Re: How Chapman-Kolmogorov implies Markov ??

On Feb 9, 7:56 am, "rayoha...@xxxxxxxxxxxx" <juanp...@xxxxxxxxx>
wrote:
It's easy to see that a stochastic process that follows the Markov
property, then follows the Chapman-Kolmogorov equation. But the visce
versa holds too, and I can find a proof of that. Some one of you can
show me a proof ??

Before you fade back into obscurity, perhaps you would help me
understand the question.

Starting with

http://en.wikipedia.org/wiki/Chapman-Kolmogorov_equation

I run into some conceptual problems. An "indexed set of random
variables" is almost an empty concept, being, I suppose, a set of
random variables each identified with an associated element of an
index set. Perhaps the notation "f_i" is supposed to imply there is
something equivalent about all of these random variables, though what
this is is undefined. Identical marginal distributions?

Now, when we begin labeling a set of the indexes, "i_1, i_2, ...,
i_n", I get lost. We have in mind again, of course, an almost empty
concept, that we index the index set iself, or else write a finite
sequence in it, explicitly using the integers as our indices.

It occurs to me now (for the first time? well... experience is
infinite/life finite), that indexing a set by the integers and writing
a sequence in it are related but distinct concepts: the sequence may
reuse elements, while indexing implies a one-to-one relation -- we
don't say "oh, x_3 and x_7 are the same variable... did I forget to
mention that"?

Which is intended?

Anyway, I get the feeling, not unknown in mathematical arguments, that
the formal exposition has run ahead of the sense -- the author knows
what he is trying to capture, I don't! Rather than trying to reverse
engineer his intentions through the ambiguities, would you possibly be
so kind as to fill in story?

Please assume I am just sophisticated enough to understand the
intention, though not conversant with it.

(P.S. Thanks for posting a real question, anyway)

.

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