existence of the strong solution to SDE .

Question

The following text are mostly from Shreve's "Brownian Motion and Stochastic Calculus",GTM113,chapter 5.2.

Consider a the following SDE: $$X_t=X_0+\int_0^tb(s,X_s)\,ds+\int_0^t\sigma(s,X_s)\,dW_s$$ (1)If $X_t$ is a strong solution ,then according to definition 2.1(page 285) ,$X_t$ should satisfy the equation above.But my question is "in what sense"? I don't think it is "almost surely" because the stochastic integral can't be defined pathwise,so if it is "a.s." the last term in the equation is meaningless.In my opinion ,it should be in this norm: $$\|X\|=\mathbb E\left[\int_0^\infty|X_s|^2\,ds\right]$$ (2) under the condition that $b,\sigma$ are global lipschitz and linear growth,thm 2.9(page 289) proves the existence of the strong solution by Picard iteration.We get the following equation: $$X_t^{k+1}=X_0+\int_0^tb(s,X_s^{k})\,ds+\int_0^t\sigma(s,X_s^{k})\,dW_s$$ the last step is to show $X_t:=\lim X_t^k$ is truly a strong solution.The author leaves it as an exercise 2.11,and the answer is in page 388.

There are four terms in the equation above.The author shows that the 1st and the 3nd term are converge a.s.,the last term converge in $L^2$.So $X_t$ is truly a strong solution.I'm not satisfied with this argument because I think we should verify the convergence in one sense,not a.s for two parts and $L^2$ for another.

Should we verify them in the norm I mentioned in question (1)? or there are something else I misunderstand?

Answer 1

That the stochastic integral $$\int_0^t \sigma(s,X_s) \, dW_s$$ is not defined as an pathwise integral does not mean that $$\int_0^t \sigma(s,X_s) \, dW_s$$ has no "a.s." meaning. Note that for each $t$ $$M_t := \int_0^t \sigma(s,X_s) \, dW_s$$ is a random variable; so in particular it does make sense to state that

$$X_t = X_0 + \int_0^t b(s,X_s) \, ds + M_t$$

almost surely. In fact, one can show that the uniqueness does not only hold "almost surely", but in a stronger sense:

Let $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ be any two solution of the SDE $$dZ_t = b(t,Z_t) \, dt + \sigma(t,Z_t) \, dW_t$$ with initial values $X_0 \in L^2$ and $Y_0 \in L^2$. Then $$\mathbb{E} \left( \sup_{t \leq T} |X_t-Y_t|^2 \right) \leq C_T \mathbb{E}(|X_0-Y_0|^2).$$ In particular if $X_0=Y_0$, we have $$\mathbb{E} \left( \sup_{t \geq 0} |X_t-Y_t|^2 \right)=0.$$

For a proof see e.g. René Schilling, Lothar Parztsch: Brownian Motion - An Introduction to Stochastic Processes.

Regarding your second question: We know that

$$Y_k := X_{t}^{k+1} - X_0 - \int_0^t b(s,X_s^k) \, ds$$

converges (for each fixed $t$) to $Y := X_t-X_0 - \int_0^t b(s,X_s) \, ds$. On the other hand, Shreve proves that

$$Y_k = \int_0^t \sigma(s,X_s^k) \, dW_s$$

converges in $L^2$ to $Z := \int_0^t \sigma(s,X_s) \, dW_s$. Then there exists a subsequence such that $Y_{n_k} \to Z$ almost surely and the uniqueness of (a.s.) limits proves $Y=Z$, i.e.

$$X_t-X_0 - \int_0^t b(s,X_s) \, ds = \int_0^t \sigma(s,X_s) \, dW_s \quad \text{a.s.}$$