1. 4.1 Mathematical Expectation
Example: Repair costs for a particular machine are represented by the following probability distribution:
What is the expected value of the repairs?
That is, over time what do we expect repairs to cost on average? x
$50
$200
$350
P(X = x)
0.3
0.2
0.5
The expected value or mean of a probability distribution is the long run theoretical average.
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2. Expected Value – Repair Costs
μ = mean of the probability distribution
For discrete variables,
μ = E(X) = ∑ x f(x)
So, for our example,
E(X) = 50(0.3) + 200(0.2) + 350(0.5) = $230
E(x) <weighted average>
E(X) = 50(0.3) + 200(0.2) + 350(0.5) = $230
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3. Another Example – Investment
By investing in a particular stock, a person can take a profit in a given year of $4000 with a probability of 0.3 or take a loss of $1000 with a probability of 0.7. What is the investor’s expected gain on the stock?
X $ $1000
P(X)
E(X) = $4000 (0.3) -$1000(0.7) = $500 X
P(X)
E(X) = 4000 (0.3) -1000(0.7) = 500
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4. Expected Value - Continuous Variables
For continuous variables,
μ = E(X) = E(X) = ∫ x f(x) dx
Vacuum cleaner example: problem 7 pg. 88
x, 0 < x < 1
f(x) = 2-x, 1 ≤ x < 2
0, elsewhere
(in hundreds of hours.) {
= 1 * 100 = hours of operation annually, on average
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5. Functions of Random Variables
Ex 4.4. pg. 111: Probability of X, the number of cars passing through a car wash in one hour on a sunny Friday afternoon, is given by
Let g(X) = 2X -1 represent the amount of money paid to the attendant by the manager. What can the attendant expect to earn during this hour on any given sunny Friday afternoon?
E[g(X)] = Σ g(x) f(x) = Σ (2X-1) f(x)
= (2*4-1)(1/12) +(2*5-1)(1/12) …+(2*9-1)(1/6) = $12.67 x 4 5 6 7 8 9
P(X = x)
1/12
1/4
1/6
Σ (2x-1) f(x) = 7(1/12) + 9(1/12) … + 17(1/6) = $12.67
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6. 4.2 Variance of a Random Variable
Recall our example: Repair costs for a particular machine are represented by the following probability distribution:
What is the variance of the repair cost?
That is, how might we quantify the spread of costs? x
$50
200
350
P(X = x)
0.3
0.2
0.5
JMB Chapter 4 Lecture 1
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7. Variance – Discrete Variables
For discrete variables,
σ2 = E [(X - μ)2] = ∑ (x - μ)2 f(x) = E (X2) - μ2
Recall, for our example, μ = E(X) = $230
Preferred method of calculation:
σ = [E(X2)] – μ2
= 502 (0.3) (0.2) (0.5) – = $17,100
Alternate method of calculation:
σ2 = E(X- μ)2 f(x)
= (50-230)2 (0.3) + ( )2 (0.2) + ( )2 (0.5) = $17,100
E(X2) – μ2 = 502 (0.3) (0.2) (0.5) – 2302 = $17,100
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8. Variance - Investment Example
By investing in a particular stock, a person can take a profit in a given year of $4000 with a probability of 0.3 or take a loss of $1000 with a probability of 0.7. What are the variance and standard deviation of the investor’s gain on the stock?
E(X) = $4000 (0.3) -$1000 (0.7) = $500
σ2 = [∑(x2 f(x))] – μ2
= (4000)2(0.3) + (-1000)2(0.7) – 5002 = $5,250,000
σ = $ X
P(X)
E(X) = 4000 (0.3) -1000(0.7) = 500
σ2 = ∑(x2 f(x)) –μ2 = (4000)2(0.3) + (-1000)2(0.7) – 5002 = $5,250,000
σ =$
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9. Variance of Continuous Variables
For continuous variables,
σ2 = E [(X - μ)2] =[∫ x2 f(x) dx] – μ2
Recall our vacuum cleaner example pr. 7 pg. 88
x, 0 < x < 1
f(x) = 2-x, 1 ≤ x < 2
0, elsewhere
(in hundreds of hours of operation.)
What is the variance of X? The variable is continuous, therefore we will need to evaluate the integral. {
E(X) = ∫x2 f(x)dx – μ2
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10. Variance Calculations for Continuous Variables
(Preferred calculation)
What is the standard deviation?
σ = hours
[∫01 x3 dx + ∫12x2 (2-x)dx] – μ2 = x4/4 |10 + (2x3/3 – x4/4)| =
σ =
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11. Covariance/ Correlation
A measure of the nature of the association between two variables
Describes a potential linear relationship
Positive relationship
Large values of X result in large values of Y
Negative relationship
Large values of X result in small values of Y
“Manual” calculations are based on the joint probability distributions
Statistical software is often used to calculate the sample correlation coefficient (r)
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12. What if the distribution is unknown?
Chebyshev’s theorem:
The probability that any random variable X will assume a value within k standard deviations of the mean is at least 1 – 1/k2. That is,
P(μ – kσ < X < μ + kσ) ≥ 1 – 1/k2
“Distribution-free” theorem – results are weak
If we believe we “know” the distribution, we do not use Chebyshev’s theorem to characterize variability
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