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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

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


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

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


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Nevermind! Read it wrongfully. – Stefan Hansen Jun 12 '12 at 9:09
Yes, this is what I meant. I edited my question for more clearness – math Jun 12 '12 at 9:10
If you're asking why the integrand in $\int_0^t b(s,0) \mathrm{d} W_s$ is predictable, then it follows by the fact that $(s,\omega)\mapsto b(s,0)$ is deterministic and hence it is adapted (you argued yourself that it is continuous). – Stefan Hansen Jun 12 '12 at 9:20
@Stefan Hansen: What exactly is meant by predictability in this case and with respect to which $\sigma$-fields? How do you show that it is predictable in the general case, i.e. $(s,\omega)\mapsto b(s,X_s)$ ? – math Jun 12 '12 at 9:26
A process $(X_t)_{t\geq 0}$ is predictable if the map $(t,\omega)\mapsto X_t(\omega)$ is measureable wrt. to the predictable $\sigma$-field, which is the $\sigma$-field on $[0,\infty)\times\mathbb{R}$ generated by all left-continuous and adapted processes. Now because $(X_t)_{0\geq t\geq T}\in R$ we have the $(X_t)$ is adapted and hence also $(b(t,X_t))$ is adapted (and hence predictable because you have assumed $b$ is continuous). – Stefan Hansen Jun 12 '12 at 9:34

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