Math 300 , Spring 2008, Day 16, M, March 3.solution links added. Hit reload

Expectation:
Reading:  M&M 4.4 pp. 291-94 (E(X)= µ_X, computation), with Ash pp.74-76, + fig. 2 p. 105. Then Moore pp 298-9, (Ash rest of 3.1+, to p.78).  Then Var(X),  Moore pp.300-5, Ash 3.4, p. 91 +p. 180. (Then Ash Sec. 3.2.  Skip 3.3)
HW  Handout "Expected values and variances" problems A thru D. 
  Similar problems were assigned in 251.  I'm not positive the numbering of the parts is identical, but much of the work is the same.
--"Handout" used in the fall:
http://aurora.wells.edu/~srs/Math251-Fall07/Algebraofmeans.htm
--Some help on the 2-sided chip and the two-urns situation:
http://aurora.wells.edu/~srs/Math251-Fall07/Algebraofmeanshelp.htm
--Solutions to all the problems given in the fall:
http://aurora.wells.edu/~srs/Math251-Fall07/AlgofmeansSols.pdf
Do  M&M hw (p. 4 of handout)
Ash p. 77 : 1, 2, 3, 5
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Poisson: questions?
An alternate formula (r = lambda) is P(X = k) =
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.  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).
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The Algebra of Means and Variances:
  (M&M Means: (cf. ex. 4.24 p. 299  for data), p. 298
         Variances: p. 302 )
  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  : wted avg 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 (Ash p.76)  using formulas on p. 36, Ash.
  Use the first formula, expanding e^x, to show the sum of all the probabilities in the Poisson distribution = 1, as it should.  e^x =  1 + x + x²/2! + x³/3! +....+ xⁿ/n! +...  (Another formula everyone should know)
Show if X is Poisson with parameter lambda, E(X ) = lambda.

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

Look thru these now:  Discuss 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: X_i =1 if i'th trial is a Success.
       (Re)read M&M pp.340-41 for derivation of Binomial mean and standard deviation.
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