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Suppose we want to study a SDE of the form

$ dX_t = a(t,X_t)dt + b(t,X_t)dW_t$

and $X_0=Y$, on a filtered probability space $(\Omega,\mathcal{F}, \mathbb{F},P)$ and where $W$ is a $(P,\mathbb{F})$ Brownian Motion. Moreover, $Y$ is a $\mathcal{F}_0$ measurable function and $a,b,:[0,\infty)\times \mathbb{R}\to \mathbb{R}$. We define a strong solution to the above SDE as a continuous adapted process $X=(X_t)_{t\ge 0}$ such that

$ \int_0^t{|a(s,X_s)|+|b(s,X_s)|^2 ds}<\infty$

and

$X_t = Y + \int_0^t{a(s,X_s)ds}+ \int_0^t{b(s,X_s) dW_s}$

P-a.s. for all $t\ge 0$.

If we assume that $E(Y^2)<\infty$ and $a,b$ continuous satisfying $|g(t,x)-g(t,y)|\le K|x-y| $ for all $t,x,y$ and $g\in \{a,b\}$ as well as $|g(x,t)| \le K(1+|x|)$ for al $t,x,y$ and $g\in \{a,b\}$. Then for each $T>0$ the above SDE has a unique strong solution in the space $R$, which is the set of all continuous adapted processes $X=(X_t)_{0\le t \le T}$ with

$\|\sup_{0\le t\le T}|X_t|\|_{L^2(P)}<\infty$

The idea of the proof is the same as in the non stochastic case (define a map from $R\to R$ and show that this is a contraction. The map $\phi$ is defined as

$\phi(X)_t:=Y+\int_0^t{a(s,X_s)ds}+ \int_0^t{b(s,X_s) dW_s}$

First we show that $\phi(0)\in R$,

$|\phi(0)|^2\le 3(Y^2 +\sup_{0\le t\le T}(\int_0^t|a(s,0)|ds)^2+\sup_{0\le t\le T}|\int_0^t b(s,0)dW_s|^2)$

Now there's a sentence which I do not understand. They claim that both integrals are well defined and that the stochastic integral is a martingale. Clearly the non stochastic integral is well defined.

Do we assume that all the processes from $R$ satisfy the following condition:

$ \int_0^t{|a(s,X_s)|+|b(s,X_s)|^2 ds}<\infty$

This is not a mathematical question, more how to understand the assumptions of the theorem. We proved the following statement about stochastic integral:

Fix a continuous local Martingale $M$ with $M_0=0$. For each $H\in L^2(M)$, where $L^2(M)$ is the set of all predictable processes $H$ satisfying $ E[\int_0^\infty H_s^2d\langle M\rangle_s]<\infty$ there exists a unique element $H\cdot M\in \mathcal{H}^{2,c}_0$ satisfying $\langle H\cdot M,N\rangle = \int Hd\langle M,N\rangle$ for all continuous local martingale $N$, null at 0.

The space $\mathcal{H}^{2,c}_0$ is the space of all continuous $(P,\mathbb{F})$ martingales $M$, null at zero and that are bounded in $L^2(P)$.

In the situation above, $W$ is clearly a continuous local martingale, null at zero. Why is $(s,\omega)\mapsto b(s,\omega)$ predictable? It is by assumption continuous, but why is it adapted? Then predictability follows. If I can assume that

$\int |b(s,X_s)|^2 ds<\infty$

then I agree that the stochastic integral is a martingale. However, since we defined the space $R$ just as the space of continuous adapted processes, I do not see why I can assume this!

Thank you for your help

math

  • 0
    yes of course :)2012-06-12

0 Answers 0