5.4. TRANSFORMATIONS OF THE WIENER PROCESS 177
Proof. To show that
Winv(t) =tW(1/t)
is a Wiener process by the four definitional properties requires another fact
which is outside the scope of the text. The fact is that any Gaussian process
X(t) with mean 0 and Cov [X(s),X(t)] = min(s,t) must be the Wiener
process. See the references and outside links for more information. Using
this information, we present a partial proof:
1.
Winv(t)−Winv(s) =tW(1/t)−sW(1/s)
which will be the difference of normally distributed random variables
each with mean 0, so the difference will be normal with mean 0. It
remains to check that the normal random variable has the correct vari-
ance.E[
(Winv(t)−Winv(s))^2]
=E
[
(sW(1/s)−tW(1/t))^2]
=E
[
(sW(1/s)−sW(1/t) +sW(1/t)−tW(1/t)−(s−t)W(0))^2]
=s^2 E[
(W(1/s)−W(1/t))^2]
+s(s−t)E[(W(1/s)−W(1/t))(W(1/t)−W(0))] + (s−t)^2 E[
(W(1/t)−W(0))^2]
=s^2 E[
(W(1/s)−W(1/t))^2]
+ (s−t)^2 E[
(W(1/t)−W(0))^2]
=s^2 (1/s− 1 /t) + (s−t)^2 (1/t)
=t−sNote the use of independence ofW(1/s)−W(1/t) fromW(1/t)−W(0)
at the third equality.- It seems to be hard to show the independence of increments directly.
Instead rely on the fact that a Gaussian process with mean 0 and
covariance function min(s,t) is a Wiener process, and thus prove it
indirectly.
Note that
Cov [Winv(s),Winv(t)] =stmin(1/s, 1 /t) = min(s,t).- By definition,Winv(0) = 0.
- The argument that limt→ 0 Winv(t) = 0 is equivalent to showing that
limt→∞W(t)/t = 0. To show this requires use of Kolmogorov’s in-
equality for the Wiener process and clever use of the Borel-Cantelli