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 ${\bf P}(X=2\cap Y=2)$, which is ${\bf P}_{XY}(2,2)$ by definition

$${\bf P}(X=2\cap Y=2)=0.100$$

There are three cases for which $X=2\cap Y\leq 2$. The additional law may be used to find

$$\begin{array}{rcl} {\bf P}(X=2\cap Y\leq 2)&=& {\bf P}(X=2\ca... ...P}_{XY}(2,1)+{\bf P}_{XY}(2,2)=0.03+0.045+0.1=0.175 \end{array}$$

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

$$\begin{array}{rcl} {\bf P}(X\leq 2\cap Y\leq 2)&=& {\bf P}_{X... ...015+0.03+0.02+0.03+0.045+0.03+0.045+0.1\\ &=&0.325 \end{array}$$

2.

To find ${\bf P}_X(x)$, sum ${\bf P}_{XY}(x,y)$ over all y values. The result are shown here

x	0	1	2	3	4
${\bf P}_X(x)$	0.180	0.195	0.250	0.195	0.180

To find ${\bf P}_Y(y)$, add probabilities down columns

y	0	1	2	3	4
${\bf P}_Y(y)$	0.150	0.225	0.250	0.225	0.150

To test independence, we must see if ${\bf P}_{XY}(x,y)={\bf P}_X(x){\bf P}_Y(y)$for every x and y. For x=0 and y=0

$${\bf P}_{XY}(0,0)=0.01 \mbox{ but } {\bf P}_X(0)*{\bf P}_Y(0)=0.027.$$

Therefore X and Y are not independent.

3.

To calculate ${\bf P}_{Y\vert X}(y\vert x)={\bf P}_{XY}(x,y)/{\bf P}_X(x)$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 $\mu_X=\mu_Y=2$.

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

$$\sigma_X=\sqrt{0^2*0.180+1^2*0.190+\cdots+4^2*0.180-2^2}=\sqrt{1.83}=1.353.$$

Similarly $\sigma_Y=1.285$.

5.

The shorter way to find ${\rm Cov}(X,Y)$ is

$$\sum_{x}\sum_{y}xy{\bf P}_{XY}(x,y)-\mu_X\mu_Y.$$

We have $\mu_X=\mu_Y=2$

$$\begin{array}{rcl} {\rm Cov}(X,Y)&=&0*0*0.01+0*1*0.015+\cdots... ...\ &+&\cdots +4*3*0.015+4*4*0.01-2*2 \\ &=&-.070 \end{array}$$

From previous exercise, $\sigma_X=1.353$ and $\sigma_Y=1.285$

$${\rm Corr}(X,Y)=\frac{{\rm Cov}(X,Y)}{\sigma_X\sigma_Y}=-0.403.$$

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

$$\mu_{Y\vert X=0}=0*0.0556+\cdots+4*0.2778=2.778$$

Similar calculations lead to the following table:

x	0	1	2	3	4
$\mu_{Y\vert X=x}$	2.778	2.359	2.000	1.641	1.222

As x increases, $\mu_{Y\vert X=x}$ decreases, as we'd expect given that X and Y have a negative correlation.

7.

The additional law can be used to find ${\bf P}_T(t)$. For example, T=X+Y=2 if $(X=0\cap Y=2)\cup(X=1\cap Y=1)\cup(X=2\cap Y=0)$. Thus

$${\bf P}_T(2)={\bf P}_{XY}(0,2)+{\bf P}_{XY}(1,1)+{\bf P}_{XY}(2,0)=0.09$$

Similar calculations yield the distribution of T

t	0	1	2	3	4	5	6	7	8
${\bf P}_T(t)$	0.010	0.035	0.090	0.205	0.320	0.205	0.090	0.035	0.010

$\mu_t=4$, by symmetry. To calculate the variance, the shortcut approach is convenient.

$$\sigma_T^2=0^2*0.01+\cdots+8^2*0.01-4^2=2.08$$

We previously have found all the relevant means, variances, and the covariance. We must have $\mu_T=\mu_X+\mu_Y=2+2=4$ and

$$\sigma_T^2=\sigma_X^2+\sigma_Y^2+2{\rm Cov}(X,Y)=2.08$$

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 ${\bf P}(X)*{\bf P}(Y)\neq{\bf P}(X,Y)$ for one set of (X,Y) pairs. Consider the pair (3,0).

$$0={\bf P}(3,0)\neq{\bf P}(X)*{\bf P}(Y)=0.03*0.52$$

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).

$${\bf E}X=0*0.37+1*0.43+2*0.17+3*0.03=0.86$$

$${\bf E}Y=0*0.52+1*0.28+2*0.16+3*0.04=0.72$$

To compute the covariance, we also need ${\bf E}XY$.

$$\begin{array}{rcl} {\bf E}XY &=&0*0*0.25+0*1*0.08+0*2*0.04+0*... ...&+&3*0*0.00+3*1*0.00+3*2*0.01+3*3*0.02 \\ &=&0.92 \end{array}$$

Therefore, ${\rm Cov}(X,Y)={\bf E}(XY)-{\bf E}(X){\bf E}(Y)=0.92-0.86*0.72=0.3008$.

$${\bf E}X^2=0^2*0.37+1^2*0.43+2^2*0.17+3^2*0.03=1.38$$

$$\sigma_X=\sqrt{{\rm Var}(X)}= \sqrt{{\bf E}(X^2)- ({\bf E}X)^2}=\sqrt{1.38-0.86^2}=0.8003$$

Similarly $\sigma_Y=0.8727$. The correlation is then

$${\rm Corr}(X,Y)=0.3008/(0.8003*0.8727)=0.4307.$$

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

$${\bf E}P_1=\frac{1}{2}(0.0097+0.0025)=0.0061, \quad {\bf E}P_2=\frac{1}{2}(0.0025+0.0170)=0.00975,$$

$${\rm Var}P_1=0.16^2/4+0.081^2/4+2*1/2*1/2*0.00285120=0.00946585,\quad \sigma_{P_1}=0.0972926,$$

$${\rm Var} P_2=0.081^2/4+0.062^2/4+2*1/2*1/2*0.00010044=0.00265147,\quad \sigma_{P_2}=0.05149243.$$