1.  We would use the absolute value rather than squaring.

2.  false

3.  (5/9 )^2 * 3.2

4. 

Let S denote the amount of time the patron holds the book, and let I be
the indicator random variable for returning to a foreign branch.
Then

B = S + 2I

Var(S) = E(S^2) - (ES)^2
ES = (0.1) 4 + (0.2) 5 + (0.3) 6 + (0.4) 7
E(S^2) = (0.1) 4^2 + (0.2) 5^2 + (0.3) 6^2 + (0.4) 7^2
Var(S) = 1

Var(B) = Var(S) + 4 * Var(I)
Var(I) =  0.5(1-0.5)
Var(B) = 1 + 1 = 2

5.  (i), (iv); note: Var(R_k) = (1/n)^2 Var(N_k) = (1/n)^2 np(1-p)

6.  

Var(IX) = E[(IX)^2] - [E(IX)]^2   # mailing tube
        = E(I^2) E(X^2) - [EI EX]^2  # indep
        = EI E(X^2) - (EI)^2 (EX)^2  # I^2 = I, algebra
        = EI [Var(X) + (EX)^2] - (EI)^2 (EX)^2  # mailing tube
        = p (r+q^2) - p^2 q^2

vary <- function(p,q,r) 
{
   p * (r+q^2) - p^2 * q^2
}

print(vary(0.4,15,6))  # 56.4


7.  (i) is true, e.g. X = Y = constant 1
   (ii) is false; the smallest X-Y could be is -1, and it would have to
   constantly be -1 to have a long-run average cube of -1; that would
   mean X = const 0 and Y = const 1; but since we are given that EX > 0,
   we could not have X = const 0

8.  

   E[(I+J)^3] = E(I^3) + 3 E(I^2 J) + 3 E(I J^2) + E(J^3)

Using the hint, this is

   EI + 3 E(IJ) + 3 E(IJ) + EJ = EI + EJ + 6 E(IJ)

Using the given information, this is

   0.8 + 6 E(IJ)
 
So we just need E(IJ).  The second given quantity is

   0.5 = E[(I-J)^2] 
       = E(I^2) - 2 E(IJ) + E(J^2)
       = EI - 2 E(IJ) + EJ
       = 0.8 - 2 E(IJ)

So, E(IJ) = 0.15, and

E[(I+J)^3] = 0.8 + 6(0.15) = 1.7

9.

varalight <- function(nreps)
{
  nstops <- 2
  alight2 <- vector(length=nreps)

  for (i in 1:nreps) {
    passengers <- 0
    for (j in 1:nstops) {
      alight <- 0
      if (passengers > 0) {
        for (k in 1:passengers) {
          if (runif(1) < 0.2) {
            alight <- alight + 1
            passengers <- passengers - 1
          }
        }
      }
      newpass <- sample(0:2,1,prob=c(0.5,0.4,0.1))
      passengers <- passengers + newpass
    }
    alight2[i] <- alight
  }
  mean(alight2^2) - (mean(alight2))^2
}

set.seed(9999);
print(varalight(10000))  # 0.1136

check:  Let A = # who alight at stop 2.  Support = {0,1,2}. 

P(A = 0) = 0.5 * 1 + 0.4 * 0.8 + 0.1 * 0.8^2 = 0.884
P(A = 1) = 0.5 * 0 + 0.4 * 0.2 + 0.1 * 2 * 0.8 * 0.2 = 0.112
P(A = 2) = 0.5 * 0 + 0.4 * 0 + 0.1 * 0.2^2 = 0.004
E(A^2) = 1^2 * 0.112 + 2^2 * 0.004 = 0.128
EA = 1 * 0.112 + 2 * 0.004 = 0.12
Var(A) = 0.128 - 0.12^2 = 0.1136

10.

et <- function(m,nreps)
{ 
   passgo <- 0
   for (i in 1:nreps) {
      pos <- 0
      for (turn in 1:m) {
         roll <- sample(1:6,1)
         newpos <- pos + roll 
         if (newpos > 7) {
            passgo <- passgo + 1
            newpos <- newpos %% 8
         }
         if (newpos == 3) {
            bonus <- sample(1:6,1)
            newpos <- newpos + bonus
            if (newpos > 7) {
               passgo <- passgo + 1
               newpos <- newpos %% 8
            }
         }
         pos <- newpos
      }
   }
   passgo / nreps
}

set.seed(9999)
print(et(3,10000))  # 1.0697