1. Understand the concepts of a random variable and a probability distribution.
2. Be able to distinguish between discrete and continuous random variables.
3. Calculate probabilities for multiple random variables based on joint probability functions
4. Be able to compute and interpret the expected value, variance, and standard deviation for a
discrete random variable.
Learning Objectives
DISCRETE RANDOM VARIABLES
:
S
→
R.
e.g., Y (T T H) = 3 ,
71
NOTE : In this example X(s) and Y (s) are actually integers .
0 if the sequence has no H , i.e., Y (T T T ) = 0 .
• Y (s) = The index of the first H ;
• X(s) = the number of Heads in the sequence ; e.g., X(HT H) = 2 ,
and examples of random variables are
S = {HHH , HHT , HT H , HT T , T HH , T HT , T T H , T T T } ,
EXAMPLE : Toss a coin 3 times in sequence. The sample space
is
X(·)
DEFINITION : A discrete random variable is a function X(s) from
a finite or countably infinite sample space S to the real numbers :
DISCRETE RANDOM VARIABLES
correspond to {HHH , HHT , HT H , T HH} .
corresponds to the event {HHT , HT H , T HH} ,
X(s) = the number of Heads ,
X
72
instead of
X(s) .
NOTATION : If it is clear what S is then we often just write
1 < X(s) ≤ 3 ,
and the values
X(s) = 2 ,
the value
with
S = {HHH , HHT , HT H , HT T , T HH , T HT , T T H , T T T } ,
EXAMPLE : For the sample space
Value-ranges of a random variable correspond to events in S .
Value-ranges of a random variable have a probability .
events in S have a probability .
P (0 < X ≤ 2) =
73
P (X ≤ −1) , P (X ≤ 0) , P (X ≤ 1) , P (X ≤ 2) , P (X ≤ 3) , P (X ≤ 4) ?
6
.
8
X(s) = the number of Heads ,
QUESTION : What are the values of
we have
with
S = {HHH , HHT , HT H , HT T , T HH , T HT , T T H , T T T } ,
EXAMPLE : For the sample space
Thus
and
Value-ranges of a random variable correspond to events in S ,
P ( {HT T , T HT , T T H} )
P ( {HHT , HT H , T HH} )
P ( {HHH} )
pX (1) ≡
pX (2) ≡
pX (3) ≡
74
pX (0) + pX (1) + pX (2) + pX (3) = 1 .
where
P ( {T T T } )
X(s) = the number of Heads ,
pX (0) ≡
we have
with
=
=
=
=
( Why ? )
1
8
3
8
3
8
1
8
S = {HHH , HHT , HT H , HT T , T HH , T HT , T T H , T T T } ,
EXAMPLE : For the sample space
NOTATION : We will also write pX (x) to denote P (X = x) .
E0
E1
TTT
THH
HTH
HHT
TTH
THT
HTT
Graphical representation of X .
E2
HHH
0
1
2
3
X(s)
75
(X : S → R must be defined for all s ∈ S and must be single-valued.)
The events E0 , E1 , E2 , E3 are disjoint since X(s) is a function !
S
E3
76
The graph of pX .
pX (x) ≡ P (X = x) ,
FX (x) ≡ P (X ≤ x) ,
( Why ? )
( Why ? )
• FX (−∞) = 0 and FX (∞) = 1 .
• P (a < X ≤ b) = FX (b) − FX (a) .
p(x) for pX (x)
77
and
F (x) for FX (x) .
NOTATION : When it is clear what X is then we also write
( Why ? )
• FX (x) is a non-decreasing function of x .
PROPERTIES :
is called the (cumulative) probability distribution function .
DEFINITION :
is called the probability mass function .
DEFINITION :
1
8
,
p(1) =
3
8
,
p(2) =
≡
≡
≡
≡
≡
F ( 0)
F ( 1)
F ( 2)
F ( 3)
F ( 4)
P (X ≤ 4)
P (X ≤ 3)
P (X ≤ 2)
P (X ≤ 1)
P (X ≤ 0)
P (X ≤ −1)
3
8
=
=
=
=
=
=
,
78
= F (2) − F (0) =
7
8
−
P (0 < X ≤ 2) = P (X = 1) + P (X = 2)
We see, for example, that
≡
F (−1)
we have the probability distribution function
p(0) =
1
8
=
1
1
1
8
4
8
7
8
0
p(3) =
6
8
1
8
.
,
S = {HHH , HHT , HT H , HT T , T HH , T HT , T T H , T T T } ,
EXAMPLE : With X(s) = the number of Heads , and
79
The graph of the probability distribution function FX .
= {H , T H , T T H , T T T H , · · · } .
1
2
,
p(2) =
n→∞
k=1
1
lim (1 − n ) = 1 .
n→∞
2
1
= 1 − n ,
2
80
EXERCISE : Draw the probability mass and distribution functions.
NOTE : The outcomes in S do not have equal probability !
k=1
, ···
( Why ? )
X(T T H) = 3
, p(3) = 18 , · · ·
n
n
1
p(k) =
k
2
k=1
k=1
and, as should be the case,
∞
n
p(k) = lim
p(k) =
1
4
X(T H) = 2 ,
and
F (n) = P (X ≤ n) =
p(1) =
X(H) = 1 ,
Then
The random variable X is the number of rolls until ”Heads” occurs :
S
Then the sample space is countably infinite , namely,
EXAMPLE : Toss a coin until ”Heads” occurs.
T H̃HT ,
T T H̃T ,
T H̃HH ,
T T H̃H ,
T T T H̃ ,
T H̃T H ,
H̃T T H ,
TTTT
T H̃T T ,
H̃T T T ,
1
2
,
pX (2) =
independent of n .
pX (1) =
1
4
,
81
pX (3) =
1
8
,
pX (4) =
··· ,
, and
}.
1
)
16
1
16
Each outcome in S4 has equal probability 2−n (here 2−4 =
where for each outcome the first ”Heads” is marked as H̃ .
H̃T HT ,
H̃T HH ,
S4 = { H̃HHH , H̃HHT , H̃HT H , H̃HT T ,
REMARK : We can also take S ≡ Sn as all ordered outcomes of
length n. For example, for n = 4,
X(s) is the number of tosses until ”Heads” occurs · · ·
For example,
(0 for T T T ) .
=
2
8
82
=
1
4
.
= P ( 2 Heads , 1st toss is Heads)
pX,Y (2, 1) = P (X = 2 , Y = 1)
pX,Y (x, y) = P (X = x , Y = y) .
Then we have the joint probability mass function
X(s) = # Heads , Y (s) = index of the first H
we let
S = {HHH , HHT , HT H , HT T , T HH , T HT , T T H , T T T } ,
EXAMPLE : Toss a coin 3 times in sequence. For the sample space
The probability mass function and the probability distribution function
can also be functions of more than one variable.
Joint distributions
For
1
8
0
0
0
1
8
2
8
1
8
4
8
0
1
8
y=3
1
8
0
2
8
1
8
0
0
0
1
8
1
8
0
y=2
pX (x)
1
1
8
3
8
3
8
1
8
83
• The marginal probability pY is the probability mass function of Y .
• The marginal probability pX is the probability mass function of X.
NOTE :
x=0
x=1
x=2
x=3
pY (y)
y=1
y=0
Joint probability mass function pX,Y (x, y)
we can list the values of pX,Y (x, y) :
X(s) = number of Heads, and Y (s) = index of the first H ,
S = {HHH , HHT , HT H , HT T , T HH , T HT , T T H , T T T } ,
EXAMPLE : ( continued · · · )
1
8
0
0
0
1
8
2
8
1
8
4
8
0
1
8
2
8
0
1
8
1
8
0
y=2
1
8
0
0
1
8
0
y=3
1
1
8
3
8
3
8
1
8
pX (x)
84
QUESTION : Are the events X = 2 and Y = 1 independent ?
• (X = 2 and Y = 1) corresponds to the event {HHT , HT H} .
• Y = 1 corresponds to the event {HHH , HHT , HT H , HT T } .
• X = 2 corresponds to the event {HHT , HT H , T HH} .
For example,
x=0
x=1
x=2
x=3
pY (y)
y=1
y=0
X(s) = number of Heads, and Y (s) = index of the first H .
EXAMPLE : ( continued · · · )
TTH
E 00
HHT
HTH
TTT
E 22 THH
E 21
E12 THT
E11
HTT
E 13
HHH
0
2
1
3
85
QUESTION : Are the events X = 2 and Y = 1 independent ?
The events Ei,j ≡ { s ∈ S : X(s) = i , Y (s) = j } are disjoint .
S
E 31
X(s)
Y(s)
≡ P (X = x , Y = y) ,
≡ P (X ≤ x , Y ≤ y) ,
and
86
F (x, y) for FX,Y (x, y) .
p(x, y) for pX,Y (x, y) ,
NOTATION : When it is clear what X and Y are then we also
write
is called the joint (cumulative) probability distribution function .
FX,Y (x, y)
DEFINITION :
is called the joint probability mass function .
pX,Y (x, y)
DEFINITION :
1
8
0
0
0
1
8
2
8
1
8
4
8
0
1
8
1
8
0
2
8
1
8
0
y=3
0
0
1
8
1
8
0
y=2
1
1
8
3
8
3
8
1
8
pX (x)
1
8
2
8
4
8
5
8
5
8
1
8
1
8
1
8
1
8
1
8
1
8
3
8
6
8
7
8
7
8
y=2
1
1
1
1
8
4
8
7
8
FX (·)
1
1
8
4
8
7
8
y=3
87
Note that the distribution function FX is a copy of the 4th column,
and the distribution function FY is a copy of the 4th row. ( Why ? )
x=0
x=1
x=2
x=3
FY (·)
y=1
y=0
Joint distribution function FX,Y (x, y) ≡ P (X ≤ x, Y ≤ y)
x=0
x=1
x=2
x=3
pY (y)
y=1
y=0
Joint probability mass function pX,Y (x, y)
EXAMPLE : Three tosses : X(s) = # Heads, Y (s) = index 1st H .
y=2
y=3
pX (x)
1
8
0
0
0
1
8
2
8
1
8
4
8
0
1
8
2
8
0
1
8
1
8
0
1
8
0
0
1
8
0
1
1
8
3
8
3
8
1
8
1
8
3
8
6
8
7
8
7
8
1
8
2
8
4
8
5
8
5
8
1
8
1
8
1
8
1
8
1
8
1
1
1
1
8
4
8
7
8
FX (·)
1
1
8
4
8
7
8
y=3
88
P (1 < X ≤ 3 , 1 < Y ≤ 3) = F (3, 3) − F (1, 3) − F (3, 1) + F (1, 1) ?
QUESTION : Why is
x=0
x=1
x=2
x=3
FY (·)
y=2
y=1
y=0
Joint distribution function FX,Y (x, y) ≡ P (X ≤ x, Y ≤ y)
x=0
x=1
x=2
x=3
pY (y)
y=1
y=0
Joint probability mass function pX,Y (x, y)
In the preceding example :
, Y = sum of the two rolls.
89
• List the values of the joint cumulative distribution function FX,Y (x, y) .
• List the values of the joint probability mass function pX,Y (x, y) .
• How many outcomes are there in S ?
• What is a good choice of the sample space S ?
X = result of the first roll
Define the random variables X and Y as
Suppose the outcome of a single roll is the side that faces down ( ! ).
Suppose each of the four sides is equally likely to end facing down.
(The sides are marked 1 , 2 , 3 , 4 .)
Roll a four-sided die (tetrahedron) two times.
EXERCISE :
G(s) = the number of green balls drawn .
90
• the marginal distribution functions FR (r) and FG (g) .
• the joint distribution function FR,G (r, g) .
• the marginal probability mass functions pR (r) and pG (g) .
• the joint probability mass function pR,G (r, g) .
List the values of
and
R(s) = the number of red balls drawn,
Define the random variables
2 red , 3 green , 4 blue balls .
Three balls are selected at random from a bag containing
EXERCISE :
for all x and y ,
for all x and y ,
=
P (Ex ) · P (Ey ) ,
for all x and y .
•
91
X −1 ({x}) ≡ { s ∈ S : X(s) = x } .
• In the current discrete case, x and y are typically integers .
NOTE :
P (Ex Ey )
are independent in the sample space S , i.e.,
Ex ≡ X −1 ({x}) and Ey ≡ Y −1 ({y}) ,
or, equivalently, if the events
pX,Y (x, y) = pX (x) · pY (y) ,
or, equivalently, if their probability mass functions satisfy
P (X = x , Y = y) = P (X = x) · P (Y = y) ,
Two discrete random variables X(s) and Y (s) are independent if
Independent random variables
TTH
E 00
HHT
HTH
TTT
E 22 THH
E 21
E12 THT
E11
HTT
E 13
HHH
0
2
1
3
Are X and Y independent ?
•
92
What are the values of pX (2) , pY (1) , pX,Y (2, 1) ?
•
Three tosses : X(s) = # Heads, Y (s) = index 1st H .
S
E 31
X(s)
Y(s)
pX,Y (k, ) = pX (k) · pY (),
93
for all 1 ≤ k, ≤ 6 ?
Y the result of the 2nd roll .
X be the result of the 1st roll ,
Are X and Y independent , i.e., is
and
Let
Roll a die two times in a row.
EXERCISE :
pX,Y (x, y) = pX (x) · pY (y) .
X(s) and Y (s) are independent if for all x and y :
RECALL :
x=0
x=1
x=2
x=3
pY (y)
1
8
0
0
0
1
8
1
8
0
2
8
1
8
1
8
1
8
1
8
2
8
1
8
4
8
0
0
0
0
0
94
y=0 y=1 y=2 y=3
1
1
8
3
8
3
8
1
8
pX (x)
Joint probability mass function pX,Y (x, y)
Are these random variables X and Y independent ?
EXERCISE :
95
QUESTION : Is FX,Y (x, y) = FX (x) · FY (y) ?
Joint distribution function FX,Y (x, y) ≡ P (X ≤ x, Y ≤ y)
y = 1 y = 2 y = 3 FX (x)
1
1
5
1
x=1
3
12
2
2
5
5
25
5
x=2
9
36
6
6
2
5
x=3
1
1
3
6
5
2
FY (y)
1
1
3
6
Joint probability mass function pX,Y (x, y)
y = 1 y = 2 y = 3 pX (x)
1
1
1
1
x=1
3
12
12
2
1
1
2
1
x=2
9
18
18
3
1
1
1
1
x=3
9
36
36
6
2
1
1
pY (y)
1
3
6
6
EXERCISE : Are these random variables X and Y independent ?
i≤k
i≤k
i≤k
96
FX (xk ) · FY (y ) .
=
i≤k
j≤
pX (xi ) } · {
j≤
pY (yj ) }
pY (yj ) }
(by independence)
for all x, y .
pX (xi ) · pY (yj )
pX,Y (xi , yj )
{ pX (xi ) ·
j≤
j≤
{
P (X ≤ xk , Y ≤ y )
=
=
=
=
FX,Y (xk , y ) =
PROOF :
FX,Y (x, y) = FX (x) · FY (y) ,
The joint distribution function of independent random variables
X and Y satisfies
PROPERTY :
and
Ey = Y −1 ({y}) ,
P (Ex |Ey ) ≡
P (Ex Ey )
P (Ey )
=
pX,Y (x, y)
.
pY (y)
97
Thus it is natural to define the conditional probability mass function
pX,Y (x, y)
pX|Y (x|y) ≡ P (X = x | Y = y) =
.
pY (y)
Then
be their corresponding events in the sample space S.
Ex = X −1 ({x})
For given x and y , let
Let X and Y be discrete random variables with joint probability
mass function
pX,Y (x, y) .
Conditional distributions
TTH
E 00
HHT
HTH
TTT
E 22 THH
E 21
E12 THT
E11
HTT
E 13
HHH
0
2
1
3
98
• What are the values of P (X = 2 | Y = 1) and P (Y = 1 | X = 2) ?
Three tosses : X(s) = # Heads, Y (s) = index 1st H .
S
E 31
X(s)
Y(s)
1
1
1
99
.
pX,Y (x,y)
pX (x)
pX,Y (x,y)
pY (y)
EXERCISE : Also construct the Table for pY |X (y|x) =
1
Conditional probability mass function pX|Y (x|y) =
y=0 y=1 y=2 y=3
x=0
1
0
0
0
2
4
x=1
0
1
8
8
4
4
x=2
0
0
8
8
2
x=3
0
0
0
8
Joint probability mass function pX,Y (x, y)
y = 0 y = 1 y = 2 y = 3 pX (x)
1
1
x=0
0
0
0
8
8
1
3
1
1
x=1
0
8
8
8
8
1
2
3
x=2
0
0
8
8
8
1
1
x=3
0
0
0
8
8
1
4
2
1
pY (y)
1
8
8
8
8
.
EXAMPLE : (3 tosses : X(s) = # Heads, Y (s) = index 1st H.)
1
1
pX,Y (x,y)
pY (y)
.
100
EXERCISE : Also construct the Table for P (Y = y|X = x) .
QUESTION : What does the last Table tell us?
1
Conditional probability mass function pX|Y (x|y) =
y=1 y=2 y=3
1
1
1
x=1
2
2
2
1
1
1
x=2
3
3
3
1
1
1
x=3
6
6
6
EXAMPLE :
Joint probability mass function pX,Y (x, y)
y = 1 y = 2 y = 3 pX (x)
1
1
1
1
x=1
3
12
12
2
1
1
2
1
x=2
9
18
18
3
1
1
1
1
x=3
9
36
36
6
2
1
1
pY (y)
1
3
6
6
≡
k
xk · P (X = xk )
=
k
xk · pX (xk ) .
E[aX + b] = a E[X] + b .
•
101
E[aX] = a E[X] ,
•
EXERCISE : Prove the following :
EXAMPLE : The expected value of rolling a die is
6
1
1
1
1
E[X] = 1 · 6 + 2 · 6 + · · · + 6 · 6 = 6 · k=1 k =
( E[X] is also called the mean of X .)
Thus E[X] represents the weighted average value of X .
E[X]
The expected value of a discrete random variable X is
Expectation
7
2
.
= {H , T H , T T H , T T T H , · · · } .
X(T H) = 2
,
n
k
lim
k
n→∞
2
k=1
X(T T H) = 3 .
102
Perhaps using Sn = {all sequences of n tosses} is better · · ·
REMARK :
1
1
1
E[X] = 1 · + 2 · + 3 · + · · · =
2
4
8
n
k
n
k/2
k=1
1 0.50000000
2 1.00000000
3 1.37500000
10 1.98828125
40 1.99999999
Then
X(H) = 1 ,
= 2.
The random variable X is the number of tosses until ”Heads” occurs :
S
EXAMPLE : Toss a coin until ”Heads” occurs. Then
k
g(xk ) p(xk ) .
E[g(X)] =
k=1
6
1
k
6
2
=
103
1 6(6 + 1)(2 · 6 + 1)
6
6
SOLUTION : Here g(X) = X 2 , and
=
91
6
∼
= $15.17 .
What should the entry fee be for the betting to break even?
The pay-off of rolling a die is $k2 , where k is the side facing up.
EXAMPLE :
E[g(X)] ≡
The expected value of a function of a random variable is
1
2
2
3
1
3
2
9
1
9
= 1·
E[XY ] = 1 ·
+ 2·
+ 3·
E[Y ]
1
3
2
9
1
9
2
3
= 1·
x=1
x=2
x=3
pY (y)
104
+ 6·
+ 4·
1
36
1
18
1
12
+ 2·
+ 2·
1
3
1
6
+ 2·
1
12
1
18
1
36
1
6
1
6
1
6
+ 9·
+ 6·
+ 3·
+ 3·
+ 3·
1
12
1
18
1
36
1
6
1
36
1
18
1
12
y=1 y=2 y=3
E[X]
EXAMPLE :
k
=
=
=
1
1
2
1
3
1
6
5
2
5
3
3
2
pX (x)
,
.
,
( So ? )
The expected value of a function of two random variables is
E[g(X, Y )] ≡
g(xk , y ) p(xk , y ) .
k
xk pX (xk )} · {
105
(by independence)
y pY (y )}
y pY (y )}
xk y pX (xk ) pY (y )
xk pX (xk )
xk y pX,Y (xk , y )
= E[X] · E[Y ] .
k
k{
k
= {
=
=
=
EXAMPLE : See the preceding example !
E[XY ]
If X and Y are independent then E[XY ] = E[X] E[Y ] .
PROOF :
•
PROPERTY :
k {xk
{y
pX,Y (xk , y )} +
pX (xk )} +
xk pX,Y (xk , y ) +
xk pX,Y (xk , y ) +
106
k
{
y
pY (y )}
k
pX,Y (xk , y )}
y pX,Y (xk , y )
y pX,Y (xk , y )
( Always ! )
k
(xk + y ) pX,Y (xk , y )
k {xk
k
k
k
= E[X] + E[Y ] .
=
=
=
=
=
NOTE : X and Y need not be independent !
E[X + Y ]
PROOF :
PROPERTY : E[X + Y ] = E[X] + E[Y ] .
E[XY ] = E[X] E[Y ] ,
107
then it does not necessarily follow that X and Y are independent !
Thus if
X and Y are not independent
E[XY ] = 16
•
E[Y ] = 8 ,
E[X] = 2 ,
•
Show that
Probability mass function pX,Y (x, y)
y = 6 y = 8 y = 10 pX (x)
1
1
2
x=1
0
5
5
5
1
1
x=2
0
0
5
5
1
1
2
x=3
0
5
5
5
2
1
2
pY (y)
1
5
5
5
EXERCISE :
k
(xk − μ)2 p(xk ) ,
We have
V ar(X)
108
= E[X 2 ] − μ2 .
= E[X 2 ] − 2μ2 + μ2
= E[X 2 ] − 2μE[X] + μ2
= E[X 2 − 2μX + μ2 ]
which is the average weighted square distance from the mean.
V ar(X) ≡ E[ (X − μ) ] ≡
2
μ = E[X] .
Then the variance of X is
Let X have mean
Variance and Standard Deviation
V ar(X) =
E[ (X − μ)2 ] =
E[X 2 ] − μ2 .
σ
k=1
[k2 ·
1
] − μ2
6
=
109
35
12
∼
= 1.70 .
7 2
1 6(6 + 1)(2 · 6 + 1)
35
− ( ) =
.
=
6
6
2
12
The standard deviation is
V ar(X) =
6
EXAMPLE : The variance of rolling a die is
which is the average weighted distance from the mean.
σ(X) ≡
The standard deviation of X is
,
E[Y ] = μY .
Cov(X, Y )
We have
110
= E[XY ] − E[X] E[Y ] .
= E[XY ] − μX μY − μY μX + μX μY
= E[XY − μX Y − μY X + μX μY ]
= E[ (X − μX ) (Y − μY ) ]
k,
Then the covariance of X and Y is defined as
Cov(X, Y ) ≡ E[ (X−μX ) (Y −μY ) ] =
(xk −μX ) (y −μY ) p(xk , y ) .
E[X] = μX
Let X and Y be random variables with mean
Covariance
111
Cov(X, Y ) < 0 .
• If X > μX when Y < μY and X < μX when Y > μY then
Cov(X, Y ) > 0 .
• If X > μX when Y > μY and X < μX when Y < μY then
Cov(X, Y ) measures ”concordance ” or ”coherence ” of X and Y :
NOTE :
k,
(xk − μX ) (y − μY ) p(xk , y )
= E[XY ] − E[X] E[Y ] .
=
Cov(X, Y ) ≡ E[ (X − μX ) (Y − μY ) ]
We defined
112
• V ar(X + Y ) = V ar(X) + V ar(Y ) + 2 Cov(X, Y ) .
• Cov(X + Y, Z) = Cov(X, Z) + Cov(Y, Z) ,
• Cov(X, cY ) = c Cov(X, Y ) ,
• Cov(cX, Y ) = c Cov(X, Y ) ,
• Cov(X, Y ) = Cov(Y, X) ,
• V ar(aX + b) = a2 V ar(X) ,
EXERCISE : Prove the following :
from which the result follows.
113
E[XY ] = E[X] E[Y ] .
and that if X and Y are independent then
Cov(X, Y ) ≡ E[ (X − μX ) (Y − μY ) ] = E[XY ] − E[X] E[Y ] ,
We have already shown ( with μX ≡ E[X] and μY ≡ E[Y ] ) that
PROOF :
If X and Y are independent then Cov(X, Y ) = 0 .
PROPERTY :
114
then it does not necessarily follow that X and Y are independent !
Cov(X, Y ) = 0 ,
X and Y are not independent
•
Thus if
Cov(X, Y ) = E[XY ] − E[X] E[Y ] = 0
E[XY ] = 16
•
E[Y ] = 8 ,
E[X] = 2 ,
•
Show that
Probability mass function pX,Y (x, y)
y = 6 y = 8 y = 10 pX (x)
1
1
2
x=1
0
5
5
5
1
1
x=2
0
0
5
5
1
1
2
x=3
0
5
5
5
2
1
2
pY (y)
1
5
5
5
EXERCISE : ( already used earlier · · · )
from which the result follows.
115
Cov(X, Y ) = 0 ,
and that if X and Y are independent then
V ar(X + Y ) = V ar(X) + V ar(Y ) + 2 Cov(X, Y ) ,
We have already shown (in an exercise !) that
PROOF :
V ar(X + Y ) = V ar(X) + V ar(Y ) .
If X and Y are independent then
PROPERTY :
for
Compute
Cov(X, Y )
E[XY ] , V ar(X) , V ar(Y )
E[X] , E[Y ] , E[X 2 ] , E[Y 2 ]
116
Joint probability mass function pX,Y (x, y)
y = 0 y = 1 y = 2 y = 3 pX (x)
1
1
x=0
0
0
0
8
8
1
3
1
1
x=1
0
8
8
8
8
1
2
3
x=2
0
0
8
8
8
1
1
x=3
0
0
0
8
8
4
2
1
1
pY (y)
1
8
8
8
8
EXERCISE :
for
Compute
Cov(X, Y )
E[XY ] , V ar(X) , V ar(Y )
E[X] , E[Y ] , E[X 2 ] , E[Y 2 ]
x=1
x=2
x=3
pY (y)
1
3
2
9
1
9
2
3
117
1
12
1
18
1
36
1
6
1
12
1
18
1
36
1
6
y=1 y=2 y=3
1
1
2
1
3
1
6
pX (x)
Joint probability mass function pX,Y (x, y)
EXERCISE :