Durretts proof of Kolmogorovs Extension Theorem
Consider Lemma A.3.2. from Durretts text, which says that if Bn A and Bn , then
P (Bn ) 0. He begins with these two sentences:
Suppose P (Bn ) > 0. By repeating sets in the sequence, we can suppose
Bn = Kn f
Random walks via sample path counting
We will draw the sample paths of a simple, symmetric random walk on Z with time on
the horizontal axis and space on the vertical axis. Dene Nn,x to be the number of paths
between (0, 0) and (n, x), so that Nm,n (y, x)
Hewitt-Savage zero-one law
Let (Xi ) be i.i.d. S-valued random variables on (, F, P ). The map
X() := (X1 (), X2 (), . . .)
is measurable from (, F) to (S N , S N ). The image measure on S N is the distribution of
the sequence (Xi ). Since (Xi ) is i.i.d
Lattice random variables.
A random variable X has a lattice distribution if there are
constants b and h > 0 so that P (X b + hZ) = 1. The largest h that works is called the
span of the distribution.
Examples:
If P (X = +1) = P (X = 1) = 1/2, then X has a
Records for random permutations
Consider a random permutation (x1 , x2 , . . . , xn ) of distinct numbers. We will create this
permutation by rst picking xn at random, and then prepending (x1 , x2 , . . . , xn1 ), an
independent random permutation of the
Recurrence
Let (Xj ) be i.i.d. random variables taking values in Rd . Our random walk is given by
S0 = 0 and Sn = X1 + + Xn for n 1. On the space Rd , we will use the norm
x = supi jxi j and for x Rd and > 0 dene the open ball of radius centered at x by
B
Skorohods representation theorem
Let S be complete, separable metric space. Assume that the probability measures in S
fn g converge weakly to 0 . Then there exist S-valued random elements fXn g on
n=0
n=1
the Lebesgue probability space, having distributio
Portmanteau Theorem.
The following are equivalent
(a) n
(b) lim inf n (G) (G) for open G
(c) lim sup n (K) (K) for closed K
(d) lim n (A) = (A) for Borel A with (A) = 0
Proof. The strategy of the proof looks like this:
(a) (b) (c)
(d)
[(a) (b)] For non-e
Stirlings formula Inspired by Patin and Michel
For t > 0, use the change of variables x = y t + t to get
xt ex dx =
(t + 1) =
0
where
y
1+
gt (y) =
t
0
t
t
e
t
ey
t
gt (y) dy,
t
if y > t
otherwise.
d
Since dt log(gt (y) = f(y/ t), where f is as in the Le
Lemma 2.
For m 2 an integer,
n=0
P ( Sn < m) (2m)d
Corollary 3.
The convergence (resp. divergence) of
of > 0 is sucient for transience (resp. recurrence).
Example.
n
n=0
P ( Sn < ).
P ( Sn < ) for a single value
If E X < and EX = = 0, then the random walk
Renewal theory
Let (i ) be i.i.d. random variables with P (0 < ) = 1 and P (0 < ) = 0, with
distribution function F . Dene renewal times T0 = 0 and Tk = k i for k 1. Then
i=1
(Tk ) is a case (ii) random walk, that is, Tk as k .
For t 0 dene Nt = inf(k 0 :