Homework 6 Solutions

W&S, p 272, #14


This problem applies some "rules" which you'll use again and again and again.  You should be able to derive these rules using the definitions of expectation and variance.  They are all lumped together in the descriptive category of "linear combinations of random variables."  Assume X1 and X2 are random variables, and a and b are constants:
i)  E(a + bX1) = a + bE(X1)
Var(a + bX1) = b^2 E(X1)

ii) E(X1 + X2) = E(X1) + E(X2) .
Var(X1 + X2) = Var(X1) + Var(X2) + 2Cov(X1, X2)
Var(X1 - X2) = Var(X1) + Var(X2) - 2Cov(X1, X2)

Note that rule (i) implies that E(X1 - X2) = E(X1 + -1 X2) = E(X1) + E(-1 X2) = E(X1) - E(X2)
Note that if X1 and X2 are indpt, then Cov(X1, X2) = 0, which simplifies rule ii for variances.

Note that (ii) for expectations can be expanded: E(X1 + X2 + .... +Xn) = E(X1) + E(X2) + ... + E(Xn).  The same works for variances ONLY if all of the RVs are indpt of each other.  If so, then Var(X1 + ... + Xn) = Var(X1) + Var(X2) + ... + Var(Xn)

Now on to the problem:

14a) Y1 = X1 + X2 + ... + X8
E(Y1) = E(X1) + ... + E(X8) = 58 + ... + 58 = 8*58 = 464.  In general: E(X1 + ... + Xn) = n*E(X) if all of the X's have the same mean.   This applies rule (ii).

Var(Y1) = Var(X1) + .... + Var(Xn) only if we assume independence.  If not indpt, then we can't really do this problem.
= 12^2 * 8
So the SD(Y1) = sqrt(12^2 * 8) = sqrt(8)*12 = 33.94.   Again, this applies rule (ii)

In general, if X1...Xn are indpt and have the same SD, then SD(X1 + ... + Xn) = sqrt(n)*SD(X)


No need to "memorize" these generalizations, though.  It is much easier to think them through each time.

14b) Y2= 8*X3
NOTE; IT IS EXTREMELY IMPORTANT THAT you understand that 8*X3 does NOT equal X1 + ... + X8.  As you're about to see, they have the same mean, but different variances.  Why? Because they represent different experiments. Y1 represents the experiment in which 8 students separately take an exam.  Y2 represents an experiment in which a SINGLE student takes the exam one time, and we replicate that student's score 8 times.
E(Y2) = E(8*X3) = 8E(X3) = 8*58 = 464.  This time we used rule (i) with a = 0, b = 8
Using Rule (i) again: Var(Y2) = 8^2 Var(X) = 64 * 144
SD(Y2) = sqrt(64*144) = 8*12 = 96.

14c) Y3 = (X1 + ... + X8)/8
E(Y3) = (1/8) E(X1 + .... + X8)  (rule i)
= (1/8) * 8*E(X)  (rule ii)
= E(X) = 58

This is a VERY important result. In general, if Xbar = (X1 + ... + Xn)/n  then E(Xbar) = E(X)

Var(Y3) = (1/8)^2 Var(X1 + .... + X8) by rule i
= (1/8)^2 * 8 * Var(X) by rule ii
= (1/8) * Var(X)

So Sd(Y3) = SD(X)/sqrt(8) = 12/sqrt(8) = 4.24

In general, SD(Xbar) = SD(X)/sqrt(n)

14d) Primarily, we've assumed that X1 through X8 are independent of each other.  This seems pretty safe if no one is cheating.  If they are, as we see by comparing (a) to (b), the variances might be different.


PAGE 317 #7

a)  E(X) = -1*(18/38) + 1(20/38)= 0.05263158
Var(X) = (-1 -  0.05263158)^2 * (18/38) + (1 - 0.05263158)^2 * (20/38)= .99723
SD(X) = sqrt(.99723) = .996814

b) (i)  Let Y = X1 + ... + X50 represent the amount the casino earns fter 50 plays.
E(Y) = 50 * .05263158 (by rule ii and the last hw problem) = 2.63
SD(Y) = sqrt(50) * .996814 (again using rule ii and the result of the last HW problem) = 7.049

c) Y = X1 + ... Xn
E(Y) = n E(X) = n*.05263158
Sd(Y) = sqrt(n) SD(X) = sqrt(n)*.99723

We haven't talked about *why* you can use the normal approximation for Y, but you can.  (The reason is  the Central Limit Theorem.)  The idea is that the exact distribution of Y is quite complicated, but we can get pretty good approximate answers if we use a normal distribution with mean E(Y) = 2.63 and SD(Y) = 7.05

d) Y represents the amount of money the CASINO makes from 50 $1 bets.  So if Y < 0, then the players won money.  So we want P(Y < 0)
Standardize:  P(Z < (0 - 2.63)/7.05 ) = P(Z < - .3730) = .356

So 35.6% of the players who make 50 1$ bets will win money.  Note that this is also the probability the casino loses money on any  given player who makes 50 $1 bits will result

e) Y represents the money the casino makes in 1000 bets.  E(Y) = 1000*0.05263158 = 52.63158  SD(Y) = sqrt(1000)*.99723= 31.53
P(Y < 0) = P(Z <  -52.631/31.53) = P(Z <  - 1.67) = 0.047

Y represents 100,000 bets.  E(Y) = 5231.58  SD(Y) = 315.3518
P(Y < 0) = P(Z <  -16.6) = 0   (based on the table in the book.  more precisely it equals 3.4 times 10 to the negative 64.

Clearly if Y is based on a million bets, the chances the casino will lose money is even smaller.

f)

	avg profit	total profit
50	-.370, .475	-18.52, 23.78
1000	-.042, .147	-42.24, 147.50
100000	.043,.062	4317.10, 6209.21
1000000	.050, .0556	49639.89, 8254.85

52.631-3*sqr50


For all of these, the average profit will be the same: E( (X1 +... + Xn)/n) = E(X) = .05263158 .  The standard deviation of the average is SD(X)/sqrt(n).  So each interval is .0526 +/- 3*.99723/sqrt(n)
But the total profit will be E(X1 + ... + Xn) = n*E(X) and the SD will be sqrt(n).99723

The theme is that the more people who gamble, the closer the casino's average winnings per person is to the expected value of 0.0526 and the less likely it is that they will lose money.  You can see there is still some chance of losing money with 1000 plays, but essentially none for 100,000 or more.