00:01
So in this question we are asked to show this proof, which says that for a sample space or state space of 1 through s, show that the m plus n step transition probability of going from i to j is equal to the summation for all k of the m step transition probability of going from i to k times the n -step transition probability of going from k to j.
00:40
And so they give us a hint in the question to start with this probability here.
00:49
So both of these represent the m plus n -step transition probability of going from i to j.
01:03
So we can also write this as the probability.
01:35
So this is the probability of going from state i to state j at n plus m transitions or at the n plus mth state but also having gone through state k at the nth state state so it's a probability of going to state j after n plus m transitions and also going to state k after n transitions given that we start at state i so we only want to show that we go to some intermediate state k and that can be any state at all in the sample space.
02:20
And so we want the probability of the union of these probabilities for all k.
02:29
Or rather, that's the union of these paths for all k.
02:33
So this is basically the probability of going through any path from i to k and then to j.
02:46
So this is k as an element of the state space.
02:51
So now since each of these paths is a unique path, no path is a subset of another path.
03:03
If you take one of the unique paths, you cannot have taken the other paths.
03:09
Which means that the probability of the union of all these paths can be written as the summation of the probabilities of the individual paths.
03:21
So this is the summation of k for all k is an element of s, probability of a right of a right.
03:32
Arriving at state j after n plus m transitions, and also arriving at state k after n transitions.
03:44
Conditional on having started at state i.
03:51
And now using the chain rule for probability, we can rewrite this as follows...