Download 5.3 EXPECTED VALUE AND VARIANCE

Section 5.3
5.3.1
Expected Value and Variance
5.3 EXPECTED VALUE AND VARIANCE
def: The expected value of a random variable
X on a probability space (S, p) is the sum
E(X) =
X(s)p(s)
s∈S
Example 5.3.1: The expected outcome of a
fair die is
1
1
1
1
1
1
21
7
1· +2· +3· +4· +5· +6· =
=
6
6
6
6
6
6
6
2
Example 5.3.2: The expected outcome of the
standard loaded die is
2
6
91
13
1
+2·
+ ··· + 6 ·
=
=
1·
21
21
21
21
3
Over a non-uniform probability space the
expected value of a random variable is the
weighted mean.
Coursenotes by Prof.
Jonathan L. Gross for use with Rosen:
Discrete Math and Its Applic., 5th Ed.
Chapter 5
5.3.2
DISCRETE PROBABILITY
Example 5.3.3: Flip a fair coin three times.
The expected number of heads is
0·
3
3
1
12
3
1
+1· +2· +3· =
=
8
8
8
8
8
2
This calculation is based on the binomial
distribution.
Example 5.3.4: Flip a standard loaded coin
three times. The expected number of heads is
12
48
64
1
+1·
+2·
+3·
125
125
125
125
300
12
0 + 12 + 96 + 192
=
=
=
125
125
5
0·
This calculation too is based on the binomial
distribution.
Coursenotes by Prof.
Jonathan L. Gross for use with Rosen:
Discrete Math and Its Applic., 5th Ed.
Section 5.3
5.3.3
Expected Value and Variance
SUMMING RANDOM VARIABLES
Theorem 5.3.1. Let X1 and X2 be random
variables on a probability space (S, p). Then
E(X1 + X2 ) = E(X1 ) + E(X2 )
Proof:
E(X1 + X2 ) =
(X1 (s) + X2 (s))p(s)
s∈S
=
X1 (s)p(s) + X2 (s)p(s)
s∈S
=
X1 (s)p(s) +
s∈S
X2 (s)p(s)
s∈S
= E(X1 ) + E(X2 )
Example 5.3.5: When two fair dice are rolled,
here are both calculations:
7 7
+ = 7 and
2 2
6
6
1 252
=7
E(X1 + X2 ) =
(j + k) =
36 j=1
36
E(X1 ) + E(X2 ) =
k=1
Coursenotes by Prof.
Jonathan L. Gross for use with Rosen:
Discrete Math and Its Applic., 5th Ed.
Chapter 5
5.3.4
DISCRETE PROBABILITY
Theorem 5.3.2. Let X1 , . . . , Xn be random
variables on a probability space (S, p). Then
E(X1 + · · · + Xn ) = E(X1 ) + · · · + E(Xn )
Proof:
By induction on n, using Thm 5.3.1. ♦
Example 5.3.6: When 100 fair coins are
tossed, the expected number of heads is
1
2 · 100 = 50
Example 5.3.7: When 100 standard loaded
coins are tossed, the expected number of heads is
0.8 · 100 = 80
Coursenotes by Prof.
Jonathan L. Gross for use with Rosen:
Discrete Math and Its Applic., 5th Ed.
Section 5.3
5.3.5
Expected Value and Variance
GEOMETRIC DISTRIBUTION
def: The geometric distribution on the
positive integers is
pr(k) = (1 − p)k−1 p
Example 5.3.8: A coin with p(H) = p is
tossed until the first occurrence of heads. Then
the probability of requiring exactly k tosses is
(1 − p)k−1 p. We observe that
∞
(1−p)k−1 p = p
k=1
∞
(1−p)k−1 =
k=1
p
=1
1 − (1 − p)
It is proved in the text that
E(X) =
∞
(1 − p)k−1 pk
k=1
d
1
−1 = p (1 − x) =
dx
p
x=1−p
Coursenotes by Prof.
Jonathan L. Gross for use with Rosen:
Discrete Math and Its Applic., 5th Ed.
Chapter 5
5.3.6
DISCRETE PROBABILITY
INDEPENDENT RANDOM VARIABLES
def: The random variables X and Y on the
probability space (S, p) are independent if for
all real numbers r1 and r2
p(X = r1 ∧ Y = r2 ) = p(X = r1 ) · p(Y = r2 )
Example 5.3.9: Suppose that X is the sum of
two fair dice and Y is the product. Then
p(X = 2) =
1
36
and
p(Y = 5) =
1
18
However,
p(X = 2 ∧ Y = 5) = 0 =
1 1
·
36 18
Thus X and Y are not independent.
Coursenotes by Prof.
Jonathan L. Gross for use with Rosen:
Discrete Math and Its Applic., 5th Ed.
Section 5.3
5.3.7
Expected Value and Variance
VARIANCE and STANDARD DEVIATION
def: The variance of a random variable X on a
probability space (S, p) is the sum
2
2
X(s) − E(X) p(s)
σ (X) = V (X) =
s∈U
def: The standard deviation of a random
variable X on a probability space (U, p) is
σ(X) = V (X)
Example 5.3.10: Flip a fair coin three times.
The variance of the number of heads is
3 2 1 3 2 3 3 2 3 3 2 1
· + 1−
· + 2−
· + 3−
·
0−
2
8
2
8
2
8
2
8
24
3
9 1 1 3 1 3 9 1
=
= · + · + · + · =
4 8 4 8 4 8 4 8
32
4
The standard deviation is
√
3
3
=
4
2
Coursenotes by Prof.
Jonathan L. Gross for use with Rosen:
Discrete Math and Its Applic., 5th Ed.
5.3.8
Chapter 5 DISCRETE PROBABILITY
Example 5.3.11: Flip a standard loaded coin
three times. The variance of the number of heads
is
12 2 1
12 2 12
0−
·
·
+ 1−
5
125
5
125
12 2 64
12 2 48
+ 3−
+ 2−
·
·
5
125
5
125
1500
12
144 + 49 · 12 + 4 · 48 + 9 · 64
= 5 =
=
55
5
25
The standard deviation is
√
2 3
12
=
25
5
CHEBYSHEV INEQUALITY
Chebyshev Inequality. Let X be a random
variable on any probability space that has a
mean and a variance. Then
1
p |X(s) − E(X)| ≥ kσ(X) ≤ 2
k
Coursenotes by Prof.
Jonathan L. Gross for use with Rosen:
Discrete Math and Its Applic., 5th Ed.