Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following from a book:

Assume that $$ P_x(\tau_C \circ \theta_{(k-1)N} > N|F_{(k-1)N}) = P_{X_{(k-1)N}}(\tau_C > N). $$ Integrating over $\{ \tau_C > (k-1)N\}$ using the definition of conditional probability we have $$ P_x(\tau_C > kN) = E_x\left(\mathbf{1}\{\tau_C \circ \theta_{(k-1)N} > N\} \cdot\mathbf{1}\{\tau_C > (k-1)N\}\right) $$

I'm a bit unsure how from that equality he gets the second equality we see. I can see that the LHS of the first equality multiplied by $\mathbf{1}\{\tau_C > (k-1)N\}$ and then taking the expectation wrt. x yields the RHS of the second equality, but how does the RHS of the first equality being "intergrated over" as claimed produce the LHS of the second equality?

Here, $\tau_C$ is the first hitting time of some set $C$.

Thanks.

share|improve this question
    
The second equality holds for every random process, be it Markov or not, since the events $\{ \tau_C > kN\}$ and $\{\tau_C \circ \theta_{(k-1)N} > N\}\cap\{\tau_C > (k-1)N\}$ coincide. –  Did Feb 28 '12 at 17:29
    
I can see that these events are equal when I write it out properly. But the author of the book seems to suggest a different method for calculating this.. –  lax Feb 28 '12 at 18:29
    
Hence: never take at face value what is written in a book... –  Did Feb 28 '12 at 18:31
    
Thanks for the help. Is there an easy way to see that this? I find having to write the first set in $\{\tau_C \circ \theta_{(k-1)N} > N\}\cap\{\tau_C > (k-1)N\}$ in terms of the state space and using the shift of the sequence not very intuitive. –  lax Feb 28 '12 at 19:33
    
@Lax: to ask anyone on this website use "@" interface like I did in the current message. Otherwise the person won't receive your comment in his/her inbox. Answering your last comment: $$ \{\omega:\tau_C(\omega)>n\} = \bigcap\limits_{i=0}^{n}\{\omega:X_i(\omega)\in C^c\} $$ and $$ \{\omega:(\tau_C\circ\theta_{j})(\omega)>n\} = \bigcap\limits_{i=j}^{n+j}\{\omega:X_i(\omega)\in C^c\} $$ –  Ilya Feb 29 '12 at 12:56
show 2 more comments

1 Answer

The second equality holds for every random process, be it Markov or not, since, for every $k\geqslant1$, the events $\{ \tau_C \gt kN\}$ and $\{\tau_C \circ \theta_{(k-1)N} \gt N\}\cap\{\tau_C \gt (k-1)N\}$ coincide.

Proof:

Introducing the canonical process $(X_i)_{i\geqslant0}$, $\{ \tau_C \gt n\}=\{X_i\notin C\ \text{for every}\ 1\leqslant i\leqslant n\}$ for every nonnegative $n$ and $\{\tau_C \circ \theta_m \gt n\}=\{X_i\notin C\ \text{for every}\ m+1\leqslant i\leqslant m+n\}$ for every nonnegative $n$ and $m$. Use the first identity for $n=kN$ and $n=(k-1)N$ and the second identity for $n=N$ and $m=(k-1)N$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.