4.1 Mathematical Expectation

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
2
Expected Value Repair Costs

• µ E(X)
• 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

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 investors
  expected gain on the stock?

X 4000 -1000 P(X) 0.3
0.7 E(X) 4000 (0.3) -1000(0.7)
500
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 lt x lt 1
• f(x) 2-x, 1 x lt 2
• 0, elsewhere
• (in hundreds of hours.)

1 100 100.0 hours of operation annually, on
average
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?
• Eg(X) S g(x) f(x) S (2X-1) f(x)
• (24-1)(1/12) (25-1)(1/12) (29-1)(1/6)
  12.67

x 4 5 6 7 8 9
P(X x) 1/12 1/12 1/4 1/4 1/6 1/6
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 define the spread of costs?

x 50 200 350
P(X x) 0.3 0.2 0.5
7
Variance Discrete Variables

• For discrete variables,
• s2 E (X - µ)2 ? (x - µ)2 f(x) E
  (X2) - µ2
• Recall, for our example, µ E(X) 230
• Preferred method of calculation
• s2 E(X2) µ2 502 (0.3) 2002
  (0.2) 3502 (0.5) 2302 17,100
• Alternate method of calculation
• s2 E(X- µ)2 f(x)
• (50-230)2 (0.3) (200-230)2 (0.2)
  (350-230)2 (0.5) 17,100

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 investors gain on the
  stock?
• E(X) 4000 (0.3) -1000 (0.7) 500
• s2 ?(x2 f(x)) µ2
• (4000)2(0.3) (-1000)2(0.7) 5002
  5,250,000
• s 2291.29

9
Variance of Continuous Variables

• For continuous variables,
• s2 E (X - µ)2 ? x2 f(x) dx µ2
• Recall our vacuum cleaner example problem 7 (pg.
  88)
• x, 0 lt x lt 1
• f(x) 2-x, 1 x lt 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.

10
Variance Calculations for Continuous Variables

• 
  (Preferred calculation)
• What is the standard deviation?
• s 0.4082 hours

11
Covariance

• 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
• Calculations based on the joint probability
  distributions

• Y

12
What if the distribution is unknown?

• Chebyshevs 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 ks lt X lt µ ks) 1 1/k2