Math 300 , Spring 2002, Day 22, M, March 25 Hit reload to get most current versionAfter class

Derivation of E(X(X-1)) for Poisson Distribution, and how to get variance from it (D&E, Day 19) (Microsoft Word file)

Midterm due  Fri. if you need the extra time

Continuing with Expectation handout:  Covariance:
 Cov(X, Y) = E[(X-E(X)) · (Y-E(Y))] (def.)    = 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.
 Var(X+Y) = E[(X-muX) + (Y-muY) ]2 =  relabeling for clarity
                     E [  (a )        +   ( b )     ]2 = E(a2 + 2ab + b2) = var(X) + 2 cov(X,Y) + var(Y)
For Var(X) we have 3 terms,  c = (Z -muZ),   E[a + b + c]2, get E( a2 + 2ab + b2+ 2ac + 2bc +c2)
   reorganized  = var(X)  + var(Y) +Var(Z)+ 2 cov(X,Y)+  2 cov(X,Z)+ 2 cov(Y,Z)
I'll go over variance of Hypergeometric, with handout, Wednesday.

HW: Ash p. 233
#8 (Note that if E(XY) = E(X)E(Y),  then. Cov(X,Y) = 0.  Use this & your results from p.223#14)
#13, #14 (finding Covariances by brute force)
A.  Cov(X,Y) = 0 does not imply that X and Y are independent.  Show this is true using this example:
  x,y pair:  (0,3)     (1,1)     (2,3)      Graph, find Cov, Show X & Y dependent.
  prob:       1/4       1/2        1/4
B.  Continuing the pattern, to find Var(X+Y+Z+W), we need to find [a + b + c +d]2
  I suggest this structure (not what I did in class for 3 terms).  Fill in the "times table", and then collect like terms.  How many Cov terms will there be?
    a   b   c   d
a |
b |
c |
d |
- - -  - - - -
 Read pp. 95-6.  Graph f(x) on p. 95 and g(x) on p. 96 --piecewise-defined functions (Prep for  Ash Ch. 4. )


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