Re: How Chapman-Kolmogorov implies Markov ??

On Feb 10, 7:13 pm, "rayoha...@xxxxxxxxxxxx" <juanp...@xxxxxxxxx>
wrote:

sorry for my notation ... i will try to be more clear ... it´s just
dificult to do it in text mode ...

No need to apologize -- you've done a yeoman effort already: I'm sorry
I induced you to write it out yet again (not the least because in my
guilt I may feel obligated to try to read it :-).

I think sometimes in text mode the clearest way to convey things _is_
text... i.e., English text, with very sparse equations, or referring
to those written on a convenient web page.

I already gave you a rough lay-opinion, which I'll rephrase slightly:
it seems probable that if you laid on enough conditions or generality,
you would be able to require a markov chain by requiring the CK
relation, though I'm not sure what this is going to do for you. For
insufficient conditions, you will be left with the situation you
already encountered: in effect trying to show an integrand was
everywhere zero from a zero integral. The relation _could_ hold in
infinite ways for a particular transition which would not imply
underlying Markov conditions. Again... my guess is that by the time
you've put in sufficient assumptions you have in effect assumed your
result to start with.

Sorry to be so glib: may I ask just what has motivated your question?
What's the context? What are you trying to accomplish?

[You will also get more professional advice in sci.math, or one of the
groups related to probability or statistics]

definition:

p_(k|r)( y_1 , t_1 ; ... ; y_k , t_k | y_(k+1) , t_(k+1) ; ... ; y_(k
+r) , t_(k+r) )

is the conditional probability that SVs (stochastic variables)
Y_1 , ... , Y_k gives values
y_1 , ... , y_k at respectives times t_1 , ... , t_k if SVs Y_(k
+1) , ... , Y_(k+r) gives values
y_(k+1) , ... , y_(k+r) at respectives times t_(k+1) , ... , t_(k+r)

also

p_n( y_1 , t_1 ; ... ; y_n , t_n )

is the joint probability that SVs Y_1 , ... , Y_n
gives values y_1 , ... , y_n at times
t_1 , ... , t_n

That seems clear. I'll have to think about the details some more.

.

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