Math 340                       Homework Set 12
____________________________________________________________________________

1. (#3, page 202)

 a. E(2X+3Y) = 2 E(X) + 3 E(Y) = 5 E(X) = 5, since X and Y are id.

 b. var(2X+3Y) = var(2X) + var(3Y), using independence
	       = 4 var(X) + 9 var(Y), using scaling rules
	       = 13 var(X), since id
	       = 26.

 c. E(XYZ) = E(X)^3 = 1, since iid

 d. var(XYZ) = E(X^2 Y^2 Z^2) - E(XYZ)^2, using def and practical formula
	     = E(X^2))^3 - 1, since iid
	     = 27 - 1, since E(X^2) = var(X) + E(X)^2 = 2 + 1 = 3.

_____________________________________________________________________________

2. (#5, page 202)

E[(X-a)^2] = E(X^2 - 2aX + a^2) = E(X^2) - 2aE(X) + a^2
	   = var(X) + mu^2 - 2a mu + a^2
	   = var(X) + (mu - a)^2.
_____________________________________________________________________________

3. (#13, page 202)

Let X be the IQ score of an individual in this population of n=10^6. Then,
we are told that E(X)=100 and sigma(X)=10. Now, we need to find P(X>130).
Namely, the proportion of the population whose IQ scores exceed 130. Once we 
know this proportion, then our estimate for the number of people whose IQ
scores exceed 130 is N ~ (10^6) P(X>130).

a. Note that the IQ scores exceeding 130 are those values of X more than 3
   standard deviations from the mean, E(X)=100.  But we know that for ANY
   r.v. the following is true (based on Chebychev Inequality):

   P(|X - E(X)| > 3 SD) < 1/9  =>  P(|X-100| > 30) < 1/9.

   Note that this probability corresponds to the TOTAL area under BOTH tails
   of the dist of X (i.e., X>130 and also X<70). Since we have no further info 
   regarding the dist of X, the best estimate we can make is

   P(X > 130) ~ 1/9.

   Therefore,  N ~ (10^6) (1/9) ~ 111,111.

b. We are told that the dist of X is symmetric. Thus,

   P(X>130) = P(X<70) => P(X>130) ~ (1/9)(1/2) => N ~ (10^6)(1/18) = 55,555.

c. Now X is normal. Hence,

   P(X > 130) = P(Z > z), where z = (130-100)/10 = 3. Thus,

   P(X >130) = P(Z > 3) = 1 - P(Z < 3) = 1 - 0.9987 = 0.0013.

   Therefore,  N ~ (10^6)(0.0013) = 1,300.

That is how additional information about a dist can be so helpful!
______________________________________________________________________________

4. (#17, page 202)

i. First, E(X)= (-1/4) + 0 + (1/2) = 1/4.
ii. Next, Var(X) = E(X^2) - E(X)^2. So, first we need to calculate E(X^2).

    R(X^2) = {0, 1}
    P(X^2 = 0) = P(X=0) = 1/4
    P(X^2 = 1) = P(X=1 or X=-1) = 3/4.

    Hence, E(X^2) = 0 + 3/4 = 3/4. Therefore,
    Var(X) = E(X^2) - E(X)^2 = 3/4 - 1/16 = 11/16.

iii. Finally, S = X_1 + ... + X_25 where X_i's are iid each having the dist 
    of X. Hence, 

    E(S) = 25 E(X) = 25/4, and
    Var(S) = 25 Var(X) = (25) (11/16), using independence.

    Now, we can answer the questions. What we need to use in this problem is
    the Central Limit Thm which says that (regardless of the dist of X) the 
    dist  of S is approximately normal: N(E(S), SD(S)).

a. (Note that the answer in the back of the book is not accurate.)

   P(S < 0) ~ P(Z < z), where z=(0-E(S))/SD(S) = -5/sqrt(11) ~ -1.51.

   Hence, P(S < 0) ~ P(Z < -1.51) = 1 - P(Z < 1.51) = 1 - 0.9345 ~ 7%.

b. P(S = 0) ~ P(z_1 < Z < z_2), where

   z_2 = (0.5 - E(S))/SD(S) = -23/5sqrt(11) ~ -1.39, and

   z_1 = (-0.5 - E(S))/SD(S) = -27/5sqrt(11) ~ -1.63.

   Hence, P(S=0) ~ P(-1.63 < Z < -1.39) = P(1.39 < Z < 1.63) = etc.

______________________________________________________________________________