DISCRETE RANDOM DISCRETE RANDOM VARIABLESVARIABLES

RANDOM VARIABLES

• A random variable assigns a numericalnumerical value to each simple event in the sample space

• Its value is determined by the outcome of a random experiment

TYPES OF RANDOM VARIABLES

• Two types -- Discrete and Continuous• Discrete

– # possible outcomes is finite – can list them all• Choose an integer between 1 and 5

– # possible outcomes is countably infinite• Choose any positive integer

• Continuous -- infinite # of possible outcomes– Outcomes are from a bounded set

• Choose any number on the real number line between 1- 5

– Outcomes are from an unbounded set• Choose any number on the real number line

DISCRETE PROBABILITY DISTRIBUTIONS

• Assigns probabilities to each possible outcome

• Probabilities– Lie between 0 and 1– Sum to 1

EXAMPLE

• Suppose Mount Vernon (MV) Ford in Alexandria, Va. has four Thunderbird automobiles in stock this weekend

• X = the number of Thunderbirds MV Ford will sell from stock this weekend

PROBABILITY DISTRIBUTION

Suppose past data indicates that the number of sales in a weekend has the following probability distribution:

Sales (x) Probability P(x)Sales (x) Probability P(x)

0 0 .05.05

11 .25.25

22 .35.35

33 .30.30

44 .05.05

PROBABILITIES

• What is the probability that, this weekend MV Ford will sell:

1) Two ThunderbirdsP(X=2) = .35

2) Between 1 and 3 ThunderbirdsP(1 X 3) = P(1) + P(2) + P(3) = .25 + .35 + .30 = .90

3) Two or more ThunderbirdsP(X 2) = P(2) + P(3) + P(4) = .35 + .30 + .05 = .70

4) Three or less ThunderbirdsP(X 3) = P(0) + P(1) + P(2) + P(3) =

.05 + .25 + .35 + .30 = .95

Cumulative Probability

X P(X) Cumulative P(Xx)

0 .05 .05

1 .25 .30

2 .35 .65

3 .30 .95

4 .05 1.00

Cumulative Probability ApproachWhat is the probability this weekend MV Ford will sell:

1) Two ThunderbirdsP(X=2) = P(X 2)- P(X 1) = .65 - .30 = .35

2) Between 1 and 3 ThunderbirdsP(1 X 3) = P(X 3) - P(X 0) because we want to include 3, 2, and 1

= .95 - .05 = .90

3) Two or more ThunderbirdsP(X 2) = 1- P(X 1) because we want all the probability except that of 0

and 1 = 1-.30 = .70

4) Three or less ThunderbirdsP(X 3) = .95 -- This is the cumulative probability.

EXPECTED VALUE

• The Expected Value E(X) of a random variable is the same as its mean or X

• E(X) = xp(x)

• What is the mean number of Thunderbirds MV Ford would sell on weekends in which they have 4 Thunderbirds available?

• E(X) = 0(.05) + 1(.25) + 2(.35) + 3(.30) + 4(.05) = 2.05

A SECOND EXAMPLE

Suppose Tustin Ford in Tustin, Ca. has 3 Thunderbirds in stock this weekend.

Y = the number of Thunderbirds Tustin Ford will sell from stock this weekend and the probability distribution for their sales is:

Sales (y) Probability P(y)Sales (y) Probability P(y)

0 0 .20.20

11 .40.40

22 .25.25

33 .15.15

Expected Sales at Tustin Ford

• E(Y) = 0(.20) + 1(.40) + 2(.25) + 3(.15) = 1.35

LAWS OF EXPECTATION

• E(Constant) = Constant

• E(cX) = cE(X)– Manufacturer pays MV Ford an incentive of

$1000 for each Thunderbird sold • E(Incentive Amount) = E(1000X) = 1000E(X)

= 1000(2.05) = $2050

LAWS OF EXPECTATION (CONT’D)

• E(X+Y) = E(X) + E(Y)– What is the expected total number of Thunderbirds sold by the

two dealerships this weekend?

• E(X+Y) = E(X) + E(Y) = 2.05 + 1.35 = 3.40

• E(X-Y) = E(X) - E(Y) – What is the expected difference in the number of Thunderbirds

sold between the two dealerships this weekend?

• E(X-Y) = E(X) - E(Y) = 2.05 - 1.35 = .70

VARIANCE OF RANDOM VARIABLES

• Suppose Wilson Ford’s distribution of Thunderbird sales this weekend is:

W P(W) 2 .95 3 .05

E(W) = 2(.95) + 3(.05) = 2.05, the same as E(X)!the same as E(X)!

But the sales at Wilson Ford (random variable W) have less variability than sales at MV Ford (the random variable X).

VARIANCE• Var(X) = E(X - )2 = (x- )2p(x) or

= E(X2) - 2 = x2p(x) - 2

X P(X) X - (X - )2 X2

0 .05 -2.05 4.2025 0

1 .25 -1.05 1.1025 1

2 .35 -0.05 0.0025 4

3 .30 0.95 0.9025 9

4 .05 1.95 3.8025 16

Var(X) and

• Var(X) = 4.2025(.05) + 1.1025(.25) + .00025(.35) + 0.9025(.30) + 3.8025(.05) = .9475

OR• Var(X) = (0(.05) + 1(.25) + 4(.35) + 9(.30) + 16(.05) )

- (2.05)2 = .9475

.973396.9475Var(X)σ

Spreadsheet for Calculating Mean and Variance of Y

=A2^2Drag down to C5

=SUMPRODUCT(A2:A5,B2:B5)

=SUMPRODUCT(C2:C5,B2:B5) - B7^2

=SQRT(B8)

LAWS OF VARIANCE• Var(Constant) = 0

• Var(cX) = c2Var(X) Thus cX = cX

• If X, Y independent

– Var(X+Y) = Var(X) + Var(Y)

X+Y = SQRT(Var(X+Y))

X+Y DOES NOT = X + Y

– Var(X-Y) = Var(X) + Var(Y) -- Yes it is a “+” sign

X-Y = SQRT(Var(X-Y))

X-Y DOES NOT = X - Y

Example

• What is the variance in the number of Thunderbirds MV Ford has in stock this weekend? Recall it has 4 available.– V(4) = 0 4 is a constant

• Given that MV Ford makes $1000 in incentives for each Thunderbird sold, what is the standard deviation of the incentive amount received by MV Ford when it has 4 Thunderbirds available?– Var(1000X) = 10002Var(X) = 1,000,000(.9475) = 947,500

$973.40947500σ1000X

Example (Cont’d)

• Suppose that the number of Thunderbirds at MV Ford on the East Coast is independent of the number of Thunderbirds sold at Tustin Ford on the West Coast. (A reasonable assumption?)

• What is the standard deviation in the total number of Thunderbirds sold by MV Ford and Tustin Ford?– Var(X+Y) = Var(X) + Var(Y) = .9475 + .9275 = 1.875

1.3691.875Y)Var(Xσ YX

Example (Cont’d)

• What is the standard deviation of the difference in the number of sales between MV Ford and Tustin Ford?– Var(X-Y) = Var(X) + Var(Y) = .9475 + .9275 = 1.875

1.3691.875Y)Var(Xσ YX

REVIEW

• Definition of Random Variables– Discrete or Continuous

• Calculating Probabilities of Random Variables– “Long Way”– Cumulative Probability Approach

• Calculation of Mean, Variance, Standard Deviation

• Laws of Expectation

• Laws of Variation