HW1
 Homework #1. Due January 30

1.Associate with the space of infinite (countable) number of coin tossings a set X of (Bernoulli) sequences x= (x1, x2, x3,....), where xi is either 1 or -1.
a. Define an elementary event (interval) In to be a set of points (sequences) x such that the nth component (n is fixed, but arbitrary) xn is the same for all points in this set. Is an elementary event an open/closed set?
b. Assume that the (probability) measure m of an elementary event is 1/2 and that for two different k,n elementary events Ik and In are independent, that is
m(Ik In) = m (Ik) m(In). Construct the sigma-algebra of events Fn that depends on the outcome of the first n coin tossings. Then F1 F2 F3... Fn ... . Consider the union of all Fi: F0 = Fi. Is F0 an algebra? Is F0 a sigma-algebra?
Hint: Consider an event (1,1,1,1,1,1,....).
c. Construct the sample space of possible events F - the (minimal) sigma-algebra that contains all elementary events and the outer (probability) measure on this sigma-algebra. Are the sets of events F and F0 the same? Using the extension theorem define the probability space (measure space).

2. Probability of long leads. The Arc-sine law.
Suppose a population of n people casts their votes for Al or Bill at random with probability 1/2. This can be viewed as a coin tossing model with a random variable (measurable function) gn(x) = x1+ x2+ x3+ ... +xn. What is the distribution of the proportion of the time k/n when Al is leading? It is generally expected that in a prolonged series of ballot castings (coin tossings) Al should lead about half the time and Bill the other half, that is with very high probability (close to 1) k/n=1/2. This statement is entirely wrong.
a. Show that gn(x) is a measurable function.
b. We say that Al is winning at time t=n if gn(x) > 0 or gn(x) = 0 and gn-1(x) > 0. Show that the event
An(k/n)= "Al is winning in less than k out of n tossings"
is a measurable event.
c. Using Matlab (or other software) and performing numerical simulations compute numerically the distribution of the function
fn(k/n)= "Probability of Al winning in less than k out of n tossings" for large n=10,000.
d. Compare (numerically) probabilities that
"Al is winning in less than .02 n out of n tossings" with the probability "Al is winning in larger .49 n but less than .51 n out of n tossings".
"Al is winning in larger than .02 n out of n tossings" with the probability "Al is winning in larger .49 n but less than .51 n out of n tossings".
"The most fortunate contender is winning in larger than .976 n out of n tossings" with the probability "The most fortunate contender is winning in less than .524 n out of n tossings".
e. Let f = limn fn. f= 2/ arcsin(x1/2) . Verify (numerically) the formula for f.
Note. fn converges to f almost everywhere. A proof of this fact is a technical point of the analysis of this problem. The idea of the proof is to consider a set (an event) of points x where fn does not converge. Then this set can be represented as a union of countable number of events "the trajectory of fn oscillates up above a rational number q1 and then down below a rational number q2, and so on, q1 > q2". Then one can show that the measure of each of these events is zero.
The explanation of the arc-sine law lies in the fact that frequently enormously many trials are required before the particle returns to the origin. Geometrically speaking, the path crosses the time axis very rarely. We feel intuitively that if Al and Bill toss a coin for a long time 2n, the number of ties should be roughly proportional to 2n. This is not so. Actually the number of ties increases in probability only as (2n)1/2. Reflection about the time axis argument shows that the number of times when the particle actually crosses the time axis is a half the number of times when the particle returns to the origin (to the time axis). Hence we wish to understand the probability of the number of returns to the origin.
f. Using Matlab (or other software) and performing numerical simulations compute numerically the distribution of the function fn(x)= "the probability that up to and including time 2n the particle returns to the origin fewer than x (2n)1/2" times for large n=10,000.
g. Let f = limn fn. f= (2/)1/2 0 x exp(-1/2 s2) ds. Verify (numerically) the formula for f.
One should be warned that the number of returns is not normally distributed.

The next three exercises are manifestations of the Littlewood's three principles
3. Prove the following theorem(see hints for this exercise in Royden problem 13 p. 64)
Let E be a given subset of R (or Rn). Assume *(E) < . It is Lebesgue measurable if and only if for every > 0, there is a closed set F and an open set O such that F E O and (O\F) < . Moreover (E) = inf{(O), E O, O - open } = sup {(F), F E, F - closed}.

4. Egorov's theorem. (see Royden problem 30 p. 74)
If < fn > is a sequence of measurable functions that converge to a real-valued function f almost everywhere on a measurable set E with finite measure (E) < , then for any > 0 there is a subset A E with (A) < such that fn converges uniformly on E\A.

5. Lusin's theorem.(see Royden problem 31 p. 74)
Let f(x) be a measurable real-valued function on [a,b], and let > >0 be arbitrary. Then there is a continuous function g(x) such that the measure of the set {x:| f(x) g(x)} is less than .

6. Continuity of the Lebesgue integral (see also Royden problem 5 p. 89)
Let f(x) be an integrable function. Show that the function F(x) defined by
F(x) = - x f(x) d x
is a continuous function.

7. Convergence theorems for "convergence in measure" (see Royden problem 21 p. 96)
Prove that Fatou's lemma and the Monotone and Lebesgue Convergence theorems remain valid if "convergence a.e." is replaced by "convergence in measure".

8. Green's function for the Poisson equation on a unit disk D in R3.
For the Poisson equation
u(x) = f(x), x=(x1,x2,x3) a solution (there are many solutions, because we don't specify the boundary conditions) is given by
u(x) = -1/(4 ) D f(y)/|x-y| dy.
The function G(x)=-1/(4 |x|) is called the Green's function for the Poisson equation on a unit disk D in R3.
a. Suppose f(x) =1. Then u(x) = -1/(4 ) D 1/|x-y| dy. Compute u(x).
Suggestion: you can use Mathematica (or other software) to compute it.
Answer: u(x) = |x|2/6 - 1/2.
b. Define a discrete Laplacian h f(x)= 1/3h2 (f(x1+h, x2, x3 )+ f(x1-h, x2, x3 )+ f(x1, x2+h, x3 )+ f(x1, x2-h, x2 )+ f(x1 , x2, x3+h )+ f(x1, x2, x3-h )- 6 f(x1, x2, x3 )). Show that for any h > 0 h u(x) = D f(y)h G(x-y) d x.
c. Show that almost everywhere G(x-y) = 0 for any fixed y. Hence D f(y) G(x-y) d x=0.
d. Using integral convergence theorems explain the "contradiction" for u(x) defined in part a. u(x) =1, but D G(x-y) d x=0.