ENGR 323
Watkins
Problem 5-5
BHW #12
Problem Statement
Determine the value of C that makes the function f(x,y) = C(x+y) a joint probability mass
function over the nine points with x = 1,2,3 and y = 1,2,3.
Problem Solution
Definition.
The joint probability mass function of the discrete random variable x and y, denoted as
(1) fxy(x,y) ≥ 0
(2) ∑x ∑y fxy(x,y) = 1
(3) fxy(x,y) = P(X =x, Y=y)
pg. 219
since f(x,y) = C(x+y) the range consist of all the combinations of x + y. therefor the range
of f(x,y) is (1+1=2, 1+2= 3, 1+3= 4, 2+1= 3, 2+2= 4, 2+3= 5, 3+1= 4, 3+2= 5, 3+3= 6)
= (2,3,4,3,4,5,4,5,6). Table #1 shows all possible values of f(x,y) = C(x+y)
Applying this to the definition:
∑x∑y f(x,y) = C(2+3+4+3+4+5+4+5+6) = 1
Solving for C
C*36 = 1
C = 1/36
Table #1 summary of values for JPMF for problem 5-5
f(x,y) 1
2
3
1
2c
3c
4c
2
3c
4c
5c
3
4c
5c
6c
Table #2 summary of calculated values for JPMF for problem 5-5
f(x,y) 1
2
3
1
2/36 3/36 4/36 = 9/36
2
3/36 4/36 5/36 =12/36
3
4/36 5/36 6/36 =15/36
sum 9/36 12/36 15/36
ENGR 323
Watkins
Problem 5-6
BHW #12
Problem Statement
Continuation of Exercise 5-5. Determine the following probabilities.
A. P(X=1,Y<4)
B. P(X=1)
C. P(X=2)
D. P(X<2,Y<2)
Problem Solution
A. P(X=1,Y<4)
We will use the solution for f(x,y) in problem 5-5. One can see that it is 1/36 * the range
of values obtained with X=1, and Y<4. In other words the probability is equal to the sum
of the probabilities f(1,1) + f1,2) + f(1,3).
Therefor the probability of P(X=1,Y<4) = C(X+Y)
= 2/36+3/36+4/36
=9/36
=¼
referring to table #2, this is the probability is the sum of the first row of values.
B. P(X=1)
The probability requested in part A. is the same probability requested in part B. This is
because for P(X=1) that X is held constant at 1, while Y is allowed to take on any value.
In symbolic terms P(X=1) = P(X=1,Y<4)
Therefor the probability of P(X=1)
= C(X+Y)
= 2/36+3/36+4/36
=9/36
=¼
C. P(X=2)
This problem is similar to parts A. and B. except for we want P(X=2) instead of P(X=1).
So this problem equates to: P(X=2) = f(2,1) + f(2,2) + f(2,3). The probability can also be
stated as P(X=2,Y<4). This is the summing of the second row in Table #2.
Therefor the probability of P(X=2)
= C(X+Y)
= 3/36+4/36+5/36
= 12/36
= 1/3
D. P(X<2,Y<2)
The probability of P(X<2,Y<2) is the same as P(X=1, Y=1) = f(1,1)
Therefor the probability of P(X<2,Y<2) = 2/36
= 1/18
ENGR 323
Watkins
Problem 5-7
BHW #12
Problem Statement
Continuation of Exercise 5-5. Determine E(X) and V(X).
Problem Solution.
Definition:
Let Rx denote the set of all points in the range of (X,Y) for which X=x. The conditional
mean of Y given X=x, E(X) = µx = ∑xfx(X) = ∑x∑yxfxy(x,y)
Doing some plugging and chugging of numbers into the definition the solution is as
follows:
1[f(1,1) + f(1,2) + f(1,3)] + 2[f(2,1) + f(2,2) + f(2,3)] + 3[f(3,1) + f(3,2) + f(3,3)]
which in turn equals:
=1[1/36 (9)] + 2[1/36 (12)] + 3[1/36 (15)]
=1(9/36) + 2(12/36) + 3(15/36)
= 13/6 = 2.167
Variance:
Knowing the mean, you can now calculate the variance by plugging the numbers into the
equation V(X) = ∑x(x- µx)^2fxy(x,y)
V(X) = (1-2.167)^2*(9/36) + (1-2.167)^2*(12/36) + (1-2.167)^2* (15/36)
V(X) = 0.639
ENGR 323
Watkins
Problem 5-8
BHW #12
Problem Statement
Continuation of Exercise 5-5.
A. Determine the marginal probability distribution of the random variable X.
B. Determine the conditional probability distribution of Y given that X = 1.
C. Determine the conditional probability distribution of X given that Y = 2.
Problem Solution
Definitions:
If X and Y are discrete random variables with joint probability mass function fxy(X,Y),
then the marginal probability mass functions of X and Y are
fx(x) = P(X=x) = ∑rx = fxy (x,y)
fy(y) = P(Y=y) = ∑ry = fxy (x,y)
Where
Rx denotes the set of all points in the range of (X,Y) for which X=x and
Ry denotes the set of all points in the range of (X,Y) for which Y=y
A. The best way to solve this problem is to use a table.
Therefor the table goes as follows:
Table #3
X
Fx(x) = Fxy(x,1) + Fxy(x,2) + Fxy(x,3)
1
(1/36)*(2 + 3 + 4)
=1/4
2
(1/36)*(3 + 4 + 5)
=1/3
3
(1/36)*(4 + 5 + 6)
=5/12
summary of the marginal probability mass functions.
9/36, x = 1
12/36, x = 2
15/36, x = 3
0 otherwise
B. Determine the conditional probability mass function
Definition:
Given discrete random variables X and Y with joint probability mass function fxy(x,y) the
conditional probability mass function of Y given X = x is
fy|x(y) = fxy(x,y)/ fx(x) for fx(x) >0
In part B. we can use the same approach is used as in part A.. That being the use of a
table.
Table #4
Y
fyx(y) = fxy(x,y)/ fx(x)
1
(2/36) / (1/4) =
2/9
2
(3/36) / (1/4) =
1/3
3
(4/36) / (1/4) =
4/9
Summary of the conditional probability mass function.
fy|x=1(y) = 2/9, y = 1
3/9, y = 2
4/9, y = 3
0 otherwise