Hi all, I have a question regarding the difference between $P(X_3=0)$ and $P(X_3=0 |T>3)$ here. Are they the same or not?
The reason I am asking this is because the transition matrix indicates that state $2$ is an absorbing state, so I thought it is enough just to calculate $P_{00}^3$ i.e. the transition matrix to the power of $3$, and then take the $(0,0)$ component to be the probability that the chain starts at 0 and ends at $0$ right after $3$ transitions without ever entering state 2?
But how come we need to divide by $P(X_3=0) + P(X_3=1)$? I thought calculating $P_{00}^3$ alone already gives the answer?
The probability $\mathbb P(X_3=0)$ is not the same as the conditional probability $\mathbb P(X_3=0\mid T>3)$. Indeed, $$ \mathbb P(X_3 = 0\mid X_0=0) = \frac{457}{1000} \approx 0.457 $$ while $$ \mathbb P(X_3 = 0\mid X_0=0, T>3) = \frac{15215}{23328}\approx 0.652221. $$ Note that we can simplify the computation of $\mathbb P(X_3=0\mid X_0=0,T>3)$ a bit by considering the transition matrix without the row/column corresponding to state $3$, rescaled so that the rows sum to one: $$ Q = \left( \begin{array}{cc} \frac{7}{9} & \frac{2}{9} \\ \frac{3}{8} & \frac{5}{8} \\ \end{array} \right). $$ Then $$ Q^3 = \left( \begin{array}{cc} \frac{15215}{23328} & \frac{8113}{23328} \\ \frac{8113}{13824} & \frac{5711}{13824} \\ \end{array} \right), $$ and the desired probability is simply the $(1,1)$-entry of this matrix.