can any explain the proof in study material section 5.3 Random Walk on s={...-2,-1,0,1,2....} The Markov property Holds: P[Xn=J|X1=i1,......Xm-1=im-1,Xm=i] P[Xm+Ym+1+...Yn=j|X1=i1,....Xm-1=im-1,Xm=i] P[Ym+1+....Yn=j-i]= P[Xn=j|Xm=i] please explain the second and third line. regards Suresh sharma 9051838188
ok please explain i am writing down the whole thing: Simple Random Walk on S ={....-2,-1,0,1,2.....} This is defined as Xn = Y1+Y2+.......+Yn where the random variable Yj (the steps of the walk ) are mutually independent with common probability distribution: P[Yj=1]=p, P[Yj=-1] =1-p The Markov Property Holds : P[Xn=J|X1=i1,......Xm-1=im-1,Xm=i] P[Xm+Ym+1+...Yn=j|X1=i1,....Xm-1=im-1,Xm=i] P[Ym+1+....Yn=j-i]= P[Xn=j|Xm=i] please explain the second and third line.
Hi Suresh, The second line is replacing Xn with Xm plus the steps Y between times m and n. X is a process that adds up the Ys The third line has subtracred Xm from the LHS and j from the RHS. This is OK because Xm=j One thing that is massively helpful for everyone is if you can give full course references when you post queries. It's great when students can help each other and also really useful for the tutors if we feel that our input is needed. I notice that you've also posted your algorithm for others to look at. This is great, well done! Good luck! John
