Solution to Homework for Section 6.1, page 398 
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1.  (#1,  page 398)

a.  X  =  number of heads in three tosses.
    Clearly, this is a Bernoulli type experiment and X
    is the total of successes in 3 trials. So X is binomila.
    =>  X has binomial (3, 1/2) distribution.

b.  P( Y = y  |  X = x)

    First note that R(X) = {0, 1, 2, 3} as we may get 0 or 1 or 2 or 3
    heads on the first three tosses. Now, consider the range of Y. To
    see this, consider an example. Say X=2. Namely, only one of the three
    coins landed tails and we toss this one coin again. Thus, the number
    of heads on the second toss can only be 0 or 1. This makes the possible
    values of Y (given X=2) either 2 or 3. Hence, for any given value of X
    the range of Y is {x, x+1,..., 3}. Since X \in {0,1,2,3}, this makes
    range of Y={0,1,2,3}.

    Now, to calculate the conditional dist of Y given X consider the following.
    Let T=number of heads on the second toss. Then, we note the following:

    a. R(T) = {x, x+1, ..., 3}
    b. T has a binomial dist B(3-x, 1/2). 
    c. Y = T + X

    Hence, 

    P(Y=y|X=x) = P(T+X=y | X=x) = P(T=y-x) = (3-x choose y-x)(1/2)^(3-x).

    This is the cond dist of Y given X, and as you can see it is "shifted"
    binomial.
    
    One example will illustrate how this problem works.

Say we want  P( Y = 2  |  X = 1).  This means that when we tossed our
three coins the first time,  we got two tails and one head.  Then,  we
picked up those two coins that landed tails,  tosses them and got exactly
one head (the second time around).  Hence,

P(Y = 2 |  X = 1)  =  the probability of getting exactly one success in a
binomial (2, 1/2) distribution.

                             =  (2 choose 1) (1/2) (1/2)

                             =  1/2

So,  as you can now tell,  the conditional distribution of Y given  X = x is
binomial  (3-x, 1/2).

c. Once you have the cond dist of Y given X, you can construct the joint dist
   of X and Y simply by using the basic def, i.e.,

   P(Y=y, X=x) = P(Y=y | X=x) P(X=x)

   Note that the numbers in the RHS of the above eqn are the ones on (a)&(b).

d. Once you have the joint dist of X&Y, you can get the marginal dist of Y
   by adding the entries in the appropriate row or column of that dist.

e. To get the cond dist of X given Y use Baye's rule:

   P(X=x | Y=y) = P(Y=y | X=x) P(X=x) / P(Y=y).

   All the numers above were found in previous parts.

f. The theoretical answer here the cond expec. of X given Y.  But since in this
   section we are not supposed to use those results, simply do as the book
   suggests.

g. Once you have done (f) you will observe a pattern. Use that pattern in
   answering this part.


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2.  (#3,  page 398)

Here is one way that you can think about this problem.  Say,

total number of families =  n  =  m_{0} + m_{1} +  ...  +  m_{4}

where,

m_{i}  =  number of families with i children,  i = 0, 1, 2, 3, 4.

Then,

0 m_{0} + 1 m_{1} + 2 m_{2} + 3 m_{3} + 4 m_{4}  =  N  = total

number of children.

a.  Clearly,  P(U=0) = 0 since we have already selected a child.

P(U=1)  =  m_{1} / N  =  (m_{1} / n) / (N / n)  

Note that m_{1} / n  =  P(T=1)  =  20%,  and  N/n = E(T) = 2.  Thus,

P(U=1)  =  m_{1} / N  =  (m_{1} / n) / (N / n)  =  20% / 2  =  10%

Next,

P(U=2)  =  2 m_{2} / N  
               =  (2 m_{2} / n) / (N / n)  
               =  (2 40%) / 2
               =  40%
etc.



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3. (#5, page 399)

   We need to show the following:

   P(X_1 = k | X_1 + X_2 = n) = (N choose k) p^k (1-p)^(N-k)

   for the, as yet, unknowns N and p. So, we start by calculating
   the L.H.S. of the above equation.

   L.H.S. = P(X_1=k, X_1+X_2=n) / P(X_1+X_2=n)

   = P(X_1=k) P(X_2 = n-k) / P(X_1 + X_2 = n)

   = [e^-(\lambda_1 + \lambda_2) / k! (n-k)!][\lambda_1^k \lambda_2^(n-k)]

   / P(X_1 + X_2 = n)

   At this point we are done with the numerator, and only calculations for
   the denomenator are left.

   P(X_1 + X_2 = n) = \sum_{0}^{n} P(X_1 = k| X_2 = n-k) P(X_2 = n-k)

   = \sum_{0}^{n} P(X_1 = k) P(X_2 = n-k), using independence again.

   = e^-(\lambda_1+\lambda_2} \sum_{0}^{n}[\lambda_1^k \lambda_2^(n-k)] /

   [k! (n-k)!] 

   = [e^-(\lambda_1 + \lambda_2)]/n! \sum_{0}^{n} [n!/k!(n-k)!]\lambda_1^k

   \lambda_2^(n-k)

   = [e^-(\lambda_1 + \lambda_2)]/n! (\lambda_1 + \lambda_2)^n, using the

   binomial theorem for expanding (A+B)^n.

   Finally, we put all of this together:

   P(X_1 = k | X_1 + X_2 = n) = 

   [n! / k!(n-k)!] [\lambda_1 / (\lambda_1+\lambda_2)]^k [\lambda_2 /

   (\lambda_1 + \lambda_2)]^(n-k)

   Note that if we let p=\lambda_1 / (\lambda_1 + \lambda_2), then
   1-p will be \lambda_2 / (\lambda_1 + \lambda_2). Hence we have
   found what we were looking for: N=n and p=\lambda_1/(\lamabda_+\lambda_2).
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