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Let $(X(t))_{t\geq0}$ be a Markov proces such that s $X(0)=x_0$. Consider the random vector $(X(t_1),\dots,X(t_n))$ with corresponding joint density $g(x_1,\dots ,x_n)$. Is it then true that $g(x_1,\dots x_n)=p(0,t_1,x_0,x_1)\cdot p(t_1,t_2,x_1,x_2) \cdots p(t_{n-1},t_n,x_{n-1},x_n)$ where $p(a,b,c,d)$ is the corresponding transition density between time a to b and state c to d.

If so I would love a proof or a reference. I know a bit of theory about Markov processes, but am not perfectly comfortable with transition densities. I tried proving it just using the Markov property, but without great success.

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The obvious: iterate this. –  Did Sep 12 '13 at 11:51
    
Will this work with an uncountable state space? –  Henrik Sep 12 '13 at 13:52
    
Why not? (Say, you erased a comment, no?) –  Did Sep 12 '13 at 13:54
    
Is the left hand side not just 0 always? I thought I understood it (indicated it in a comment), worked with it and suddenly I felt indicating that I understood it was a bit misleading :) –  Henrik Sep 12 '13 at 14:04
    
Hmmm... how do you define p(t,s,x,y) in the first place? The trouble you have seems to come from undigested aspects of THAT. –  Did Sep 12 '13 at 14:07
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1 Answer

The question might be based on a confusion between transition probabilities and transition densities.

In the first case the conditional distribution of $X(t+s)$ conditionally on $\mathcal F_s=\sigma(X(u);u\leqslant s)$ is such that $$ P[X(t+s)=y\mid X(s)=x,\mathcal F_s]=p(s,t+s,x,y), $$ for every states $(x,y)$. Thus, every entry $p(s,t+s,x,y)$ is nonnegative and, for every $x$, $$ \sum_yp(s,t+s,x,y)=1. $$ This implies that, for every $x$, the set $\{y\mid p(s,t+s,x,y)\ne0\}$ is at most countable.

In the second case the conditional distribution of $X(t+s)$ conditionally on $\mathcal F_s$ is such that $$ P[X(t+s)\in B\mid X(s)=x,\mathcal F_s]=\int_Bp(s,t+s,x,y)\mathrm dy, $$ for every state $x$ and every (measurable) set $B$. Since the conditional distribution of $X(t+s)$ is absolutely continuous, for every $y$, $$ P[X(t+s)=y\mid X(s)=x,\mathcal F_s]=0, $$ but it may well happen that $\{y\mid p(s,t+s,x,y)\ne0\}$ is the whole state space, even when said state space is uncountable.

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