2x2
The marginal p.d.f. of X is f1(x) = f(x, y) dy =
–
1
3
y2 – 2y
———
4x2
1
= —
x2
y=1

E(X) =
1

1
x — dx =
x2
1
=

1
— dx = ln(x) = 
x
x=1
if
1<x

E(X2)
=

x2
1
Var(X) = 
1
— dx
x2
dx = 
=
1

8. - continued

y–1
—— dx
2x2
The marginal p.d.f. of Y is f2(y) = f(x, y) dx =
–

1–y
——
2x
1
y–1
= ——
2
if
1<y<3
x=1
3
E(Y) =
1
3
y–1
y —— dy =
2
1
3
y2 – y
—— dy =
2
=
y3
y2
7
— – — = —
3
6
4
y=1
3
E(Y2) =
3
y–1
2
y —— dy
2
1
2
Var(Y) = —
9
=
1
3
y3 – y2
—— dy =
2
y4
y3
17
— – — = —
3
8
6
y=1
8. - continued
f(x, y) = f1(x)f2(y)
Since _________________________,
then the random variables X
are
and Y _______________
independent
3 

y–1
dx dy =
xy ——
2
2x
E(XY) =
1 1
3
y2 – y
—— dy
2
1
—
x
1
dx =
1

7
—
3
Cov(X,Y) =
=
1
1
— dx = 
x
The least squares lines for predicting Y from X is
The least squares lines for predicting X from Y is
For 1 < x , the conditional p.d.f. of Y | X = x is
For 1 < y < 3, the conditional p.d.f. of X | Y = y is
For 1 < x , E(Y | X = x) =
For 1 < x , Var(Y | X = x) =
For 0 < y < 3, E(X | Y = y) =
For 0 < y < 3, Var(X | Y = y) =