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TA: Yury Petrachenko, CAB 484, [email protected], http://www.ualberta.ca/∼yuryp/
Bivariate and Multivariate Probability Distributions
5.1 Contracts for two construction jobs are randomly assigned to one or more of three firms, A,
B, and C. Let Y 1 denote the number of contracts assigned to firm A, and Y 2 the number of contracts assigned to firm B. Recall that each firm can receive 0, 1, or 2 contracts.
(a) Find the joint probability function for Y 1 and Y 2.
(b) Find F (1, 0).
Solution. (a) Let’s list all nine possible assignments of construction jobs to the three firms: AA, AB, AC, BA, BB, BC, CA, CB, CC. (The first symbol signifies the selection for the first job, the second — for the second one). These assignments are equally likely. Hence, the probability of each of them is 1/9. Now, let’s determine the values of Y 1 and Y 2 in each case: Assignment (y 1 , y 2 ) p(y 1 , y 2 ) AA (2, 0) 1 / 9 AB (1, 1) 1 / 9 AC (1, 0) 1 / 9 BA (1, 1) 1 / 9 BB (0, 2) 1 / 9 BC (0, 1) 1 / 9 CA (1, 0) 1 / 9 CB (0, 1) 1 / 9 CC (0, 0) 1 / 9 To find the joint probability function for Y 1 and Y 2 , we need to rearrange these data into the following table: y 1 0 1 2 0 1 / 9 2 / 9 1 / 9 y 2 1 2 / 9 2 / 9 0 2 1 / 9 0 0 (b) Compute
F (1, 0) = P (Y 1 ≤ 1 , Y 2 ≤ 0) = p(0, 0) + p(1, 0) =
5.6 Let Y 1 and Y 2 have the joint probability density function given by
f (y 1 , y 2 ) =
k y 1 y 2 , 0 ≤ y 1 ≤ 1 , 0 ≤ y 2 ≤ 1 , 0 , elsewhere.
(a) Find the value of k that makes this a probability density function. (b) Find the joint distribution function for Y 1 and Y 2. (c) Find P (Y 1 ≤ 1 / 2 , Y 2 ≤ 3 /4).
Solution. (a) We must have ∫ (^) ∞
−∞
−∞
f (y 1 , y 2 ) dy 1 dy 2 = 1.
Let’s compute:
1 = k
0
0
y 1 y 2 dy 1 dy 2 = k
0
y 2
y^21 2
1 y 1 =
dy 2 =
k 2
0
y 2 dy 2 =
k 4
So, k = 4 makes f a probability density function. (b) By definition, if 0 ≤ y 1 ≤ 1 and 0 ≤ y 2 ≤ 1:
F (y 1 , y 2 ) =
∫ (^) y 1
0
∫ (^) y 2
0
4 t 1 t 2 dy 2 dy 1 = 4
∫ (^) y 1
0
t 1 dy 1
∫ (^) y 2
0
t 2 dy 2 = 4 y 12 2
y^22 2 = y^21 y 22.
So, overall
F (y 1 , y 2 ) =
1 , y 1 ≥ 1 , y 2 ≥ 1 , y 12 y 22 , 0 ≤ y 1 ≤ 1 , 0 ≤ y 2 ≤ 1 , y 12 , 0 ≤ y 1 ≤ 1 , y 2 ≥ 1 , y 22 , 0 ≤ y 2 ≤ 1 , y 1 ≥ 1 , 0 , y 1 ≤ 0 , y 2 ≤ 0.
(c) We have P (Y 1 ≤ 1 / 2 , Y 2 ≤ 3 /4) = F (1/ 2 , 3 /4) =
2
4
Solution. This is almost a pure integration technique problem. For the first probability consider the integral ∫ (^1)
1 / 4
0
(y 1 + y 2 ) dy 1 dy 2 =
1 / 4
0
y 1 dy 1 dy 2 +
1 / 4
0
y 2 dy 1 dy 2
0
y 1 dy 1 +
1 / 4
y 2 dy 2 =
Now, the second probability: ∫ (^1)
0
∫ (^1) −y 2
0
(y 1 + y 2 ) dy 1 dy 2 =
0
(y 2 1 2
1 −y 2 y 1 =
dy 2 =
0
((1 − y 2 ) 2 2
dy 2
0
y 22
dy 2 =
y 2 − y 23 3
1 0
Marginal and Conditional Probability Distributions
5.17 Continuing Exercise 5.1.
(a) Find the marginal probability distribution of Y 1. (b) According to results in Chapter 4, Y 1 has a binomial distribution with n = 2 and p = 1/3. Is there any conflict between this result and the answer you provided in (a)?
Solution. (a) Refer to the table in the solution to Exercise 5.1 (a) above. To find the marginal probability distribution of Y 1 , we just need to add up numbers in respective columns of that table. We have
P (Y 1 = 0) = P (Y 1 = 0, Y 2 = 0) + P (Y 1 = 0, Y 2 = 1) + P (Y 1 = 0, Y 2 = 2)
= p(0, 0) + p(0, 1) + p(0, 2) =
P (Y 1 = 1) =
, P (Y 2 = 2) =
(b) If Y 1 is binomial with such parameters, then it must be p(y 1 ) =
y 1
3
)y 1 ( 2 3
) 2 −y 1
. We can easily check that this equality holds for y 1 = 0, 1 , 2. Hence, no conflict. §
5.22 Continuing Exercise 5.6.
(a) Find the marginal density functions for Y 1 and Y 2. (b) Find P (Y 1 ≤ 1 / 2 | Y 2 ≥ 3 /4). (c) Find the conditional density function of Y 1 given Y 2 = y 2. (d) Find the conditional density function of Y 2 given Y 1 = y 1. (e) Find P (Y 1 ≤ 3 / 4 | Y 2 = 1/2).
Solution. (a) Let’s integrate for all y 2 :
f 1 (y 1 ) =
0
4 y 1 y 2 dy 2 = 2y 1 , 0 ≤ y 1 ≤ 1.
Because the joint density is symmetric in y 1 and y 2 , the other marginal density is
f 2 (y 2 ) = 2y 2 , 0 ≤ y 2 ≤ 1.
(b) Let’s use the definition of the conditional probability:
P (Y 1 ≤ 1 / 2 | Y 2 ≥ 3 /4) =
P (Y 1 ≤ 1 / 2 , Y 2 ≥ 3 /4)
P (Y 2 ≥ 3 /4)
0
∫^3 /^4 4 y^1 y^2 dy^2 dy^1 1 3 / 4 2 y^2 dy^2
=
0 2 y^1 dy^1
3 / 4 2 y^2 dy^2
3 / 4 2 y^2 dy^2
0
2 y 1 dy 1 = y^21
∣∣^1 /^2
0
(c)&(d) Again, by definition
f 1 (y 1 |Y 2 = y 2 ) = f 1 (y 1 |y 2 ) = f (y 1 , y 2 ) f (y 2 )
4 y 1 y 2 2 y 2 = 2y 1.
f 2 (y 2 |Y 1 = y 1 ) = f 2 (y 2 |y 1 ) = f (y 1 , y 2 ) f (y 1 )
4 y 1 y 2 2 y 1 = 2y 2.
(e) Since we already know the conditional density f 1 (y 1 |y 2 ), we can use it now:
P
Y 1 ≤
| Y 2 =
0
f 1
y 1
∣∣ Y
dy 1 =
0
2 y 1 dy 1 =
