Math 300 , Spring 2004, Day 16, M, March 8After classHit reload to get most current version

Expectation:

Mean of Poisson, using exponential series: (as in Ash)
Mean of geometric, using formulas on p. 36, Ash.  Let X be geometric, parameter p of success.
    P(X=k) = q^k-1p, k = 1,2,3,4,....
   E(X) = 1p+2q¹p+3q²p +4q³p+ ...
            = p(1 + 2q +3q² +4q³+ ....)
            = p[ deriv with respect to q of (1+q + q² + q³ + q⁴+ ..)]
            = p[deriv of (1/(1-q))]  Take the derivative with respect to q, then let 1-q = p and simplify:
            = 1/p    Expected time to first success is 1/prob of success on a single trial.
                          If prob of "2" on a die is 1/6, expected times to throw a "2" is 6.
Got to HERE
  Handout, p1 arrows-->. If W is a function of X (W=g(X)), you can find E(W) either by
  >> summing g(x_i)·P(X=x_i)   for all the values of x_i,    or by
   gathering all the probabilities of  the x's which have the same w-value, thus finding the distribution of W, and
 >>  summing w_j·P(W=w_j)   for all the values of w_j.
  Usually we use the first way, as when we find Var(X) = E(X - E(X))² , but both work.

Prove: Prove E(kX) = kE(X).   (Use the above.)
E(kX) = Sum_i[(kx_i) p_i] = Sum_i[k(x_i p_i)] = kSum_i[(x_i p_i)] (pulling k out of sum: distributive law)
           = kE(X) by def. of E(X).
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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)

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+...+X_n then E(X) = E(X₁)+E(X₂)+...+E(X_n). (Ash 3.2)
        (If the X_i's  are independent (not usually true), then the variance can be found the same way.)
         If the X_i's take on the values 0 or 1, they are called indicator random variables, and E(X_i) = P(X_i = 1)
    Binomial, n drawings without replacement (mean only): X_i =1 if i'th trial is a Success. E(X_i) = p
                E(X) = Sum[ E(X_i)] = np
(Re)read Moore pp.371-3 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) =  sum_{i=1 to infinity}  (P(X > i))  =
                                             p₁ +  p₂ +  p₃ +......
                                                  +  p₂ +  p₃ +.....
                                                           +  p₃ +......
                                           1p₁ +2 p₂ + 3p₃ +....      = sum _{i = 0 to infinity}( i·p_i)
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Need to do:  proof of E (X+Y) = E(X) + E(Y)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Read Ash Sec 3.2, 3.4.  Skip 3.3.   Next, Ash Ch. 7, pp. 220-235 (ignore anything with integral signs)
HW:   Finish handout, M&M problems
A.Find the variance of a Bernoulli random variable (M&M say "Similarly" ...sigma²_S = p(1-p), p. 372, just below µ_S formula..)  Find it using the distribution.)
B. (new) Finish the Expected value of Geometric (above) from : p[deriv of (1/(1-q))] Take the derivative of (1/(1-q)) with respect to q, then let 1-q = p and simplify the formula.
Postpone:  B. a. Prove E(kX) = kE(X) as above, only when n = 3.  Write it out with +'s.
                Start E(kX) = (kx₁)p₁+(kx₂)p₂+(kx₃)p₃ =...
        b.  Prove E (a+X) = a+E(X), first with n = 3 and +'s, then in general with i's and Sum (big Sigma) notation.
Begin looking at: From Ash (These are all tricks of one sort or another.  Don't be discouraged if you can't do most of them without looking.)
Use p. 77 #5 to find E(X) for a geometric distribution (another way)
   p. 84, 1 + See  addition on Alg of exp. handout  p. 4
      3, 4, 5,
      6 optional
  p. 92, 7, 10, 17.
     Others Optional, but good: see the list on the Alg. of exp.  handout p. 4.