Statistics 101: Homework 4

Exercise 1

1.
First of all, note that:
• the probability that an applicant will pass the test is .5
• the conditional probability that an applicant is qualified given that an applicant has passed the test is equal to .9
• the conditional probability that an applicant is unqualified given that an applicant has failed the test is equal to .9
Thus, the decision tree is as shown below:
2.
No. The best strategy is to hire everyone. The expected company profit is 250 (if an applicant is given the screening test, the expected profit is equal to 245).
3.
In case of the perfect test the expected company profit is equal to 400-P. So, the perfect screen test is worth 150.
4.
This screen test is worth 95.

Exercise 2

1.
We want , which is by definition

There are three cases for which . The additional law may be used to find

For this part of the problem, there are nine cases having X and Y both less than or equal to 2. Add probabilities for all cases satisfying this event. So

2.
To find , sum over all y values. The result are shown here
 x 0 1 2 3 4 0.18 0.195 0.25 0.195 0.18
To find , add probabilities down columns
 y 0 1 2 3 4 0.15 0.225 0.25 0.225 0.15

To test independence, we must see if for every x and y. For x=0 and y=0

Therefore X and Y are not independent.
3.
To calculate we must divide the joint probabilities by appropriate marginal probability for given x value. The result is conveniently shown by a table
 y x 0 1 2 3 4 0 0.0556 0.0833 0.1667 0.4167 0.2778 1 0.1026 0.1538 0.2308 0.3077 0.2051 2 0.1200 0.1800 0.4000 0.1800 0.1200 3 0.2051 0.3077 0.2308 0.1538 0.1026 4 0.2778 0.4167 0.1667 0.0833 0.0556
If X and Y were independent, the conditional probabilities for Y would be the same. regardless of the x value. But in this exercise, the conditional probability of any particular y value changes as the x value changes. Thus X and Y are not independent.
4.
The marginal probability distribution of X and Y are both symmetric around the value 2. Therefore .

Using the shortcut method for calculating variances, we square values, weight by probabilities and sum, subtracting the square of the mean at the end.

Similarly .
5.
The shorter way to find is

We have

From previous exercise, and

As X increases, Y tends to decrease, as shown by the negative correlation. X and Y can't be independent, because independent random variables have 0 correlation.
6.
The conditional expectation of Y given X=x is calculated like any expected value, but using conditional probabilities. That is

Similar calculations lead to the following table:
 x 0 1 2 3 4 2.778 2.359 2 1.641 1.222
As x increases, decreases, as we'd expect given that X and Y have a negative correlation.

7.
The additional law can be used to find . For example, T=X+Y=2 if . Thus

Similar calculations yield the distribution of T
 t 0 1 2 3 4 5 6 7 8 0.01 0.035 0.09 0.205 0.32 0.205 0.09 0.035 0.01

, by symmetry. To calculate the variance, the shortcut approach is convenient.

We previously have found all the relevant means, variances, and the covariance. We must have and

Exercise 3

1.
The marginal probabilities of X and Y are summarized in the table below:
 y x 0 1 2 3 P(X) 0 0.25 0.08 0.04 0.00 0.37 1 0.22 0.15 0.05 0.01 0.43 2 0.05 0.05 0.06 0.01 0.17 3 0.00 0.00 0.01 0.02 0.03 P(Y) 0.52 0.28 0.16 0.04 1.00

To show that X and Y are not independent, we only need to show that for one set of (X,Y) pairs. Consider the pair (3,0).

Therefore, X and Y are not independent.
2.
Since the majority of probability mass falls along the (0,0) to (3,0) diagonal, we would expect the correlation should be positive. This because it appears that X and Y move together (as X gets smaller, Y gets smaller, vice verse).

To compute the covariance, we also need .

Therefore, .

Similarly . The correlation is then

Exercise 4

1.
One dollar invested in the Ford stock is the portfolio with the highest expected return.
2.
USAir stock is the riskiest. It is true for this particular example, but in general it is not.
3.
Here is the covariances matrix:
 Company Bristol Myers Ford IBM Merck USAir Bristol Myers 0.00384400 0.00057288 0.00010044 0.000279 0.0019840 Ford 0.00057288 0.00592900 0.00168399 0.000154 -0.0001232 IBM 0.00010044 0.00168399 0.00656100 0.002673 0.0028512 Merck 0.00027900 0.00015400 0.00267300 0.002500 0.0016800 USair 0.00198400 -0.00012320 0.00285120 0.001680 0.0256000

The pair (USAir, IBM) have the highest covariance 0.00285120. The pair (IBM, Bristol Myers) have the lowest covariance 0.00010044.

4.
Let P1 denote the portfolio with equal amounts invested in the USAir and IBM stocks, and P2 - in the IBM and Bristol Myers. Then we have that