Download Lecture 4: Probability Distributions and Probability Densities

Multivariate Distributions
Marginal Distributions
Conditional Distributions
Lecture 4: Probability Distributions and
Probability Densities - 2
Assist. Prof. Dr. Emel YAVUZ DUMAN
MCB1007 Introduction to Probability and Statistics
İstanbul Kültür University
Multivariate Distributions
Marginal Distributions
Outline
1
Multivariate Distributions
2
Marginal Distributions
3
Conditional Distributions
Conditional Distributions
Multivariate Distributions
Marginal Distributions
Outline
1
Multivariate Distributions
2
Marginal Distributions
3
Conditional Distributions
Conditional Distributions
Multivariate Distributions
Marginal Distributions
Conditional Distributions
In this section we shall concerned first with the bivariate case,
that is, with situation where we are interested at the same time in
a pair of random variables defined over a joint sample space that
are both discrete. Later, we shall extend this discussions to the
multivariate case, covering any finite number of random variables.
If X and Y are discrete random variables, we write the probability
that X will take on the value x and Y will take on the value y as
P(X = x, Y = y ). Thus, P(X = x, Y = y ) is the probability of
the intersection of the events X = x and Y = y . As in the
univariate case, where we dealt with one random variable and
could display the probabilities associated with all values of X by
means of a table, we can now, in the bivariate case, display the
probabilities associated with all pairs of the values of X and Y by
mean of a table.
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Example 1
Two caplets are selected at a random form a bottle containing
three aspirin, two sedative, and four laxative caplets. If X and Y
are, respectively, the numbers of the aspirin and sedative caplets
included among the two caplets drawn from the bottle, find the
probabilities associated with all possible pairs of values of X and Y .
Solution. The possible pairs are (0, 0), (0, 1), (1, 0), (1, 1), (0, 2),
and (2, 0). So we obtain the following probabilities:
324
P(X = 0, Y = 0) =
0
09
2
=
6
,
36
=
8
,
36
=
12
,
36
3224
P(X = 0, Y = 1) =
0
19
1
3224
P(X = 1, Y = 0) =
1
09
2
1
Multivariate Distributions
Marginal Distributions
Conditional Distributions
324
P(X = 1, Y = 1) =
1
19
0
=
6
,
36
=
1
,
36
=
3
.
36
3224
P(X = 0, Y = 2) =
0
29
0
3224
P(X = 2, Y = 0) =
2
09
2
0
Therefore, we have the following table:
H
HH x
HH
y
H
0
1
2
0
1
2
6/36
8/36
1/36
12/36
6/36
3/36
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Definition 2
If X and Y are discrete random variables, the function given buy
f (x, y ) = P(X = x, Y = y )
for each pair of values (x, y ) within the range of X and Y is called
the joint probability distribution of X and Y .
Theorem 3
A bivariate function can serve as the joint probability distribution
of a pair of discrete random variables X and Y if and only if its
values f (x, y ) satisfy the conditions
1
2
f (x, y ) ≥ 0 for each pair of values (x, y ) within its domain;
x
y f (x, y ) = 1, where the double summation extends
over all possible pairs (x, y ) within its domain.
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Suppose that X can assume any one of m values x1 , x2 , · · · , xm
and Y can assume any one of n values y1 , y2 , · · · , yn . Then the
probability of the event that X = xj and Y = yk is given by
P(X = xj , Y = yk ) = f (xj , yk ).
A joint probability function for X and Y can be represented by a
joint probability table as in the following:
H
HH y
HH
x
H
y1
y2
···
yn
Totals ↓
x1
x2
..
.
f (x1 , y1 )
f (x2 , y1 )
..
.
f (x1 , y2 )
f (x2 , y2 )
..
.
···
···
f (x1 , yn )
f (x2 , yn )
..
.
g (x1 )
g (x2 )
..
.
xm
Totals →
f (xm , y1 )
h(y1 )
f (xm , y2 )
h(y2 )
f (xm , yn )
h(yn )
g (xm )
1
···
···
···
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Example 4
Determine the value of k for which the function given by
f (x, y ) = kxy for x = 1, 2, 3; y = 1, 2, 3
can serve as a joint probability distribution.
Solution. Substituting the various values of x and y , we get
f (1, 1) = k, f (1, 2) = 2k, f (1, 3) = 3k, f (2, 1) = 2k, f (2, 2) = 4k,
f (2, 3) = 6k, f (3, 1) = 3k, f (3, 2) = 6k, f (3, 3) = 9k.
To satisfy the first condition of Theorem 3, the constant k must be
nonnegative, and to satisfy the second condition
k + 2k + 3k + 2k + 4k + 6k + 3k + 6k + 9k = 1
so that 36k = 1 and k = 1/36.
Multivariate Distributions
Marginal Distributions
Conditional Distributions
f (x, y ) = kxy for x = 1, 2, 3; y = 1, 2, 3
The joint probability function for X and Y can be represented by a
joint probability table as in the following:
H
HH y
HH
x
H
1
2
3
Totals ↓
1
2
3
Totals →
k
2k
3k
6k
2k
4k
6k
12k
3k
6k
9k
18k
6k
12k
18k
36k = 1
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Example 5
If the values of the joint probability distribution of X and Y are as
shown in the table
HH
x
HH
y
H
H
0
1
2
0
1
2
3
1/12
1/4
1/8
1/120
1/6
1/4
1/20
1/24
1/40
find
(a) P(X = 1, Y = 2);
(b) P(X = 0, 1 ≤ Y < 3);
(c) P(X + Y ≤ 1);
(d) P(X > Y ).
Multivariate Distributions
H
HH
y
0
1
2
3
x
HH
H
Marginal Distributions
0
1
2
1/12
1/4
1/8
1/120
1/6
1/4
1/20
1/24
1/40
(a) P(X = 1, Y = 2) =
Conditional Distributions
1
20 ;
(b) P(X = 0, 1 ≤ Y < 3) = f (0, 1) + f (0, 2) =
1
4
(c) P(X + Y ≤ 1) = f (0, 0) + f (1, 0) + f (0, 1) =
(d) P(X > Y ) = f (1, 0) + f (2, 0) + f (2, 1) =
+
1
8
= 38 ;
1
1
1
1
12 + 6 + 4 = 2 ;
1
1
1
7
6 + 24 + 40 = 30 .
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Example 6
If the joint probability distribution of X and Y is given by
f (x, y ) = c(x 2 + y 2 ) for x = −1, 0, 1, 3, y = −1, 2, 3
find the value of c.
Solution. Since
H
HH x
HH
y
H
−1
2
3
Totals →
−1
0
1
3
Totals ↓
2c
5c
10c
17c
1c
4c
9c
14c
2c
5c
10c
17c
10c
13c
18c
41c
15c
27c
47c
89c = 1
then we have that c =
1
89 .
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Example 7
Show that there is no value of k for which
f (x, y ) = ky (2y − x) for x = 0, 3, y = 0, 1, 2
can serve as the joint probability distribution of two random
variables.
Solution. Since
H
HH y
HH
x
H
0
1
2
Totals ↓
0
3
Totals →
0
0
0
2k
−k
k
8k
2k
10k
10k
k
11k = 1
then we find that k = 1/11. But in this case, f (3, 1) differs in sign
from all other terms.
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Example 8
Suppose that we roll a pair of balanced dice and X is the number
of dice that come up 1, and Y is the number of dice that come up
4, 5, or 6.
(a) Construct a table showing the values X and Y associated
with each of the 36 equally likely points of the sample space.
(b) Construct a table showing the values the joint probability
distribution of X and Y .
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Solution.
(a) If X is the number of dice that come up 1, and Y is the
number of dice that come up 4, 5, or 6, then we have
PP
PP Roll 2
PP
Roll 1
PP
P
1
2
3
4
5
6
1
2
3
4
5
6
(2,0)
(1,0)
(1,0)
(1,1)
(1,1)
(1,1)
(1,0)
(0,0)
(0,0)
(0,1)
(0,1)
(0,1)
(1,0)
(0,0)
(0,0)
(0,1)
(0,1)
(0,1)
(1,1)
(0,1)
(0,1)
(0,2)
(0,2)
(0,2)
(1,1)
(0,1)
(0,1)
(0,2)
(0,2)
(0,2)
(1,1)
(0,1)
(0,1)
(0,2)
(0,2)
(0,2)
(b) Then, we simply count the number of times we have each of
the possible (x, y ) values, and divide by 36.
x
y
Prob.
0
0
4/36
0
1
12/36
0
2
9/36
1
0
4/36
1
1
6/36
2
0
1/36
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Definition 9
If X and Y are discrete random variables, the function given by
f (s, t) for − ∞ < x, y < ∞
F (x, y ) = P(X ≤ x, Y ≤ y ) =
s≤x t≤y
where f (s, t) is the value of the joint probability distribution of X
and Y at (s, t), is called the joint distribution function, or the
joint cumulative distribution, of X and Y .
Theorem 10
If F (x, y ) is the value of the joint distribution function of two
discrete random variables X and Y at (x, y ), then
(a) F (−∞, −∞) = 0;
(b) F (∞, ∞) = 1;
(c) if a < b and c < d , then F (a, c) ≤ F (b, d).
Multivariate Distributions
Marginal Distributions
Example 11
With reference to Example 1, find F (1, 1).
H
HH x
HH
y
H
0
1
2
0
1
2
6/36
8/36
1/36
12/36
6/36
3/36
Solution.
F (1, 1) = P(X ≤ 1, Y ≤ 1)
= f (0, 0) + f (0, 1) + f (1, 0) + f (1, 1)
8
12
6
6
+
+
+
=
36 36 36 36
32
=
36
Conditional Distributions
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Example 12
With reference to Example 5, find the following values of the joint
distribution function of the two random variables:
(a) F (1.2, 0.9) (b) F (−3, 1.5) (c) F (2, 0) (d) F (4, 2.7).
HH
x
HH
y
H
H
0
1
2
0
1
2
3
1/12
1/4
1/8
1/120
1/6
1/4
1/20
1/24
1/40
Solution.
(a) F (1.2, 0.9) = P(X ≤ 1.2, Y ≤ 0.9)
1
+ 16 =
= f (0, 0) + f (1, 0) = 12
(b) F (−3, 1.5) = P(X ≤ −3, Y ≤ 1.5) = 0
1
4
Multivariate Distributions
H
HH
y
0
1
2
3
x
HH
H
Marginal Distributions
0
1
2
1/12
1/4
1/8
1/120
1/6
1/4
1/20
1/24
1/40
Conditional Distributions
(c) F (2, 0) = P(X ≤ 2, Y ≤ 0)
1
1
7
+ 16 + 24
= 24
= f (0, 0) + f (1, 0) + f (2, 0) = 12
1
(d) F (4, 2.7) = P(X ≤ 4, Y ≤ 2.7) = 1 − f (0, 3) = 1 − 120 =
119
120 .
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Example 13
If two cards are randomly drawn (without replacement) from an
ordinary deck of 52 playing cards, Z is the number of aces
obtained in the first draw and W is the total number of aces
obtained in both draws, find F (1, 1).
Solution. Let X be the number of aces obtained in the first draw,
and Y be the number of aces obtained in the second draw. So, we
have f (x, y ), the joint probability distribution:
f (0, 0) =
188
48 47
·
=
,
52 51
221
f (1, 0) =
4 48
16
·
=
,
52 51
221
f (0, 1) =
16
48 4
·
=
,
52 51
221
f (1, 1) =
4 3
1
·
=
.
52 51
221
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Since
H
HH w
HH
z
H
0
1
0
1
2
f (0, 0)
f (0, 1)
f (1, 0)
f (1, 1)
where z = x and w = x + y , then we have
HH
w
HH
z
H
H
0
1
0
1
2
188/221
16/221
16/221
1/221
Therefore, we obtain that
F (1, 1) =
16
16
1
220
188
+
+
=1−
=
.
221 221 221
221
221
Multivariate Distributions
Marginal Distributions
Conditional Distributions
All the definitions in this section can be generalized to the
multivariate case, where there are n random variables.
Corresponding to Definition 2, the values of the joint probability
distribution of n discrete random variables X1 , X2 , · · · , and Xn are
given by
f (x1 , x2 , · · · , xn ) = P(X1 = x1 , X2 = x2 , · · · , Xn = xn )
for each n-tuple (x1 , x2 , · · · , xn ) within the range of the random
variables; and corresponding to Definition 9, the values of their
joint distribution function are given by
F (x1 , x2 , · · · , xn ) = P(X1 ≤ x1 , X2 ≤ x2 , · · · , Xn ≤ xn )
for −∞ < x1 < ∞, −∞ < x2 < ∞, · · · , −∞ < xn < ∞.
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Example 14
If the joint probability distribution of three discrete random
variables X , Y , and Z is given by
f (x, y , z) =
(x + y )z
for x = 1, 2; y = 1, 2, 3; z = 1, 2
63
find P(X = 2, Y + Z ≤ 3).
Solution.
P(X = 2, Y + Z ≤ 3) = f (2, 1, 1) + f (2, 1, 2) + f (2, 2, 1)
6
4
3
+
+
=
63 63 63
13
= .
63
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Example 15
Find c if the joint probability distribution of X , Y and Z is given by
f (x, y , z) = kxyz for x = 1, 2; y = 1, 2, 3; z = 1, 2
and determine F (2, 1, 2) and F (4, 4, 4)
Solution. Since
1=
3 2
2 x=1 y =1 z=1
=f (1, 1, 1) + f (1, 1, 2) + f (1, 2, 1) + f (1, 2, 2) + f (1, 3, 1) + f (1, 3, 2)
+f (2, 1, 1) + f (2, 1, 2) + f (2, 2, 1) + f (2, 2, 2) + f (2, 3, 1) + f (2, 3, 2)
=k(1 + 2 + 2 + 4 + 3 + 6 + 2 + 4 + 4 + 8 + 6 + 12) = 54k
then we have that k = 1/54.
Multivariate Distributions
Marginal Distributions
Conditional Distributions
f (x, y , z) = kxyz for x = 1, 2; y = 1, 2, 3; z = 1, 2
Also,
F (2, 1, 2) = P(X ≤ 2, Y ≤ 1, Z ≤ 2)
= f (1, 1, 1) + f (1, 1, 2) + f (2, 1, 1) + f (2, 1, 2)
9
1
1
=
= (1 + 2 + 2 + 4) =
54
54
6
F (4, 4, 4) = P(X ≤ 4, Y ≤ 4, Z ≤ 4) = 1.
Multivariate Distributions
Marginal Distributions
Outline
1
Multivariate Distributions
2
Marginal Distributions
3
Conditional Distributions
Conditional Distributions
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Marginal Distributions
To introduce the concept of a marginal distribution, let us
consider the following example.
Example 16
In Example 1 we derived the joint probability distribution of two
random variables X and Y , the number of aspirin and the number
of sedative caplets included among two caplets drawn at random
from a bottle containing three aspirin, two sedative, and four
laxative caplets. Find the probability distribution of X alone and
that of Y alone.
Solution. The results of Example 1 are shown in the following
table, together with the marginal totals, that is, the totals of the
respective rows and columns:
Multivariate Distributions
Marginal Distributions
Conditional Distributions
HH
x
HH
y
H
H
0
1
2
h(y )
0
1
2
g (x)
6/36
8/36
1/36
15/36
12/36
6/36
3/36
18/36
3/36
21/36
14/36
1/36
1
The column totals are the probabilities that X will take on the
values 0, 1, and 2. In other words, they are the values
g (x) =
2
f (x, y ) for x = 0, 1, 2
y =0
of the probability distribution of X .
⎧
⎪
⎨15/36
g (x) = 18/36
⎪
⎩
3/36
for x = 0
for x = 1
for x = 2
Multivariate Distributions
Marginal Distributions
Conditional Distributions
HH
x
HH
y
H
H
0
1
2
h(y )
0
1
2
g (x)
6/36
8/36
1/36
15/36
12/36
6/36
3/36
18/36
3/36
21/36
14/36
1/36
1
By the same token, the row totals are the values
h(y ) =
2
f (x, y ) for y = 0, 1, 2
x=0
of the probability distribution of Y .
⎧
⎪
⎨21/36
h(y ) = 14/36
⎪
⎩
1/36
for y = 0
for y = 1
for y = 2
Multivariate Distributions
Marginal Distributions
Conditional Distributions
We are thus led to the following definition:
Definition 17
If X and Y are discrete random variables and f (x, y ) is the value
of their joint probability distribution at (x, y ), the function given by
f (x, y )
g (x) =
y
for each x within the range of X is called the marginal
distribution of X . Correspondingly, the function given by
f (x, y )
h(y ) =
x
for each y within the range of Y is called the marginal
distribution of Y .
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Example 18
Two textbooks are selected at random from a shelf contains three
statistics texts, two mathematics texts, and three physics texts. If
X is the number of statistics texts and Y the number of
mathematics texts actually chosen
(a) construct a table showing the values of the joint probability
distribution of X and Y .
(b) Find the marginal distributions of X and Y .
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Solution. (a) Since
33
23
3
3
9
6
2
1 1
f (0, 0) = 8 = , f (1, 0) = 8 = , f (0, 1) = 181 = ,
28
28
28
2
2
2
32
f (1, 1) =
8
1
1
2
3
2
6
3
1
2
= , f (2, 0) = 8 = , f (0, 2) = 28 = ,
28
28
28
2
2
we have the following join probability distribution table:
H
HH x
HH
y
H
0
1
2
h(y )
0
1
2
g (x)
3/28
6/28
1/28
10/28
9/28
6/28
3/28
15/28
3/28
15/28
12/28
1/28
1
Multivariate Distributions
Marginal Distributions
Conditional Distributions
HH
x
HH
y
H
H
0
1
2
h(y )
0
1
2
g (x)
3/28
6/28
1/28
10/28
9/28
6/28
3/28
15/28
3/28
15/28
12/28
1/28
1
(b) The marginal distributions of X and Y
⎧
⎧
⎪
⎪
⎨10/28 for x = 0
⎨15/28
g (x) = 15/28 for x = 1 h(y ) = 12/28
⎪
⎪
⎩
⎩
3/28
for x = 2
1/28
respectively.
for y = 0
for y = 1
for y = 2
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Example 19
Given the values of the joint probability distribution of X and Y
shown in the table
HH
x
H
H
y
H
H
−1
0
1
−1
1
1/8
0
1/8
1/2
1/4
0
find the marginal distribution of X and the marginal distribution of
Y.
Solution.
g (x) =
1
8
1
2
+
+
1
8
1
4
=
=
1
4
3
4
⎧
1
⎪
⎨8 +
for x = −1
h(y ) = 14
⎪
for x = 1
⎩1
8
1
2
=
5
8
for y = −1
for y = 0
for y = 1
Multivariate Distributions
Marginal Distributions
Outline
1
Multivariate Distributions
2
Marginal Distributions
3
Conditional Distributions
Conditional Distributions
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Conditional Distributions
We defined the conditional probability of an event A, given event
B, as
P(A ∩ B)
P(A|B) =
P(B)
provided P(B) = 0. Suppose now that A and B are the events
X = x and Y = y so that we can write
P(X = x|Y = y ) =
f (x, y )
P(X = x, Y = y )
=
P(Y = y )
h(y )
provided P(Y = y ) = h(y ) = 0, where f (x, y ) is the value of the
joint probability distribution of X and Y at (x, y ) and h(y ) is the
value of the marginal distribution of Y at y . Denoting the
conditional probability f (x|y ) to indicate that x is a variable and y
is fixed, let us make the following definition:
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Definition 20
If f (x, y ) is the value of the joint probability distribution of the
discrete random variables X and Y at (x, y ), and h(y ) is the value
of the marginal distribution of Y at y , the function given by
f (x|y ) =
f (x, y )
h(y ) = 0
h(y )
for each x within the range of X , is called the conditional
distribution of X given Y = y . Correspondingly, if g (x) is the
value of the marginal distribution of X at x, the function
w (y |x) =
f (x, y )
g (x) = 0
g (x)
for each y within the range of Y , is called the conditional
distribution of Y given X = x.
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Example 21
With reference to Example 1 and Example 16, find the conditional
distribution of X given Y = 1. f (x|1) = f (x, 1)/h(1)
H
HH
y
x
HH
H
0
1
2
g (x)
0
1
2
h(y )
6/36
8/36
1/36
15/36
12/36
6/36
3/36
18/36
3/36
21/36
14/36
1/36
1
Solution. Substituting the appropriate values from the table
above, we get
f (0|1) =
8/36
8
f (1, 1)
6/36
6
f (0, 1)
=
= , f (1|1) =
=
= ,
h(1)
14/36
14
h(1)
14/36
14
f (2|1) =
0
f (2, 1)
=
= 0.
h(1)
14/36
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Example 22
With reference to Example 18 find the conditional distribution of
Y given X = 0. w (y |0) = f (0, y )/g (0)
H
HH x
HH
y
H
0
1
2
h(y )
0
1
2
g (x)
3/28
6/28
1/28
10/28
9/28
6/28
3/28
15/28
3/28
15/28
12/28
1/28
1
Solution. Substituting the appropriate values from the table
above, we get
w (0|0) =
3/28
3
f (0, 1)
6/28
6
f (0, 0)
=
= , w (1|0) =
=
= ,
g (0)
10/28
10
g (0)
10/28
10
w (2|0) =
1/28
1
f (0, 2)
=
= .
g (0)
10/28
10
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Definition 23
Suppose that X and Y are discrete random variables. If the events
X = x and Y = y are independent events for all x and y , then we
say that X and Y are independent random variables. In such
case,
P(X = x, Y = y ) = P(X = x) · P(Y = y )
or equivalently
f (x, y ) = f (x|y ) · h(y ) = g (x) · h(y ).
Conversely, if for all x and y the joint probability function f (x, y )
can be expressed as the product of a function of x alone and a
function of y alone (which are then the marginal probability
functions of X and Y ), X and Y are independent. If, however,
f (x, y ) cannot be so expressed, then X and Y are dependent.
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Example 24
Determine whether the random variables of Example 1 are
independent.
Solution. With reference to Example 1, we have the following
joint distribution table:
H
HH x
HH
y
H
0
1
2
h(y )
0
1
2
g (x)
6/36
8/36
1/36
15/36
12/36
6/36
3/36
18/36
3/36
21/36
14/36
1/36
1
Since
15 14
8
=
·
= g (0) · h(1)
36
36 36
we see that X and Y are dependent.
f (0, 1) =
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Example 25
A box contains three balls labeled 1, 2 and 3. Two balls are
randomly drawn from the box without replacement. Let X be the
number on the first ball and Y the number on the second ball.
(1) Find the joint probability distribution of X and Y .
(2) Find the marginal distributions of X and Y .
(3) Find the conditional distribution of X given Y = 1.
(4) Find the conditional distribution of Y given X = 2.
(5) Determine whether the random variables X and Y are
independent.
Multivariate Distributions
Marginal Distributions
Conditional Distributions
Solution.
(1) Since
P(X = 1, Y = 2) = P(X = 1, Y = 3) = P(X = 2, Y = 1) =
P(X = 2, Y = 3) = P(X = 3, Y = 1) = P(X = 3, Y = 2) =
1 1
1
3 · 2 = 6,
this joint probability distribution can be expressed by the
following table:
HH
y
HH
x
H
H
1
2
3
h(y )
1
1/6
1/6
2/6
2
3
g (x)
1/6
1/6
1/6
2/6
2/6
2/6
1
1/6
2/6
2/6
Multivariate Distributions
Marginal Distributions
HH
y
HH
x
H
H
1
2
3
h(y )
(2)
1
1/6
1/6
2/6
⎧
⎪
⎨2/6,
g (x) = 2/6,
⎪
⎩
2/6,
Conditional Distributions
2
3
g (x)
1/6
1/6
1/6
2/6
2/6
2/6
1
1/6
2/6
x = 1,
x = 2,
x = 3,
2/6
⎧
⎪
⎨2/6,
h(y ) = 2/6,
⎪
⎩
2/6,
y = 1,
y = 2,
y = 3.
Multivariate Distributions
H
HH
x
y
HH
H
1
2
3
h(y )
Marginal Distributions
1
1/6
1/6
2/6
Conditional Distributions
2
3
g (x)
1/6
1/6
1/6
2/6
2/6
2/6
1
1/6
2/6
2/6
(3) The conditional distribution of X given Y = 1:
f (x|1) =
f (x, 1)
⇒
h(1)
0
f (1, 1)
=
= 0,
h(1)
2/6
1/6
1
f (2, 1)
=
= ,
f (2|1) =
h(1)
2/6
2
1/6
1
f (2, 1)
=
= .
f (3|1) =
h(1)
2/6
2
f (1|1) =
Multivariate Distributions
H
HH
x
y
HH
H
1
2
3
h(y )
Marginal Distributions
1
1/6
1/6
2/6
Conditional Distributions
2
3
g (x)
1/6
1/6
1/6
2/6
2/6
2/6
1
1/6
2/6
2/6
(4) The conditional distribution of Y given X = 2:
w (y |2) =
f (2, y )
⇒
g (2)
1/6
1
f (2, 1)
=
= ,
g (2)
2/6
2
0
f (2, 2)
=
= 0,
w (2|2) =
g (2)
2/6
1/6
1
f (2, 3)
=
= .
w (3|2) =
g (2)
2/6
2
w (1|2) =
Multivariate Distributions
H
HH
x
y
HH
H
1
2
3
h(y )
Marginal Distributions
1
1/6
1/6
2/6
Conditional Distributions
2
3
g (x)
1/6
1/6
1/6
2/6
2/6
2/6
1
1/6
2/6
2/6
(5) Since
f (1, 1) = 0 = g (1) · h(1) =
we see that X and Y are dependent.
2 2
·
6 6
Multivariate Distributions
Marginal Distributions
Thank You!!!
Conditional Distributions