LEGACY CONTENT.
If you are looking for Voteview.com, PLEASE CLICK HEREThis site is an archived version of Voteview.com archived from University of Georgia on
May 23, 2017. This pointintime capture includes all files publicly linked on Voteview.com at that time. We provide access to this content as a service to ensure that past users of Voteview.com have access to historical files. This content will remain online until at least
January 1st, 2018. UCLA provides no warranty or guarantee of access to these files.
 Definition: Expected Value
E(X) = å_{i=1,n} x_{i}f(x_{i}) for discrete
E(X) = ò_{¥}^{+¥} xf(x)dx for continuous
 Example: Discrete Uniform Distribution: Roll one die
æ 1/6 x=1,2,3,4,5,6
f(x) = ç
è 0 otherwise
E(X) = å_{i=1,6} x_{i}f(x_{i}) = 1/6(1 + 2 + 3 + 4 + 5 + 6) = 21/6 = 3.5
 E(X) is the mean of f(x). In the simple example of one die,
E(X) is quite literally the arithmetic average or center of
mass or center of balance  note that 3.5 is the "midpoint" between
1 and 6.
 Example: Continuous Uniform Distribution
æ 1/(b  a) a < x < b
f(x) = ç
è 0 otherwise
E(X) = ò_{¥}^{+¥} xf(x)dx =
ò_{a}^{b} x[1/(ba)]dx = [1/(ba)](x^{2}/2)_{a}^{b} =
[1/(ba)][(b^{2}  a^{2})/2] = (b+a)/2
 Example: Bernoulli Distribution
æ p x = 1
ç
f(x) = ç 1p x = 0
ç
è 0 otherwise
E(X) = å_{i=1,2} x_{i}f(x_{i}) = 1*p + 0*(1  p) = p
 Example:
æ 3x^{2} 0 < x < 1
f(x) = ç
è 0 otherwise
E(X) = ò_{¥}^{+¥} xf(x)dx =
ò_{0}^{1} x[3x^{2}]dx =
[3x^{4}/4]_{0}^{1} = 3/4
 Expected Value of a Function of a Random Variable
Let r(X) be any function of X. For example:
r(X) = aX + b; where a and b are constants;
r(X) = X^{1/4}
r(X) = e^{X}; and so on.
Then
E(r(X)) = å_{i=1,n} r(x_{i})f(x_{i}) for discrete
E(r(X)) = ò_{¥}^{+¥} r(x)f(x)dx for continuous
 Example: E(r(X)) = X^{2} for Discrete Uniform Distribution (roll
one dice)
E(X^{2}) = å_{i=1,6} (x_{i}^{2})*(1/6) = (1/6)(1 + 4 + 9 + 16 + 25 + 36) = 91/6
 Example: E(r(X)) = X^{2} for Continuous Uniform Distribution
E(X^{2}) = ò_{a}^{b}(x^{2})(1/[ba])dx =
(1/[ba])ò_{a}^{b}(x^{2})dx =
(1/[ba])(x^{3}/3)_{a}^{b} =
(1/3[ba])(b^{3}  a^{3}) = (b^{2} + ab + a^{2})/3
 Example: E(r(X)) = X^{2} for Bernoulli Distribution
E(X^{2}) = (1^{2})*p + (0^{2})*(1p) = p
 Example: E(r(X)) = X^{2} for
æ 3x^{2} 0 < x < 1
f(x) = ç
è 0 otherwise
E(X^{2}) =
ò_{0}^{1}(x^{2})3x^{2}dx =
ò_{0}^{1}3x^{4}dx =
(3x^{5}/5)_{0}^{1} = 3/5
 Two Important Properties of Expected Values
 E(aX + b) = aE(X) + b
To see this, let r(X) = aX + b, then:
E(r(X)) = E(aX + b) =
ò_{¥}^{+¥} (ax + b)f(x)dx =
ò_{¥}^{+¥} axf(x)dx +
ò_{¥}^{+¥} bf(x)dx =
a(ò_{¥}^{+¥} xf(x)dx) +
b(ò_{¥}^{+¥} f(x)dx) =
aE(X) + b
because the the term,
ò_{¥}^{+¥} xf(x)dx,
is just the definition of E(X), and
ò_{¥}^{+¥} f(x)dx = 1
by the 2nd Axiom of Probability.
Note that this shows that the expected value of a constant is a constant!
That is, E(b) = b, where b is a constant. Because E(X) is a
constant, this means that:
E[E(X)] = E(X)!
 E(X_{1} + X_{2} + ... + X_{n}) = E(X_{1}) + E(X_{2}) + ... + E(X_{n})
Proof: This will be shown in either Notes #7 or Notes #8.
 Variance
Variance is a measure of the dispersion of a random variable.
Definition: Variance: VAR(X) =
s^{2} = E{[X  E(X)]^{2}}
Simplifying: VAR(X) =
s^{2} = E{X^{2} 2XE(X) + [E(X)]^{2}} =
E(X^{2}) 2E(X)E(X) + [E(X)]^{2} = E(X^{2})  [E(X)]^{2}
Note that we used (12A) to derive this.
 Standard Deviation
The standard deviation of a random variable is just the square root of
the variance. Namely:
s =
[VAR(X)]^{1/2}
The standard deviation is in the same units as the random variable!
This is not true of the variance!
 Example: VAR(X) for Discrete Uniform Distribution (roll one dice)
From (2) and (8), VAR(X) = s^{2} =
91/6  (21/6)^{2}
 Example: VAR(X) for Continuous Uniform Distribution
From (4) and (9), VAR(X) =
s^{2} = (b^{2} + ab + a^{2})/3  [(b + a)/2]^{2} =
[(b  a)^{2}]/12
 Example: VAR(X) for Bernoulli Distribution
From (5) and (10), VAR(X) =
s^{2} = p  p^{2} = p(1  p)
 Example: VAR(X) for:
æ 3x^{2} 0 < x < 1
f(x) = ç
è 0 otherwise
From (6) and (11), VAR(X) = 3/5  (3/4)^{2} = 3/80
 Problem 3.18 p.87
Let X = "daily sales in dollars". Clearly the range of X is
$0, $50,000, $100,000.
To get P(X = 100,000) note that the salesman would have to contact
two customers and sell both customers the heavyequipment.
Hence: P(X = 100,000) = (2/3)*(1/10)*(1/10) = 2/300
To get P(X = 0) the salesman could either contact one customer
and make no sale, or contact two customers and make no sales.
Hence: P(X = 0) = (1/3)*(9/10) + (2/3)*(9/10)*(9/10) = 252/300
To get P(X = 50,000) the salesman could either contact one customer
and make a sale, or contact two customers and make only one sale.
Hence: P(X = 50,000) = (1/3)*(1/10)+ (2/3)*[(1/10)*(9/10) + (9/10)*(1/10)] = 46/300
This allows us to state the distribution of X:
æ 252/300 x = 0
ç
ç 46/300 x = $50,000
f(x) = ç
ç 2/300 x = $100,000
ç
è 0 otherwise
 To get the mean of daily sales:
E(X) = å_{i=1,3} x_{i}f(x_{i}) =
0*(252/300) + 50,000*(46/300) + 100,000*(2/300) = $8333.00

To get the standard deviation, s, of daily sales, we calculate the variance
and take its square root.
E(X^{2}) =
å_{i=1,3} (x_{i})^{2}f(x_{i}) = (0^{2})*(252/300) + (50,000^{2})*(46/300) + (100,000^{2})*(2/300) = 450,000,000
VAR(X) = s^{2} = 450,000,000  (8333^{2}) = 380,561,111
So that: std. dev. = [VAR(X)^{1/2}] =
s = $19,507.98
Recall that the standard deviation is in the original units of X
and can be so interpreted! This is not true of VAR(X).
 Two Important Properties of Variances
 VAR(aX + b) = (a^{2})VAR(X)
= (a^{2})s^{2}
To see this, let r(X) = aX + b, then:
VAR[r(X)] = E{[aX + b  E(aX + b)]^{2}} = E{[aX + b  aE(X)  b]^{2}} =
E{[a^{2}][X  E(X)]^{2}} = (a^{2})E{[X  E(X)]^{2}} = (a^{2})VAR(X)
= (a^{2})s^{2}
Note that (12a) of notes #5 was used to derive this.
 If X_{1}, X_{2}, ... , X_{n},
are independent random variables, then:
VAR(a_{1}X_{1} + a_{2}X_{2} + ... + a_{n}X_{n} + b) =
å_{i=1,n} (a_{i})^{2}VAR(X_{i}) =
å_{i=1,n} (a_{i})^{2}
s_{i}^{2}
See discussion in Notes #8.
 Problem 4.19 p.149
By inspection, we see that this is a continuous uniform distribution
with parameters:
b = 61 and a = 59. Hence:
E(X) = (b + a)/2 = 60 and VAR(X) =
s^{2} = [(b  a)^{2}]/12 = 4/12 = 1/3
 Problem 4.20 p.149
æ 2y 0 < y < 1
f(y) = ç
è 0 otherwise
 Find the mean and the variance:
E(Y) = ò_{0}^{1} y*2ydy =
(2(y^{3})/3)_{0}^{1} = 2/3
E(Y^{2}) = ò_{0}^{1} (y^{2})*2ydy =
(2y^{4}/4)_{0}^{1} = 1/2
Hence: VAR(Y) =
s^{2} = (1/2)  (2/3)^{2} = 1/18
 Let the profit be equal to: X = 200Y  60. Find the mean and
the variance of X.
E(X) = E(200Y  60) = 200E(Y)  60 = 200*(2/3)  60 = 220/3
VAR(X) = VAR(200Y  60) = (200^{2})VAR(Y) = 20,000/9
 To solve c) we need the Tchebysheff Inequality. Since I am not teaching it,
we will skip this!