# question on Kolmogorov's extension theorem

I'm reading Bernt Oksendal's "Stochasticc Differential Equations" and got confused at the Kolmogorov's extension theorem.

In chapter 2, page 11 (sixth edition) it says:

Theorem 2.1.5 (Kolmogorov's extension theorem)

For all $t_1, \cdots , t_k \in T$, $k \in \mathbb{N}$ let $\nu_{t_1,\cdots,t_k}$ be probability measures on $\mathbb{R}^{nk}$ s.t. $$\nu_{t_{\sigma(1)},\cdots,t_{\sigma(k)}} (F_1 \times \cdots \times F_k) = \nu_{t_1,\cdots,t_k} (F_{\sigma^{-1}(1)} \times \cdots \times F_{\sigma^{-1}(k)}) \tag{K1}$$

for all permutations $\sigma$ on $\{1,2, \cdots,k\}$ and $$\nu_{t_1,\cdots,t_k}(F_1\times \cdots \times F_k) = \nu_{t_1,\cdots,t_k, t_{k+1}, \cdots, t_{k+m}}(F_1\times \cdots \times F_k\times \mathbb{R}^n \times \cdots \times\mathbb{R}^n )\tag{K2}$$ for all $m \in \mathbb{N}$, where (of course) the set on the right hand side has a total of $k + m$ factors.

Then there exists a probability space $(\Omega,\mathscr{F},P)$ and a stochastic process $\{X_t\}$ on $\Omega$, $X_t:\Omega \rightarrow \mathbb{R}^n$, s.t. , $$\nu_{t_1,\cdots,t_k} (F_1 \times \cdots \times F_k) = P[X_{t_1} \in F_1, \cdots , X_{t_k} \in F_k]$$

for all $t_i \in T$, $k\in \mathbb{N}$ and all Borel sets $F_i$.

I don't understand here why (K1) is necessary?

(K2) is clearer to me: it simply says, given $k$ observation, $t_1$ to $t_k$ and $F_1$ to $F_k$, calculate the chance $\nu_{t_1,\cdots,t_k} (F_1 \times \cdots \times F_k)$, then it is equal to a broader observation of $n+m$. This is to ensure the consistency.

But what does (K1) mean?

If I choose $k=2$ and the permutation $\sigma$ that $\sigma(1)=2, \sigma(2)=1$, (K1) becomes $$\nu_{t_2, t_1} (F_1 \times F_2) = \nu_{t_1, t_2} (F_2 \times F_1) \tag{K1.k=2}$$

But since $\nu_{t_i}(F_i) =P(X_{t_i} \in F_i)$, isn't this (K1.k=2) obvious? Since $$\nu_{t_1}(F_2) =P(X_{t_1} \in F_2)$$ $$\nu_{t_2}(F_1) =P(X_{t_2} \in F_1)$$ so, $$\nu_{t_2, t_1} (F_1 \times F_2) \\ = P(X_{t_2} \in F_1)* P(X_{t_1} \in F_2) \\ =P(X_{t_1} \in F_2) * P(X_{t_2} \in F_1) \\ = \nu_{t_1, t_2} (F_2 \times F_1) \tag{K1.k=2}$$

This is a trivial conclusion? Or, did I mistake here -- that (K1.k=2) means the measure $\nu_1, \nu_2$ are independent, and this is not obvious?

-

Actually, for $k=2$, (K1) does not say what is in the question but that, for every measurable $F_1$ and $F_2$, one has $\nu_{t_2, t_1} (F_2 \times F_1) = \nu_{t_1, t_2} (F_1 \times F_2)$. This is a nonempty condition on $\nu_{t_2, t_1}$ and $\nu_{t_1, t_2}$.

More generally, (K1) says that each $\nu_{t_1,\cdots,t_k}$ determines completely $k!-1$ other probability measures, namely, the measures $\nu_{t_{\sigma(1)},\cdots,t_{\sigma(k)}}$ for every $\sigma$ in $\mathfrak S_k\setminus\{\mathrm{Id}\}$.

Edit: The revised version assumes that random variables $(X_t)$ exist with marginals $\nu_{t_1,\cdots,t_k}$. This is taking the conclusion as a hypothesis: to ensure the existence of such random variables is the whole point of conditions (K1) and (K2) hence to notice that if $(X_t)$ exist then (K1) holds is moot.

Additionally, the post seems to assume that $(X_t)$ is independent, hence that $\nu_{t_1,\cdots,t_k}$ is always the product of the probability measures $\nu_{t_i}$. The setting of Kolmogorov extension theorem is much wider than the independent case (in fact the independent case does not require this theorem) hence all the computations after "But what does (K1) mean?" are moot as well.

-
i think i didn't put the question clearly -- i just updated it. – athos Sep 23 '13 at 8:03
"i just updated it"... And you should not have! – Did Sep 23 '13 at 8:21
Thanks @Did , now my understanding is, the Kolmogorov's extension theorem is to guarantee that, if there's a good (fulfill K1 and K2) family of finite-dimensional measures defined on $\mathbb{R}^{nk}$, such measure could be extended to infinite dimensional case of $(R^n)^T$. My part after "but what does (K1) mean?" assumes 1. the independence of $X_{t_i}$ and 2. the existence of individual measures $\nu_{t_i}$. This is moot. Kolmogorov's theorem is to ensure the extension without such 2 assumptions. – athos Sep 23 '13 at 8:53
thanks again for your time, measure theory is a new language to me. so sometimes i struggle not only trying understand the book, but also express my misunderstandings clearly. – athos Sep 23 '13 at 8:55