Transcript Proposition 1.1 De Moargan’s Laws

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
Girls, G
Boys, B
0
1
2
3
Total
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
24xy 0  x  1,0  y  1,x  y  1
fX,Y (x,y)  
else
 0
fX(x) = 12x (1 – x)2, fY(y) = 12y (1 – y)2
What is the Cov(X,Y)?
Example: Covariance (2)
a) Let X be uniformly distributed over (0,1) and
Y= X2. Find Cov (X,Y).
1 0  x  1
fX (x)  
else
0
b) Let X be uniformly distributed over (-1,1) and
Y = X2. Find Cov (X,Y).
1
1  x  1

fX (x)   2
 0
else
Example: Correlation (1)
a) Let X be uniformly distributed over (0,1) and
Y= X2. Find Cov (X,Y).
1 0  x  1
fX (x)  
else
0
b) Let X be uniformly distributed over (-1,1) and
Y = X2. Find Cov (X,Y).
1
1  x  1

fX (x)   2
 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
24xy 0  x  1,0  y  1,x  y  1
fX,Y (x,y)  
else
 0
fX(x) = 12x (1 – x)2, fY(y) = 12y (1 – y)2, Cov(X,Y)=-0.0267
What is the (X,Y)?
Example: Correlation (4)
a) Let X be uniformly distributed over (0,1) and
Y= X2. Find (X,Y).
1
1
1 0  x  1
2
E(X) 
E(X ) 
fX (x)  
3
2
0
else

b) Let X be uniformly distributed over (-1,1) and
Y = X2. Find Cov (X,Y).
1
1  x  1

fX (x)   2
 0
else
Table : Conditional PMF of Y (Boys) for each
possible value of X (Girls)
Girls, G
Boys, B
0
1
2
3
pX(x)
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
Example: Conditional Expectation
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
24xy 0  x  1,0  y  1,x  y  1
fX,Y (x,y)  
else
 0
fX(x) = 12x (1 – x)2, fY(y) = 12y (1 – y)2
What is the conditional expectation of Y given X = x?
Table : Conditional PMF of Y (Boys) for each
possible value of X (Girls)
Girls, G
Boys, B
0
1
2
3
pX(x)
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 Variance
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
24xy 0  x  1,0  y  1,x  y  1
fX,Y (x,y)  
else
 0
fX(x) = 12x (1 – x)2, fY(y) = 12y (1 – y)2
What is the conditional variance of Y given X = x?
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.