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.

- Stark and Woods #4.37. A straightforward central limit theorem problem.
- Stark and Woods #4.16. Show x,y JGRV implies f
_{y|x}(Y|X) is a Gaussian pdf.

- Let x
_{1}...x_{n} be iidrv with Pr[x=1]=Pr[x=-1]=½.
Show that the characteristic function of the

*normalized* sum
y=(x_{1}+...+x_{n})/n^{½} is e^{n logcos(w/n½)}.
Compute the limit of this for large n.

- 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 y_{n} be the total number of letters received after n days. Note y_{n} is an even number.
Use a *correction* to the Demoivre-Laplace correction (!) in the following:
- For what value of n is Pr[y
_{n}> 39]=0.919?
HINT: Apply Demoivre-Laplace to z_{n}=y_{n}/2.
- Compute Pr[95 < y
_{50} < 103]. HINT: Again use z_{n}, which takes on integer values.
- Compute the limit of Pr[y
_{n} > 2n + 1.414n^{½}] for large n.

HINT: Reformulate *this entire problem* in terms of z_{n}=y_{n}/2,

since y_{n} is always an *even* integer, while z_{n} is always an integer.

- 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 x_{i} with Pr[x_{i}=1]=S, Pr[x_{i}=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.

*2-D Gaussian pdf when the covariance matrix is singular:*
- 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?
- SKETCH the 2-D pdf f
_{x,y}(X,Y). Do NOT determine an expression for it.
- Now change the variances of x and y to 4.01 and 1.01 (covariance is still 2).

Determine a mathematical expression for f_{x,y}(X,Y).

Use MATLAB to make a 3-D mesh plot of f_{x,y}(X,Y).

"An optimist is an accordion player with a beeper"--Ted Koppel.