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In this post user SM2 provides a way of constructing the Wiener measure $\nu$ on the measurable space $(C([0,\infty),\mathcal{B}(C([0,\infty))))$ where $\mathcal{B}(C([0,\infty)))$ is the Borel $\sigma$-algebra of $C([0,\infty))$ induced by the topology of uniform convergence on compact sets.

In 3), for a Wiener process $\{W_t\}_{t\in[0,\infty)}$ on the probability space $(\Omega,\mathcal{F},\mathbb{P})$, SM2 defines the measurable map $g:\Omega\to C([0,\infty))$ which I'm guessing is just $$\omega\in\Omega\mapsto g(\omega)\in C([0,\infty))$$ with $g(\omega)$ given by $$t\in[0,\infty)\mapsto g(\omega)(t)=W_t(\omega).$$ At the end SM2 defines $\nu:\mathcal{B}(C([0,\infty)))\to[0,1]$ by setting $\nu(A)=\mathbb{P}(g^{-1}(A))$ for $A\in\mathcal{B}(C([0,\infty))))$ so that $\nu$ is the Wiener measure.

My question is, if we don't work with $g$ but instead define $$\nu^*:\mathcal{B}(C([0,\infty)))\to[0,1]$$ by specifying $\nu^*(\{f\in C([0,\infty)):f(t)\leq a\})=\mathbb{P}(\{\omega\in\Omega:W_t(\omega)\leq a\})$ for each $t\in[0,\infty)$ and $a\in\mathbb{R}$ then will $\nu^*$ and $\nu$ coincide?

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  • $\begingroup$ Did you mention the present post to SM2? $\endgroup$
    – Did
    Commented Oct 5, 2016 at 10:05
  • $\begingroup$ @Did When I checked SM2's profile, their last seen time was late April so I figured that the account is not active. $\endgroup$
    – user375366
    Commented Oct 5, 2016 at 10:15

1 Answer 1

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This does not uniquely define a measure on $\mathcal B(C([0,\infty)))$. The problem is that the marginals of a process at a single time are not enough to determine the full process. For a play example, instead of a stochastic process in contininuous time let us look at a process $X$ in discrete time $\{0,1,\ldots,N\}$. It may be tempting to say the law of $X$ is fully determined by looking at the law of $X_n$ for $n=0,1,\ldots,N$, but really we know this is not true - for a couple of extreme examples, suppose $X_m$ and $X_n$ are identically distributed for all $m,n$, then we could have $X_m=X_n$ for all $m,n$ or $X_m,X_n$ independent for $m\neq n$, and looking at each $X_n$ individually will in no way distinguish these extremely different cases.

For the finite case there is an easy work around (namely, looking at the joint distribution of $X=(X_0,\ldots,X_N)$ which is a random variable in $\mathbb R^{N+1}$), but what about the infinite case? Thankfully, the Kolmogorov consistency theorem tells us that if we know all of the finite dimensional distributions, that is, the law of $(X_{t_1},\ldots,X_{t_n})$ for any finite subset $\{t_1,\ldots,t_n\}\subset T$ (where $T=\mathbb R$ or $\mathbb N$, whatever our time scale is), and these are consistent with one another, then this uniquely determines a measure on the relevant space. For continuous processes such as the Wiener process, we typically need a slight modification of this argument to guarantee a continuous version, but it still works. This is why SM2's construction is the correct one.

Let's end with a concrete counterexample, that is, a measure on $C([0,\infty))$ with the same marginals as $\nu$ which nonetheless defines a different process. Let $Z:\Omega\longrightarrow\mathbb R$ be any measurable function which is standard normal with respect to $\mathbb P$ and define $\Gamma_t:=\sqrt{t}Z$. Then $\Gamma:\Omega\longrightarrow C([0,\infty))$ is measurable and $$ \mathbb P(\Gamma_t\le a)=\frac1{\sqrt{2\pi t}}\int_{-\infty}^ae^{-x^2/2t}dx=\mathbb P(W_t\le a) $$ for any $t\ge0$ and $a\in\mathbb R$. However, $\Gamma$ and $W$ have different laws. For example, $\mathbb P(\Gamma_1> 1,\Gamma_2\le 0)=0$ but $\mathbb P(W_1>1,W_2\le0)>0$. So $\nu^*$ in your question could be equal to either $\nu$ or the law of $\Gamma$.

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  • $\begingroup$ Thanks. Your counterexample in the end convinced me that $\nu^*$ would not recover $\nu$. In your second paragraph you mentioned that if we were to know all finite dimensional distributions then the measure would be uniquely determined. Just to confirm, does that mean if we set $\nu^*(\{f\in C([0,\infty)):(f_{t_1},\dots,f_{t_n})\in B^n\})=\mathbb{P}(\{\omega\in\Omega:(W_{t_1}(\omega),\dots,W_{t_n}(\omega))\in B^n\})$ for any $n\in\mathbb{N}$, $t_1,\dots,t_n\in[0,\infty)$ and $B^n\in\mathcal{B}(\mathbb{R}^n))$ then would $\nu^*$ now correctly give the Wiener measure? $\endgroup$
    – user375366
    Commented Oct 7, 2016 at 10:27
  • $\begingroup$ Almost - you need $\nu^*(\{f\in C([0,\infty))\,:\,f(t_1)\in B_1,\ldots,f(t_n)\in B_n\})=\mathbb P(W_{t_1}\in B_1,\ldots,W_{t_n}\in B_n)$ for any $B_1,\ldots,B_n\in\mathcal B(\mathbb R)$. $\endgroup$
    – Jason
    Commented Oct 7, 2016 at 14:43
  • $\begingroup$ I'm having difficulty seeing why working with $\mathcal{B}(\mathbb{R}^n)$ instead of $\mathcal{B}(\mathbb{R})$ produces a different result. Is it not the case that $\mathbb{P}(\{\omega\in\Omega:(W_{t_1}(\omega),\dots,W_{t_n}(\omega)\in B^n\})=\mathbb{P}(\{\omega\in\Omega:W_{t_1}(\omega)\in B_1,\dots,W_{t_n}(\omega))\in B_n\})$ when $B^n=B_1\times\dots\times B_n$?. $\endgroup$
    – user375366
    Commented Oct 7, 2016 at 15:37
  • $\begingroup$ I'm sorry, I thought by $B^n$ you meant $B\times B\times\ldots\times B$, but as written you are completely correct! $\endgroup$
    – Jason
    Commented Oct 7, 2016 at 18:40
  • $\begingroup$ Thanks. I found your answer and comments very helpful. $\endgroup$
    – user375366
    Commented Oct 8, 2016 at 0:11

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