Simple Random Walk?

Question

Yn is the position at time n of a particle executing a simple random walk with partially reflecting barriers at 0 and a so that P{Yn = y +1|Yn-1 = y}= p for y = 0, 1, 2, ..,.a-1, and
P(Yn = y-1| Yn-1 = y) = q = 1- p for y 1, 2, .., .a-1, a..
Also P( Yn= a | Yn-1= a) = p and P( Yn= 0 | Yn-1= 0) = q

If П = [П0, П1, П2, . . . . Пa ] is defined to be the stationary distribution for this process

П_i=pП_(i-1)+ qП_(i+1) for i=1,2,….a-1

Show that
П_1= ((p/q)*)П_0 ,П_2=((p/q)^2)*П_0 and in general П_a=((p/q)^a)*П_0

Answer 1

OK let's get some definitions in order first.
Πj = ∑{i} Πi * Pij
Pij = P(X1 = j | X0 = i)

Generally Πi = p*Πi-1 + q*Πi+1
But for i = 0, Π0 = q*Π0 + qΠ1
So Π1 = (1-q)/q * Π0 = p/q * Π0

Now Π1 = p/q * Π0 = p*Π0 + q*Π2
qΠ2 = Π0 *(p/q - pq/q) = p^2/q * Π0
Π2 = (p/q)^2 * Π0

The rest can be proved by induction.
If Πn = (p/q)^n * Π0 and Π(n+1) = (p/q)^(n+1) * Π0
then Π(n+1) = p*Πn + q*Π(n+2)
q*Π(n+2) = Π0 * [(p/q)^(n+1) - p*(p/q)^n]
= Π0 * [(p/q)^(n+1) * (1 - q) ]
= Π0 * [(p/q)^(n+1) * p ]
Π(n+2) = (p/q)^(n+2) * Π0

Now we need to verify the endpoint which behaves differently.
Πa = Πa * p + Π(a-1)*p
Πa (1-p) = Πa* q = (p/q)^(a-1) *Π0 * p
Πa = (p/q)^a *Π0

So this relation holds for all n = {1,2,...,a}

Answer 2

Do you want sample of walking stick

Answer 3

I started to feel like that n was growing bigger n bigger sure it started to walk :) lol sorry i don't know the answer but i cudn't just resist not answering so had to tell u that i went nuts reading thru ur question wat is it? If u'd like to tell me i'd love to know :) calculus? or wat ?