Chapters 7 and 10: Expected Values of Two or More Random Variables

http://blogs.oregonstate.edu/programevaluation/2011/02/18/timely-topic-thinking-carefully/

Covariance

Joint and marginal PMFs of the discrete r.v. X (Girls) and Y (Boys) for family example

Boys, B

0 1 2 3 Total

Girls, G

0 0.15 0.10 0.0867 0.0367 0.3734

1 0.10 0.1767 0.1133 0 0.3900

2 0.0867 0.1133 0 0 0.2000

3 0.0367 0 0 0 0.0367

Total 0.3734 0.3900 0.2000 0.0367 1.0001

Example: Covariance(1)A nut company markets cans of deluxe mixed nuts

containing almonds, cashews and peanuts. Suppose the net weight of each can is exactly 1 lb, but the weight contribution of each type of nut is random. Because the three weights sum to 1, a joint probability model for any two gives all necessary information about the weight of the third type. Let X = the weight of almonds in a selected can and Y = the weight of cashews. The joint PDF is

fX(x) = 12x (1 – x)2, fY(y) = 12y (1 – y)2

What is the Cov(X,Y)?

X,Y

24xy 0 x 1,0 y 1,x y 1f (x,y)

0 else

Example: Covariance (2)a) Let X be uniformly distributed over (0,1) and

Y= X2. Find Cov (X,Y).

b) Let X be uniformly distributed over (-1,1) and Y = X2. Find Cov (X,Y).

X

11 x 1

f (x) 20 else

X

1 0 x 1f (x)

0 else

Example: Correlation (1)a) Let X be uniformly distributed over (0,1) and

Y= X2. Find Cov (X,Y).

b) Let X be uniformly distributed over (-1,1) and Y = X2. Find Cov (X,Y).

X

11 x 1

f (x) 20 else

X

1 0 x 1f (x)

0 else

Example: Correlation (3)A nut company markets cans of deluxe mixed nuts

containing almonds, cashews and peanuts. Suppose the net weight of each can is exactly 1 lb, but the weight contribution of each type of nut is random. Because the three weights sum to 1, a joint probability model for any two gives all necessary information about the weight of the third type. Let X = the weight of almonds in a selected can and Y = the weight of cashews. The joint PDF is

fX(x) = 12x (1 – x)2, fY(y) = 12y (1 – y)2, Cov(X,Y)=-0.0267

What is the (X,Y)?

X,Y

24xy 0 x 1,0 y 1,x y 1f (x,y)

0 else

Example: Correlation (4)a) Let X be uniformly distributed over (0,1) and

Y= X2. Find (X,Y).

b) Let X be uniformly distributed over (-1,1) and Y = X2. Find Cov (X,Y).

X

11 x 1

f (x) 20 else

X

1 0 x 1f (x)

0 else

1E(X)

2 2 1

E(X )3

Table : Conditional PMF of Y (Boys) for each possible value of X (Girls)

Boys, B

0 1 2 3 pX(x)

Girls, G

0 0.4017 0.2678 0.2322 0.0983 0.3734

1 0.2564 0.4531 0.2905 0 0.3900

2 0.4335 0.5665 0 0 0.2000

3 1 0 0 0 0.0367

pY(y) 0.3734 0.3900 0.2000 0.0367

Determine and interpret the conditional expectation of the number of boys given the number of girls is 2?

Example: Conditional ExpectationA nut company markets cans of deluxe mixed nuts

containing almonds, cashews and peanuts. Suppose the net weight of each can is exactly 1 lb, but the weight contribution of each type of nut is random. Because the three weights sum to 1, a joint probability model for any two gives all necessary information about the weight of the third type. Let X = the weight of almonds in a selected can and Y = the weight of cashews. The joint PDF is

fX(x) = 12x (1 – x)2, fY(y) = 12y (1 – y)2

What is the conditional expectation of Y given X = x?

X,Y

24xy 0 x 1,0 y 1,x y 1f (x,y)

0 else

Table : Conditional PMF of Y (Boys) for each possible value of X (Girls)

Boys, B

0 1 2 3 pX(x)

Girls, G

0 0.4017 0.2678 0.2322 0.0983 0.3734

1 0.2564 0.4531 0.2905 0 0.3900

2 0.4335 0.5665 0 0 0.2000

3 1 0 0 0 0.0367

pY(y) 0.3734 0.3900 0.2000 0.0367

Example: Double Expectation (2)

A quality control plan for an assembly line involves sampling n finished items per day and counting X, the number of defective items. Let p denote the probability of observing a defective item. p varies from day to day and is assume to have a uniform distribution in the interval from 0 to ¼.

a) Find the expected value of X for any given day.

Example: Conditional VarianceA nut company markets cans of deluxe mixed nuts

containing almonds, cashews and peanuts. Suppose the net weight of each can is exactly 1 lb, but the weight contribution of each type of nut is random. Because the three weights sum to 1, a joint probability model for any two gives all necessary information about the weight of the third type. Let X = the weight of almonds in a selected can and Y = the weight of cashews. The joint PDF is

fX(x) = 12x (1 – x)2, fY(y) = 12y (1 – y)2

What is the conditional variance of Y given X = x?

X,Y

24xy 0 x 1,0 y 1,x y 1f (x,y)

0 else

Example: Law of Total Variance

A fisherman catches fish in a large lake with lots of fish at a Poisson rate (Poisson process) of two per hour. If, on a given day, the fisherman spends randomly anywhere between 3 and 8 hours fishing, find the expected value and variance of the number of fish he catches.