Solution to Homework for Section 6.2, page 406 
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1.  page 406

    You need to first construct the joint dist of (X,Y). 
    I handed out this joint dist in class several weeks ago. So
    you already have it. Using the joint dist of (X,Y) we can get
    the cond dist of Y given X. Then, the formula for the cond expec.
    is 

    E(Y|X=x) = \sum_{y=1}^{6} y P(Y=y|X=x)

    So, for example for X=1 you would get

    E(Y|X=1)=(1)(1/36)/(11/36) + ... + (6)(2/36)/(11/36) = 41/36.
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3. page 407

REMARKS:
   a. In this problem you will need to recall one of Euler's sum
      formulas, i.e., \sum_{k=1}^{k=n} k = k(k-1)/2.


First we need to construct the joint distribution of (X_1,X_2). Since,
drawings are without replacement, there are n(n-1) possibles outcomes
for this random vector. Further, the joint distribution of (X_1,X_2)
is uniform, namely, P(X_1 = k, X_2 = l) = 1/n(n-1), for all k,l=1,...,n.
Next we find the joint distribution of (X,Y). As a consequence of their
def given that drawings are without replcement, it follows that the
joint distribution of (X,Y) is:

   R(X) = {1, ..., n-1}
   R(Y) = {2, ..., n}
   P(X=x, Y=y) = 0,         if x > y
   P(X=x, Y=y) = 2/n(n-1),  if x < y
   and X can never equal Y.

Now we can calculate the conditional distribution of Y given X=x. By def,

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

As you can see, we need the distribution of X. This distribution can be
obtained from the joint dist of (X,Y) which was found above. An easy
calculation shows that for example

   P(X=1) = [2/n(n-1)](n-1)
   P(X=2) = [2/n(n-1)](n-2)
   etc
   Hence,  P(X=x) = [2/n(n-1)] (n-x)

Thus,

   P(Y=y | X=x) = [2/n(n-1)] / {[2/n(n-1)] (n-x)} = 1/(n-x) , if y > x
		= 0,                                          if y < x.

Finally, by def

   E(Y | X=x) = \sum_{y=2}^{y=n} y P(Y=y | X=x)
       
	      = \sum_{y>x}^{y=n} y [1/(n-x)]

	      = 1/(n-x) \sum_{y>x}^{y=n} y

	      = 1/(n-x) [n(n+1)/2 - x(x+1)/2]

	      = [(n^2-x^2) + (n-x)] / 2(n-x)

	      = (n+x+1)/2.
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13.  page 408

For this problem you want to read Example 4 on page 404  and 
Example and the discussion on page 125 of the text.

a.  P( S_{m} = j  |  S_{n} = k )  

    =  P( S_{m} = j,  S_{n} = k )  /  P( S_{n} = k )

   = (m choose j)p^{j} q^{m-j}(n-m choose k-j)p^{k-j}q^{n-m-k+j}

         /   (n choose k) p^{k} q^{n-k}

    =  (m choose j) (n-m choose k-j) p^{k} q^{n-k}

         /   (n choose k) p^{k} q^{n-k}

    =  (m choose j) (n-m  choose  k-j)  /  (n  choose  k)

b.  E(S_{m}  |  S_{n} = k}  
  
      =   sum_{j = 0}^{m}  j P(S_{m} = j  |  S_{n} = k)

      =  m  k / n