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How do you show that $\{\exp(B_t(\omega))\}_{0 \le t \le T} \in H^2$ where $B_t$ is a standard wiener process $H^2=\{f\in L^2(P\times m):f~~\text{adapted}\}$ and $P\times m : {\cal F} \times {\cal B}[0,T] \to \cal R $.

I'm mostly just confused on how to even go about showing something like this. Any help would be greatly appreciated. Thanks.

Did
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ActuariallyImpaired
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1 Answers1

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We have by Tonelli's theorem

$$\int_{\Omega \times [0,T]}\exp(B_t(\omega))^2d(P(\omega)\times m(t)) = E[\int_0^T \exp(B_t)^2 dt] = \int_0^TE[\exp(2B_t)]dt $$

and note that $E[\exp(2B_t)]$ is the moment generating function of a $N(0,t)$ random variable evaluated at $x=2$. Recall (or calculate) $E[\exp(x B_t)] = \exp(x^2t/2)$ for all $x \in \mathbb{R}$. Thus the above integral is $$ = \int_0^T\exp(2^2 t/2)dt = \int_0^T \exp(2t)dt = \frac12(\exp(2T)-1) < \infty. $$ Thus $(\omega,t)\longmapsto \exp(B_t(\omega)) \in L^2(\Omega \times [0,T])$. Furthermore, $t \longmapsto \exp(t)$ is a homeomorphism so it preserves the adaptivity of $B$, and the result follows.

nullUser
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  • Thanks! But I'm sort of confused with the first step, what's the idea behind that? (I feel like this is a basic result but measure theory is not my forte.) – ActuariallyImpaired Mar 10 '13 at 20:47
  • To show an element $f \in H^2$ you must show $f \in L^2(P \times m, \Omega \times [0,T])$ for all $T$. That is, we must check $\int_{\Omega \times [0,T]} |f|^2 d(P \times m) < \infty$. That is the first step. – nullUser Mar 10 '13 at 20:52
  • A common way to show this is via Fubini's Theorem or Tonelli's Theorem which say that $\int_{X \times Y} f d(x,y) = \int_X\int_Y f dydx = \int_Y\int_X f dx dy$. Fubini's theorem says this holds when $f \in L^1(X \times Y)$, and Tonelli's theorem says this holds if $f \geq 0$. We use Tonelli's Theorem since $\exp$ is nonnegative. – nullUser Mar 10 '13 at 20:55
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    Also recall that $\int_\Omega f dP = E[f]$ by definition, so that the second integral $E[\int_0^T \exp(B_t)^2 dt]$ means exactly the same thing as $\int_{\Omega}\int_0^T \exp(B_t)^2dt dP$. – nullUser Mar 10 '13 at 20:57