I have a question concerning the definition of a family of probability measures for the solutions to an Ito diffusion $$X_t^x = x + \int_0^tb(X_s^x)ds + \int_0^t \sigma(X_s^x)dB_s$$ as it's given in Oksendal's Stochastic Differential Equations on p. 111 (A PDF of Oksendal can be found here: http://th.if.uj.edu.pl/~gudowska/dydaktyka/Oksendal.pdf).

Oksendal defines $\mathcal{M}_\infty$ to be the $\sigma$-algebra generated by the random variables $\omega\to X_t(\omega) = X_t^y(\omega)$, where $t\geq 0$, $y\in\mathbb{R}^n$. Already I'm a little confused, because he seems to be saying that for fixed $t\geq 0$, $\sigma(X_t^y) = \sigma(X_t^x)$ for all $x,y\in\mathbb{R}^n$. So my first question is: Is my interpretation correct? If so, why is it true that these random variables all generate the same $\sigma$-algebra?

He goes on to define the probability measures $P^x$ on members of $\mathcal{M}$ (he's now dropped the $\infty$ subscript for some reason) by $$P^x[X_{t_1} \in E_1,...,X_{t_k}\in E_k] = P[X_{t_1}^x\in E_1,...,X_{t_k}^x\in E_k]$$ where $E_1,...,E_k$ is any collection of Borel sets and $t_1,...,t_k\in [0,\infty)$. What he seems to be saying here is that we are just going to replace any initial condition $y\in\mathbb{R}^n$ with $x$ under $P^x$, so that $$P^x[X_{t_1}^y \in E_1,...,X_{t_k}^y\in E_k] = P[X_{t_1}^x\in E_1,...,X_{t_k}^x\in E_k].$$ So my second question is again: Is this interpretation correct? If so, it doesn't seem obvious that this definition would be consistent, i.e., it seems possible that one set could be assigned two different probabilities. For example, if $\{X_t^y \in A\} = \{X_s^z \in B\}$, but $P(X_t^x\in A) \neq P(X_s^x\in B)$.

Edit: I now realize this interpretation makes no sense, as up to null sets we have $$\{X_0^x = x\} = \{X_0^y = y\} = \Omega,$$ but $$P^x[\Omega] = P^x[X_0^y = y] = P[X_0^x = y] = 0$$ and $$P^x[\Omega] = P^x[X_0^x = x] = P[X_0^x = x] = 1.$$


Already I'm a little confused, because he seems to be saying that for fixed $t\geq 0$, $\sigma(X_t^y) = \sigma(X_t^x)$ for all $x,y\in\mathbb{R}^n$.

No. $t$ and $x$ aren't fixed in the definition, i.e. $$ \mathcal M_\infty = \sigma\bigl(X_t^x, t\ge 0,x\in \mathbb R\bigr). $$

He goes on to define the probability measures $P^x$ on members of $\mathcal{M}$

No. $P^x$ is already defined as distribution of the Brownian motion $B$ started from $x$. (And it indeed can be regarded as the push-forward of $P^0$, which is the distribution of $B$ started from $0$, not that of $X$, as gigaster writes, under the shift $\theta_x$; although this is not very relevant to the discussion.)

What Øksendal defines here, is $Q^x$, not $P^x$.

What he seems to be saying here is that we are just going to replace any initial condition $y\in\mathbb{R}^n$ with $x$ under $P^x$, so that [...]

No, again. Actually, the use of $X$ in the definition of $$Q^x[X_{t_1} \in E_1,...,X_{t_k}\in E_k] = P^0[X_{t_1}^x\in E_1,...,X_{t_k}^x\in E_k]$$ may be indeed quite misleading. The measures $Q^x$ are defined on (the Borel subsets of) $C([0,T])$, the space of continuous functions, by $$ Q^x(A) = P^0(X_\cdot^x \in A), A\in \mathcal B\bigl( C([0,T])\bigr). $$

Unfortunately, this part in Øksendal's book is written very sloppily (despite having undergone six editions). Say, in Theorem 7.1.2, which follows the definition of $Q^x$, $E^x$ is used to denote the expectation w.r.t. $Q^x$; this is admitted in the book immediately, but creates some confusion anyway, as $E^x$ was earlier used to indicate expectation w.r.t. $P^x$.

But there is a much bigger problem: Øksendal somehow omits the very intricate measurability questions. E.g., when he says that $X_r = F(X_t, t,r,\omega), r\ge t$ "by uniqueness", this is not so straightforward as it seems. Indeed, imagine that we want to argue by uniqueness and plug $F(X_t, t,r,\omega)$ as a candidate for solution into the equation, but why this process is adapted? (Adaptedness is crucial for the integration w.r.t. $B$.) Why is it even measurable? To claim this, one first need to show the joint measurability of $F(x,t,r,\omega)$ w.r.t. $x,t,r,\omega$.

This part, I hope, is better explained in our textbook with Yuliya Mishura (I can send you the corresponding chapters on request).

  • $\begingroup$ From the statement that $P^0$ denotes the distribution of $B$ starting at $0$ I understand that $P^0$ is a measure defined on the Borel sets of $C[0,T]$. But then the set $\{X^x\in A\}:=\{\omega\in\Omega\;|\; X^x(\omega)\in A\}$ is not a Borel subset of $C[0,T]$, rather the pre-image of one. Does this mean that the expression $P^0(X^x\in A)$ is only formal? If yes, then shouldn't $P^0$ be replaced by $\mathbb{P}$ ? $\endgroup$ – gigaster Apr 21 at 6:04
  • $\begingroup$ @gigaster, yes, this is a problem with Øksendal's writing. From my perspective, it seems that he is working canonically, i.e. $\Omega = C([0,T])$, $B_t(\omega) = \omega(t)$, and $\mathbb P = P^0$. The problem is that the precise framework is never written explicitly, you can just try to guess that from the places like this. $\endgroup$ – zhoraster Apr 21 at 7:33
  • $\begingroup$ @zhoraster thank you for your answer. yes, it would be great if you could send me the chapters from Mishura’s book that you reference. I have yet to find a good resource on this. $\endgroup$ –  mheldman Apr 21 at 10:45
  • $\begingroup$ @zhoraster Also, my first question "he seems to be saying that for fixed $t\geq 0$, $x,y\in\mathbb{R}^n$, $\sigma(X^y_t) = \sigma(X^x_t)$" was referencing the fact that he writes $\omega \to X_t(\omega) = X_t^y(\omega)$, where $X_t(\omega)$ has no superscript. I was trying to understand why he did not put a superscript on $X_t(\omega)$. Usually this would be because writing $X_t(\omega)$ instead of $X_t^y(\omega)$ is somehow unambiguous in this situation. $\endgroup$ –  mheldman Apr 21 at 13:41

One way to make this precise is the following:

Let $(\Omega,\mathcal{F}, \mathcal{F}_t, \mathbb{P})$ be a complete, filtered probability space. For simplicity, I assume that $b$ and $\sigma$ take values on $\mathbb{R}^n$ and $\mathbb{R}^{n\times n}$ respectively. I will also restrict the time domain to $[0,1]$ (the general case follows similarly). Let $X^x:=\{X^x_t\}_{0\leq t\leq 1}$ be the strong solution of the Ito SDE, with initial value $X_0^x=x\in\mathbb{R}^n$, $\mathbb{P}$-a.s. Define $Q^x$ as the law of $X^x$ on the "path-space" $C([0,1];\mathbb{R}^n)$ (the space of $\mathbb{R}^n$-valued continuous functions defined on [0,1], endowed with the topology of uniform convergence). In other words, consider $X^x$ as a random variable $\Omega\ni\omega\mapsto X^x(\omega)\in C([0,1];\mathbb{R}^n)$ and define $Q^x$ to be $Q^x[A]=\mathbb{P}[X^x\in A]$, for $A\in\mathscr{B}\big(C([0,1];\mathbb{R}^n)\big)$.

For example, let $t_1\in[0,1]$, $B^n$= the unit ball on $\mathbb{R}^n$ and $A=\{\omega\in C([0,1];\mathbb{R}^n)\;| \; \omega(t_1)\in B^n\}\in \mathscr{B}\big(C([0,1];\mathbb{R}^n)\big) $. Then $Q^x[A]=\mathbb{P}\big[X^x_{t_1}\in B^n\big]$.

Now, fix $x\in\mathbb{R}^n$, and observe that since $X^x$ is a strong solution, it is adapted to the natural filtration of $B$. Thus, there exists a measurable map $F_{x}: C([0,1];\mathbb{R}^n)\rightarrow C([0,1];\mathbb{R}^n)$ such that for $\mathbb{P}$-almost all $\omega\in\Omega$ $$ X^x(\omega)=F_x(B(\omega))$$

By letting $\theta$ be the Wiener measure on $C([0,1];\mathbb{R}^n), \mathscr{B}\big(C([0,1];\mathbb{R}^n)\big)$ (i.e. the law of the standard Wiener process $B$, starting at $0$) we can express the law of $\{X^x\}_{x\in\mathbb{R}^n}$ on path-space by the push-forward $$Q^x=\theta_{*F_x}$$

$$Q^x[A]=\theta_{*F_{x}}[A]=\theta[F^{-1}_x(A)]=\theta\big[\{\omega\in C([0,1];\mathbb{R}^n)| F_x(\omega)\in A\}\big]$$ $$=\mathbb{P}\big[\{\omega\in\Omega | F_x(B(\omega))\in A\}\big]=\mathbb{P}\big[X^x\in A\big]$$

Note 1: In the above, $\omega$ has been used to denote both a function on $C([0,1];\mathbb{R}^n)$ and an element of the initial probability space $\Omega$. In each case, I have explicitly stated the space in which $\omega$ lives to clarify this use of notation.

Note 2: If we use the notation of the book you refer to, then we have to use $Q^x$ instead of $P^x$, since $P^x$ is reserved for the law of a Wiener process starting at $x\in\mathbb{R}^n$ (see index). Furthermore, I have used $\theta$ instead of $P^0$. Note that the measures $\{P^x\}_{x}, \{Q^x\}_{x} $ can only be defined on the path space $C([0,1];\mathbb{R}^n)$.

  • $\begingroup$ If you define $P^x$ as the pushforward under $\theta_x$, then $X^0$ under $P^x$ will net solve the SDE. $\endgroup$ – zhoraster Apr 21 at 4:48
  • $\begingroup$ @zhoraster Thank you for letting me know. I was not being careful. I edited my answer to fix this. $\endgroup$ – gigaster Apr 21 at 5:28
  • $\begingroup$ It's better now, yes. But, although this is admitted in your post, there is an additional source of confusion, as you use $P^x$ for different things. $\endgroup$ – zhoraster Apr 21 at 5:31
  • $\begingroup$ @gigaster so if the family $Q^x$ is only defined on the path space, then what is meant by things like $Q^x[X_0 = x] = 1$? I interpret this as "$Q^x$ gives density 1 to the paths $\omega$ such that $\omega(0) = x$." $\endgroup$ –  mheldman Apr 21 at 12:09
  • $\begingroup$ @gigaster For example, what does it mean to write $E^x[X_t]$? $\endgroup$ –  mheldman Apr 21 at 13:30

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