Suppose that a pair of random variables have the same ...
Handout 5 (Chapter 5): Joint Probability Distributions and Random Samples
Some more rules of expected value and variance to use in this handout.
i) For constants a, b, and c and the random variables X and Y,
E(a(X ( b(Y ( c) = a(E(X) ( b(E(Y) ( c
ii) For constants a, b and c and the random variables X1 and X2,
Var(a( X1 ( b( X2 ( c) = a2(Var(X1)+ b2(Var(X2) ( 2( a(b(Cov(X1, X2)
iii) For constants a1 to an and the random variables X1 to Xn, [pic]
By using the following examples, the joint probability mass function for two discrete random variables and their properties, their marginal probability mass functions, the case for independent and dependent variables, their conditional distributions, expected value, variance, covariance, and correlation will be demonstrated. All the formulas and explanations will be introduced in class and you already have them in chapter 5 of the textbook.
Example 1: Quality audit records are kept on numbers of major and minor failures of circuit packs during-in of large electronic switching devices. They indicate that for a device of this type, the random variables X (the number of major failures) and Y (the number of minor failures) can be described at least approximately by the accompanying joint distribution.
| x|0 |1 |2 |Total |
|y | | | | |
|0 |0.15 |0.05 |0.01 |0.21 |
|1 |0.10 |0.10 |0.04 |0.24 |
|2 |0.10 |0.14 |0.04 |0.28 |
|3 |0.10 |0.13 |0.04 |0.27 |
|Total |0.45 |0.42 |0.13 |1 |
a. Is the table above a valid joint pmf for X and Y? Yes
b. Find the marginal probability mass functions for X and Y.
x 0 1 2 otherwise
p(x) 0.45 0.42 0.13 0
y 0 1 2 3 otherwise
p(y) 0.21 0.24 0.28 0.27 0
c. Are X and Y independent? No
d. Find the expected value and the variance of X.
E(X)=0.68 E(X2)= 0.94 Var(X)=0.4776
e. Find the expected value and the variance of Y.
E(Y)=1.61 E(Y2)= 3.79 Var(Y)=1.1979
f. Find the Cov(X,Y) and Corr(X,Y).
E(XY)=1.25 Cov(X,Y)=0.1552 Corr(X,Y)=0.2052
g. Find the conditional probability function for Y given that X=0 that is there are no circuit pack failures.
y 0 1 2 3 otherwise
P(Y=y | X=0) 0.3333 0.2222 0.2222 0.2222 0
h. What is the expected number of minor failures given that there were no major failures? E(Y | X=0) =1.3333
i. Suppose that demerits are assigned to devices of this type according to the formula D=2X+Y. Find the marginal probability mass function for D.
d 0 1 2 3 4 5 6 7 otherwise
P(D=d) 0.15 0.10 0.15 0.20 0.15 0.17 0.04 0.04 0
d=0 only when X=0,Y=0 then P(d=0)=P(X=0,Y=0)
d=3 only when X=0,Y=3 or X=1,Y=1 then P(d=3)=P(X=0,Y=3)+P(X=1,Y=1)
j. Suppose that demerits are assigned to devices of this type according to the formula D=2X+Y. Find the expected value and the variance of D.
E(D)=2.97 E(D2)=12.55 Var(D)=3.7291
k. Suppose that demerits are assigned to devices of this type according to the formula U=Min(X,Y). Find the mean value and the variance of U.
u 0 1 2 otherwise
P(U=u) 0.51 0.41 0.08 0
u=0 only when X=0,Y=0 or X=0,Y=1 or X=0,Y=2 or X=0,Y=3 or X=1,Y=0 or X=2,Y=0 then
P(U=0)=P(X=0,Y=0)+P(X=0,Y=1)+P(X=0,Y=2)+P(X=0,Y=3)+P(X=1,Y=0)+P(X=2,Y=0)
E(U)=0.57 E(U2)=0.73 Var(U)=0.4051
By using the following example, the joint probability density function for two continuous random variables and their properties, their marginal probability density functions, the case for independent and dependent variables, their conditional distributions, expected value, variance, covariance, and correlation will be demonstrated.
Example 2: Suppose that a pair of random variables, X and Y have the same joint probability density
[pic]
a. Find the marginal probability density functions for X and Y.
[pic] and [pic]
b. Are X and Y independent? No
c. Find the expected value and variance of X.
E(X)=7/12 E(X2)= 5/12 Var(X)=11/144
d. Find the expected value and variance of Y.
E(Y)=7/12 E(Y2)= 5/12 Var(Y)=11/144
e. Find the conditional probability density function of x given y=0.6.
[pic]. When y=0.6, f(x|y)=[pic] for 0 ................
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