Math 300 , Spring 2004, Day 15, F, March 5After class Hit reload to get most current version

Friday Symposium?  Who wants class canceled?

Poisson: questions?
The alternate formula (r = lambda) is
e^(rt)(rt)^k / k!, i.e. using lambda as the rate per unit interval, t as the size of the interval.
So their lambda times t = our lambda.
In order to cut down confusion, do this:  If you are using the ( rate per unit interval times t) formula, use r for that value, not
lambda.  (i.e.Old notes, lambda--> r)  This corresponds to the applet's notation also.
Use lambda only for the expected number of hits in the whole interval under consideration (The mean, in the applet).

Expectation:
Reading:  M&M 4.4 pp. 318-321 (E(X)= µ_X, computation), with Ash pp.74-76, + fig. 2 p. 105. Then Moore pp 326-8, (Ash rest of 3.1, to p.85).  Then Var(X),  Moore pp.328-33, Ash 3.4, p. 91 +p. 180. (Then Sec. 3.2.  Skip 3.3)

The Algebra of Means and Variances:
  (M&M Means: (cf. ex. 4.23 p. 327 for data), p. 326
         Variances: p. 330 )
  Rules 1 are for a linear transformation a+bX of one R.V.
  Rules 2(&3) for a linear combination X + Y

E(X)= µ_X : weighted average of X-possibilities, weighted by probability: E(X) = SUM(x_ip_i)
(cf. mean for data, each value added the number of times it occurs, divide sum by total number of occurrences.  Same as weighting each value by proportion of times it occurs)
  Draw 1 from Urn with 100 balls, 50 labeled 1, 25 each labeled 2, 4 (or urn with 4 balls, proportionate)

X= x	x1 = 1	x2 = 2	x3 = 4	SUM
P(x)	p1= 1/2 = 50/100	p2= 1/4 = 25/100	p3= 1/4 = 25/100	1
product (weighted)	x1p1= 1x1/2= .5	x2p2= 2x1/4 = .5	x3p3= 4x1/4 = 1	.5+.5+1= 2 = E(X)=µX
deviation from mean	(x1- µ) = 1-2 = -1	(x2- µ) = 0	(x3- µ) = 2
squared deviation	(x1- µ)2 = 1	(x2- µ)2= 0	(x3- µ)2 = 4
sq'd dev weighted	(x1- µ)2 p1= 1x1/2 = .5	(x2- µ)2p2= 0x1/4 = 0	(x3- µ)2p3= 4x1/4 = 1	.5+ 0 + 1 = 1.5 = Var(X)

Var(X) =E[X - E(X)]² =E[X - µ_X ]²= sigma_X ²  : sum of squared deviations from the mean.

E is a linear operator (actually affine)--constants and -, + pass across E.
Var(X) =E[X - E(X)]² is not.

mean of Poisson next time, using formulas on p. 36, Ash.
  I used the first formula, expanding e^x, to show the sum of all the probabilities in the Poisson distribution = 1, as it should.

We'll develop the rules further, prove the ones we haven't.
Handout (homework): Joint distribution, checking Moore's rules, above rule.

Look thru these now:  Discussed the first one with example W = X³
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.
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Next: 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.)  This is needed to prove
         Var(X+Y) = Var(X) + Var(Y).
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Next: 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, drawing without replacement (mean only): X_i =1 if i'th trial is a Success.
       (Re)read M&M pp.371-3 for derivation of Binomial mean and standard deviation.
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HW for today:  Do Handout, start M&M  at least. (Know the Rules 1 and 2 and have worked at least 1 example for each)
Handout "Expected values and variances" problems A thru D.
Do  M&M hw (p. 4 of handout)
Ash p. 77 : 1, 2, 3, 5