probability question

[anja...@googlemail.com]

Hello everyone,

I am trying to figure out some overall possibilities of some states
for a certain configuration and need a bit of help with probability
theory.

So, assume that there is an object that can take 3 states {0, 1, 2}
and the object can only change the state by doing one jump at a time.
So, from 0 to 1 and then from 1 to 2.

Initially the object is in state 0 and an event happens and the object
can either stay in state 0 or move to state 1.

So, we have P_1(0) and P_1(1)

Now, consider the following scenario: object stayed in state (0). We
have another event:

Again, we have a new set of probabilities
P_2(0) and P_2(1)

Now, after these two events, what is the overall probability that the
object stays in state 0 or state 1?

Would appreciate any help you can give me.

Thanks,
Anja

[Ray Koopman]

On Feb 3, 2:10 pm, "anja.e...@googlemail.com"

Draw the tree: time is horizontal, state is vertical. Each pair of
steps from a state at one time to the possible states at the next
time corresponds to a pair of conditional transition probabilities
whose sum is 1. The marginal probabiity of a given state at a given
time is the sum, over all paths, of the product of the conditional
probabilities of each step on that path.

0 - 0 - 0
\ \
1 - 1
\
2

(view in a fixed-pitch font such as Courier)

[anja...@googlemail.com]

Thanks for the reply.

to observe state (1) at end is calculated as follows:

> 0 - 0 - 0
> \ \
> 1 - 1
> \
> 2
>

Probability of going from state 0 to state 1 and staying at state (1):
P(0, 1) * P(1, 1). Then we also have probability of staying at
state(0) and then jumping to state(1): So P(0, 0) * P(0, 1).

So, overall probability of seeing state(1) is: P(0, 1) * P(1, 1) +
P(0, 0) * P(0, 1).

I guess this needs to be normalized by the sum of individual
probabilities:

(P(0, 1) * P(1, 1) + P(0, 0) * P(0, 1)) /(P(0, 1) + P(1, 1) + P(0, 0)
+ P(0, 1))...Should be 2 in this case but anyway..

Have I understood this correctly?

Thanks,

Anja

[Ray Koopman]

On Feb 4, 3:22 am, "anja.e...@googlemail.com"
<anja.e...@googlemail.com> wrote:

You shouldn't normalize. The numerator by iteelf is the probability
of being in state 1 at time 2.

A possible simplification: I interpreted your question as allowing
the transition probabilities to be different at different times and
states. If the probabiity of going from state x at time n to state
x+1 at time n+1 is a constant, say p, regardless of x and n, then
the probability of staying in state x at time n+1 is 1-p, and the
probability of going from state 0 at time 0 to state x at time n
is just the binomial probability C(n,x) * p^n * (1-p)^(n-x), where
C(n,x) is the binomial coefficient "n_choose_x" = n!/(x!*(n-x)!).