Math 300 , Spring 2008, Day 18, F, March 7 Hit reload ...

E (X+Y) = E(X) + E(Y) handout,
Handout on Bird Calls.
Read Ash Sec 3.2, 3.4.  Skip 3.3.   Next, Ash Ch. 7, pp. 220-235 (Variance. Ignore anything with integral signs)
HW:  A). Add to your "known distributions" page the means found in sec's 3.1 and 3.2.
B). Use p. 77 #5 to find E(X) for a geometric distribution (another way) 
     Hint:   p3 + p4 + p5 +...=P(X>2); you have a simple formula for that.
  From Ash (repeated from Day 17)
    p. 84 (sec. 3.2), 1 + See  addition on Alg. of exp. Handout   p. 4
      3, 4, 5,
      6 optional
  p. 92, 7+, 10, 17+. +See notes on handout. (These 3 to hand in) 
     Others Optional, but good: see the list on the handout p. 4.

D) Derivatives review: Find the derivative with respect to w:
  (w3- h)/(3+w + 2w2),    exp(-w2),        ln(2 + 3w2 )
   Find the second derivative with respect to q:    q,  q2, q3, q4, q5, q6,   qx

E) Graph the four points in the x,y plane (0,2), (1,0), (1,4), (2,2).  Let each point be equally likely (prob = 1/4).  Let X be the value on the x-coordinate, and Y be the value on the Y coordinate.
  a) Find E(X + Y) by evaluating x+y for each of the 4 points, multiplying by the probability, and summing.
  b) Find E(XY) by evaluating xy for each of the 4 points, multiplying by the probability, and summing.
  c) Find the probability distribution of X.  Find the probability distribution of Y.  Find E(X).  Find E(Y).
  d) Check if E (X+Y) = E(X) + E(Y).
  e)  Check if E(X · Y) =  E(X) · E(Y)
  f)  Check if X and Y are independent.  (For independence, for any x,y pair, P(x,y) = P(X=x) · P(Y=y).  So to show non-independence, you only need to find one x,y for which that is not true.)

F)  Another Data problem:  Handout on Bird Calls.  Do all the questions (including 1.6.4, frizzle fowl).  Recopy table 1 to have enough room to fill it in.  Est. rel. freq. is just the probabilities.
= = = = = = = = = = = = = = = = = = =
still  Expectation: 
Working to show:  E(a+bX) = a+bE(X), E(X+Y) = E(X)+E(Y), 
  then Var(a+bX) = b2Var(X) ; If X and Y independent, then Var(X+Y) = Var(X) + Var(Y).

Proved E(kX) = kE(X), last class. 
Proved E (a+X) = a+E(X)  HW [??],   
Put together, get E(a+bX) = a+E(bX) = a+bE(X)     
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Need to do:  proof of E (X+Y) = E(X) + E(Y) (handout, will go over in class)
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Def: X and Y are independent random variables if
       P(X = x and Y = y) = P(X = x) ·P(Y = y) for  every possible  pair x and y.
  What happens on X has no effect on the probabilities of Y:  P(Y = y) = P(Y = y | X = x), any x,y.

If X and Y are independent, then E(X · Y) =  E(X) · E(Y)  (will prove, next week.)  This is needed to prove
         Var(X+Y) = Var(X) + Var(Y).
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Finding E's in complex cases:  If X = X1+ X2+...+Xn then E(X) = E(X1)+E(X2)+...+E(Xn). (Ash 3.2)
        (If the Xi's  are independent (not usually true), then the variance can be found the same way.)
         If the Xi's take on the values 0 or 1, they are called indicator random variables, and E(Xi) = P(Xi = 1)  
    Binomial, n drawings without replacement (mean only):
       Using M&M notation of Xi =Si, Si =1 if i'th trial is a Success. E(Si) = p
                E(X) = Sum[ E(Si)] = np
                Var(X) = Sum[Var(Si)]
                     (remaining: variance of one Bernoulli trial S--was HW)  E(S)= p
                     Var(S) = E(S - E(S))2 = (0-p)2P(S=0)+ (1-p)2P(S=1) = p2q + q2p = pq(p+q) = pq
              Var(X) = npq
              (Moore pp.340-41 for derivation of Binomial mean and standard deviation.)

Related idea: p. 77 #5:  If the sample space is {0, 1, 2, 3, 4,...} (or {1, 2, 3, 4,...})
   E(X) =  sumi=1 to infinity  (P(X > i))  =
                                             p1 +  p2 +  p3 +......
                                                  +  p2 +  p3 +.....
                                                           +  p3 +......
                                       =    1p1 +2 p2 + 3p3 +....      = sum i = 0 to infinity( i·pi)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Variances:  much more....
New this semester:
   VarX = E(X2) - (E(X))2= E(X2) - (µx)2   (p. 225) We can turn this around and find E(X2) from VarX and E(X). 
Proof:  VarX = E[(X - µx)2] = E[ X2 - 2µx X - (µx)2 ] =  E( X2) - E(2µx X) + E[(µx)2 ]
  =  E[ X2] - 2µxE[X] + (µx)2 =  E(X2) - 2(µx)2 + (µx)2 (since E(X) =  µx)
 =  E(X2) - (µx)2

   Covariance Cov(X, Y) = E[(X-E(X)) · (Y-E(Y))] =  E[(X- µx) · (Y- µy)](def.)
              (in class?) = E(X · Y) - E(X) · E(Y)
  Var(X+Y) = Var(X) + Var(Y) + 2Cov(X, Y)    (a cov term for every pair, if summing more than 2)
           If X and Y are independent, Cov(X,Y) = 0, and we get our familiar Var sum.
    Correlation "rho"of X, Y is Cov "standardized" by dividing by both standard deviations (p. 235).
            is Theoretical version of correlation coefficient r.
           Cov(X,Y) = rho · sigmax · sigmay   (cf. M&M  rule 3, p.330)
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