My question (basically) says:
"The weather condition is either sunny (S) or rainy (R). The weather condition on day 'n + 1' depends on the day 'n' and 'n-1' only"
I also have a table of probabilities:
$$\begin{matrix} \mathrm{\underline{Yesterday}} & \mathrm{\underline{Today}} & \mathrm{\underline{Tomorrow}} & \mathrm{\underline{Prob.}} \\ R & R & R & 0.6 \\ S & S & S & 0.8 \\ S & R & R & 0.5 \\ R & S & S & 0.75 \end{matrix}$$
I have to find the tranisiton matrix, $P_{i,j}$.
What my lecturer has done is this:
"Let $Y_n = (X_{n-1}, X_n)$, which is the weather over two days. This then gives us a Markov chain depending on $Y_n$ and not $Y_{n-1}$ as well. We can then define a state space as:
S = {1, 2, ,3 ,4} where 1 = (R, R), 2 = (R, S), 3= (S,R) and 4 = (S, S)
In lectures we worked out the value for $P_{1,1}$ by doing:
$$\begin{align*}P_{1,1} &= P(Y_{n + 1} = 1 | Y_n = 1) \tag{1} \\ &= P(X_n = R, X+{n+1} = R | X_{n - 1} = R, X_n= R) \\ &= \frac{ P(X_n = R, X_{n + 1} = R, X_{n -1} = R)}{P(X_{n-1} = R, X_n = R)} \\ &= P(X_{n + 1} = R | X_{n - 1} = R, X_n = R) \\ &= 0.6 \end{align*} $$
I understand all this bit, (except the bit in the fraction), because what they are saying is that "what is the probability it will rain tomorrow, given that it rains today and yesterday", which can just be read off the table.
So if I try and calculate the probability of $P_{1,2}$, this is what I get:
$$\begin{align*}P_{1,2} &= P(Y_{n + 1} = 2 | Y_n = 1) \tag{2} \\ &= P(X_ n = S, X_{n+1} = S | X_n= S , X_{n - 1} = R) \\ &= \mathrm{(Not\, sure\, what\, the\, whole\, division\, thing\, comes\, from\, or\, what\, it\, means)} \\&= P(X_{n+1} = S | X_{n - 1} = R, X_n = S) \\ &= 0.75 \end{align*} $$
But my lecturer says it should equal 0.4. How do they get to this?
EDIT: I noticed a mistake I made, second line of $P_{1,2}$ should read
$$ P(X_n = S, X_{n+1} = S| X_n = R, X_{n - 1} = R) $$
But I still don't get what to do from here.
Also, why is it not possible to get $P_{1,3}$?
