-1
$\begingroup$

Suppose that $X_n$ is a Markov Chain.Then for $m,n \in N$ such that $m<n$
$Pr[X_n=j_n|X_m=j_m,X_{m-1}=j_{m-1},...=X_0=j_0]=Pr[X_n=j_n|X_m=j_m]$

When proving for n=3,m=1 case we have to show that $Pr[X_3=j_3|X_1=j_1,X_0=j_0]=Pr[X_3=j_3|X_1=j_1] $

I don't understand the following step.

$Pr[X_3=j_3|X_1=j_1 and X_0=j_0]= \sum_{j_2}Pr[(X_3=j_3)\cap (X_2=j_2)|(X_1=j_1)\cap(X_0=j_0)]$
I don't understand how this $j_2$ terms are brought in and why?

Improve this question
$\endgroup$
1
  • $\begingroup$ Is this self-study? You should add the tag then. $\endgroup$
    – Xi'an
    Apr 26, 2015 at 10:43

1 Answer 1

Reset to default
0
$\begingroup$

you'd use law of total probability.

$Pr(A)=\sum\limits_nPr(A\cap B_n)$

$A=Pr[X_3=j_3|X_1=j_1 and X_0=j_0],~B=(X_2=j_2),~n=j_2$

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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