EECS 501__________________________PROBLEM SET #6__________________________Fall 2001
ASSIGNED: October 19, 2001. Read Stark and Woods pp. 226-263 (not needed this week).
DUE DATE: October 26, 2001. THIS WEEK: Central limit theorem and related subjects.
  1. Stark and Woods #4.37. A straightforward central limit theorem problem.
  2. Stark and Woods #4.16. Show x,y JGRV implies fy|x(Y|X) is a Gaussian pdf.
  3. Let x1...xn be iidrv with Pr[x=1]=Pr[x=-1]=½. Show that the characteristic function of the
    normalized sum y=(x1+...+xn)/n½ is en logcos(w/n½). Compute the limit of this for large n.
  4. Every day Anne receives 0,2 or 4 letters, with respective probabilities ¼,½,and ¼.
    The number received on any day is independent of the number received on any other day.
    Let yn be the total number of letters received after n days. Note yn is an even number.
      Use a correction to the Demoivre-Laplace correction (!) in the following:
    1. For what value of n is Pr[yn> 39]=0.919? HINT: Apply Demoivre-Laplace to zn=yn/2.
    2. Compute Pr[95 < y50 < 103]. HINT: Again use zn, which takes on integer values.
    3. Compute the limit of Pr[yn > 2n + 1.414n½] for large n.
      HINT: Reformulate this entire problem in terms of zn=yn/2,
      since yn is always an even integer, while zn is always an integer.
  5. We wish to sample randomly (with replacement) to estimate what fraction S of people smoke.
    How many people must we sample so that our estimate is within ± 0.02 of the actual fraction S
    with probability 0.95. That is, find n so that Pr[|((#smokers counted)/n)-S| < 0.02] > 0.95.
    NOTE: This is what is meant by "95% confidence" that the estimate of S is off by < 0.02.
    When you see polls cited on TV or in magazines, the usual (but unstated) confidence is 95%.
    HINT: #smokers counted is sum of iidrv xi with Pr[xi=1]=S, Pr[xi=0]=1-S. Worst case is S=½:
    Variance[#smokers] depends on S, with maximum value at S=½ and minimum values at S=0 or 1.
    6. 2-D Gaussian pdf when the covariance matrix is singular:
    1. JGRVs x,y have zero means, variances 4 and 1, and covariance 2, respectively.
      Given x=3, what, with probability one, is the value of y?
    2. SKETCH the 2-D pdf fx,y(X,Y). Do NOT determine an expression for it.
    3. Now change the variances of x and y to 4.01 and 1.01 (covariance is still 2).
      Determine a mathematical expression for fx,y(X,Y).
      Use MATLAB to make a 3-D mesh plot of fx,y(X,Y).
    "An optimist is an accordion player with a beeper"--Ted Koppel.