When wishing to prove existence of Brownian motion, most authors take the route of defining the desired finite dimensional distributions, showing the family of defined finite dimensional distributions is "consistent" and then applying the Daniell-Kolmogorov extension theorem to claim there exists a probability measure on the measurable space $(\mathbb{R}^{[0,\infty)},\mathcal{F})$ where $\mathbb{R}^{[0,\infty)}$ is the space of real valued functions on $[0,\infty)$ and $\mathcal{F}$ is the sigma algebra generated by the algebra of finite dimensional cylinder sets in $\mathbb{R}^{[0,\infty)}$ . I have found some inconsistencies between authors on how they define "consistent" and I feel uneasy about how the extension theorem is applied in certain situations.

First I will make clear what I mean by a family of finite dimensional distributions. Let $T$ set of finite sequences of the form $t=(t_1,t_2,\ldots,t_k)$ of distinct nonnegative numbers whose length $k$ takes values in the set of positive integers. Then suppose for each $t$ of length $k$ we have a probability measure $\mathbb{P}_t$ on $(\mathbb{R}^k,\mathcal{B}(\mathbb{R}^k))$. We then call $(\mathbb{P}_t)$ for all $t\in T$ a family of finite dimensional distributions.

Karatzas and Shreve say such a family is consistent if the following two conditions hold,

(i.) If $s=(t_{i_1},t_{i_2},\ldots,t_{i_k})$ is a permutation of $t=(t_1,t_2,\ldots,t_k)$, then for any collection of Borel sets $(B_i)_{i=1}^k$ we have

$$\mathbb{P}_{t}(B_1 \times B_2 \times \ldots \times B_k)=\mathbb{P}_{s}(B_{i_1} \times B_{i_2} \times \ldots \times B_{i_k})$$

(ii.) If $t=(t_1,t_2,\ldots,t_k)$ with $k \geq 1$, $s=(t_1,t_2,\ldots,t_{k-1})$ and $\{B_i\}_{i=1}^{k-1}$ is a collection of Borel sets then,

$$ \mathbb{P}_{t}(B_1 \times \ldots \times B_{k-1} \times \mathbb{R}) = \mathbb{P}_{s}(B_{1} \times B_{2} \times \ldots \times B_{{k-1}}).$$

Now in the book by Kallianpur and Sundar they only insist on the second condition and state that the set of indices $t=(t_1,\ldots,t_k)$ must be strictly ordered. Both authors then apply the extension theorem abiding by their definition of a consistent family of finite dimensional distributions. Now for simplicity let $k=2$ and define the finite dimensional distribution of Brownian motion for any arbitrary $x \in \mathbb{R}$ and $t=(t_1,t_2)$ where $0\leq t_1 \leq t_2$ as,

$$ \mathbb{P}_{t}(B_1 \times B_2) = \int_{B_1 \times B_2} \frac{\exp(-\frac{|x-y|^2}{2t_1})}{\sqrt{2 \pi t_1}} \cdot \frac{\exp(-\frac{|y-z|^2}{2(t_2-t_1)})}{\sqrt{2 \pi(t_2- t_1)}} dydz $$.

I believe the first condition will not be satisfied as you cannot permute $t_1$ and $t_2$. You need the strict ordering. My questions is then am I missing something trivial, as pretty much all famous books use the conditions as in Karatzas and Shreve.

The second book is Stochastic analysis and diffusion processes by Gopinath Kallianpur and Padmanabhan Sundar.

  • $\begingroup$ Thanks for the comments, in the case of Brownian motion the densities depend on the parameters $t_i$. So I guess my question really is, when you define the finite dimensional distribution of Brownian motion do you ignore the permutation of the $t_i$'s ? In my example above $t=(t_1,_2)$ if you consider the permutation $s=(t_2,t_1)$ per the definition above you would be taking the difference $t_1-t_2$ in the denominator which causes issues? $\endgroup$
    – Emmet
    Sep 30 at 20:49
  • $\begingroup$ Apologies but I'm still confused. Then why is there a permutation in the time variables in the definition of a consistent family of FDDs? $\endgroup$
    – Emmet
    Sep 30 at 20:53
  • $\begingroup$ Thanks for the link, I understand what that post is saying. Why do we not say, $$ \mathbb{P}_{t_2,t_1}(B_2 \times B_1) = \int_{B_2 \times B_1} \frac{\exp(-\frac{|x-y|^2}{2t_2})}{\sqrt{2 \pi t_2}} \cdot \frac{\exp(-\frac{|y-z|^2}{2(t_1-t_2)})}{\sqrt{2 \pi(t_1- t_2)}} dydz$$ ? Of course this is wrong but what I'm missing is why we just ignore the permutation when writing the density ? $\endgroup$
    – Emmet
    Sep 30 at 21:16
  • $\begingroup$ It has to be relevant, you need to construct the explicit probability measure on $\mathbb{R}^k$ before you apply the extension theorem, otherwise you don't even have a stochastic process in the first place? $\endgroup$
    – Emmet
    Sep 30 at 21:22

2 Answers 2


$\newcommand\P{\mathbb P}\newcommand\R{\mathbb R}\newcommand\F{\mathscr F}$It is not true that "in the book by Kallianpur and Sundar they only insist on the second condition".

The consistency condition in the book by Kallianpur and Sundar is the following strengthened version of the second condition of Karatzas and Shreve: instead of the condition $$\P_t(B_1\times\cdots\times B_{k-1}\times\R) =\P_s(B_1\times\cdots \times B_{k-1}) \tag{KSh2}\label{KSh2}$$ of Karatzas and Shreve, Kallianpur and Sundar have (in Karatzas and Shreve's terms) the following condition: $$\P_t(B_1\times\cdots B_{j-1}\times\R\times B_{j+1}\times\cdots\times B_k) \\ =\P_s(B_1\times\cdots\times B_{j-1}\times B_{j+1}\times\cdots\times B_k) \tag{KSu}\label{KSu}$$ for all $j\in\{1,\dots,k\}$, where $t:=(t_1,\dots,t_k)$ with $t_1<\dots<t_k$ and $s:=(t_1,\ldots,t_{j-1},t_{j+1},\ldots,t_k)$; so, $s$ is obtained from $t$ by dropping the $j$th coordinate $t_j$ of the vector $t$.

So, \eqref{KSh2} is the special case of \eqref{KSu}, with $j=k$.

It is easy to see that the Kallianpur--Sundar consistency condition is equivalent to the conjunction of parts (i) and (ii) of the Karatzas--Shreve condition. In particular, to get \eqref{KSu} from \eqref{KSh2}, one needs the "permutation-invariance" condition, which is part (i) of the Karatzas--Shreve condition.

However, I do not like either the Kallianpur--Sundar consistency condition or the Karatzas--Shreve one. Here is what I think is a better condition, actually good for any "time set" $T$:

Let $T$ be any set. Let $\F_T$ denote the set of all finite subsets of $T$. Suppose that for each $F\in\F_T$ we have a probability measure $P_F$ over $\R^F$. (Because the linear space $\R^F$ for $F\in\F_T$ is finite dimensional, it is linearly isomorphic to $\R^k$ for some $k$, so that a probability measure $\mu_F$ can be constructed if we can construct probability measures over $\R^k$.) Let us then say that the family $(\mu_F)_{F\in\F_T}$ is consistent if for each $F\in\F_T$ and each $G\subseteq F$ we have $P_G=(p^F_G)_{\#}P_F$, where $(p^F_G)_{\#}P_F$ is the pushforward of the measure $P_F$ under the projection map $p^F_G\colon\R^F\to\R^G$ given by the formula $p^F_G x:=x|_G$ for all $x\in\R^F$, where $x|_G$ is the restriction of the function $x$ to $G$.

With the latter consistency condition, one does not need to enumerate or order or permute elements of the "time set" $T$. Yet, in particular cases such as when $T=[0,\infty)$, the latter consistency condition is equivalent to the Kallianpur--Sundar consistency condition and to the Karatzas--Shreve one.

  • $\begingroup$ I guess your consistency condition says that $(\mathbb{R}^F, P_F)$ forms a presheaf of probability spaces over the poset of finite subsets of $T$, which is nicely intuitive. $\endgroup$ Oct 1 at 23:37
  • $\begingroup$ @QiaochuYuan : Thank you for your comment. $\endgroup$ Oct 2 at 1:20

Though I'm not a probabilist, I'll try to explain how I think of condition (i), in the hope that it may be useful. Looking ahead to the goal of the discussion, consider a Brownian motion $X(t)$. The left side of (i) is intended to be the probability of the event that $X(t_1)\in B_1$ and $X(t_2)\in B_2$ and $\dots$ and $X(t_k)\in B_k$. The right side of (i) is the probability of the event that $X(t_{i_1})\in B_{i_1}$ and $X(t_{i_2})\in B_{i_2}$ and $\dots$ and $X(t_{i_k})\in B_{i_k}$. Because the $t$'s and the $B$'s have been permuted the same way, these two events are identical.

As Thomas Kojar wrote in the comments, this has nothing to do with Brownian motion; it works just as well for any stochastic process $X$.

Also note that, since the two events above are identical, we might just as well adopt the convention to write the constituents $X(t_i)\in B_i$ in order of increasing $t_i$'s, and that's what Kallianpur and Sundar do. Once they adopt this convention (their "strictly ordered"), they have non further use for (i), since the convention prohibits any non-identity permutations.

  • 1
    $\begingroup$ As the OP wrote in the comment, "you need to construct the explicit probability measure on $\mathbb R^k$ before you apply the extension theorem, otherwise you don't even have a stochastic process in the first place". So, an answer should not appeal to any stochastic process $X$. Rather, it should just consider how the Kallianpur--Sundar consistency condition relates to the Karatzas--Shreve one(s). $\endgroup$ Oct 1 at 1:49
  • $\begingroup$ @IosifPinelis I agree that the stochastic process $X$ in my answer (whether Brownian or not) would need more justification. I was thinking of $X$ as just motivation for the rest of the explanation. Formally, when matching up the two definitions for any particular $t_i$'s, one only needs the finitely many random variables $X(t_i)$, not the whole process. $\endgroup$ Oct 1 at 2:03
  • $\begingroup$ Even for the restrictions of a random function $X$ to finite sets, there is a little bit of friction between such sets and the finite sequences (and their permutations) in (say) the Karatzas--Shreve condition(s). Of course, I am not saying that there some substantial difficulties here. $\endgroup$ Oct 1 at 3:19

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