I'm trying to revise for a probability exam in a few weeks and still getting really confused :( so I'd be really grateful for a push in the right direction:
Consider a random walker on f0; 1; 2g who moves as follows:
- if at 0, probability 1/2 of staying at 0 and 1/2 of moving to 1;
- if at 1, probability 1/2 of moving to 0 and 1/2 of moving to 2;
- if at 2, probability 1/2 of moving to 1 and 1/2 of staying at 2.
Let the location of the random walker after k steps be Xk. Let X0 = 0.
(a) Let N be the number of steps until the first visit to 1. Consider
pk = P(N = k) and qk = P(Xk = 1).
(i) Explain why pk [tex]\geq[/tex] qk for all k > 0.
(ii) Calculate pk and qk for all k [tex]\geq[/tex] 0. (Hint: show qk+1 = 1/2(1-qk)
The Attempt at a Solution
i) It is asking me to explain why the probability that the walker takes k steps until the 1st visit to is less than or equal to the probability that the kth location of the walker is 1.
If the walker takesk steps until the 1st visit to 1 then his kth location is 1?
ii) I think I need to condition on something, I think the 1st step?
P(N=k | X0=0) = P(N=k| X0 = 0 and X1=0)P(X1=0) + P(N=k| X0 = 0 and X1=1)P(X1=1)
gives me pk = 1/2 pk + 1/2 pk+1
P(Xk=1| X0=0) = P(Xk=1 | X0=0 and X1=0)P(X1=0) + P(Xk=1 | X0=0 and X1=1)P(X1=1)
gives qk = 1/2 qk+1 + 1/2 I think
This is not what I had to show?! Once I have the recurrence relations I think I can solve them.