I am answering a practice question, and in the first part I have calculated the 2nd state probabilities of the following transition matrix:
I caluclated from this P(X2=j), j=1,2,3,4,5.
I am now being asked: P(X2=2,X4=5), which I know is equivalent to, P(X4=5 | X2=2)P(X2=2)
However this is where I run into trouble, my solution tells me I can write:
P(X4=5 | X2=2)P(X2=2) = P(X2=5 | X0=2)P(X0=2)
I know the Chapman Kolmogorov proof, but I don't understand how to apply it here? Or do I need to use another theorem.
Thanks for any help in advance.
We have \begin{align} \mathbb P(X_2=2,X_4=5) &= \mathbb P(X_4=5\mid X_2=2)\mathbb P(X_2=2)\\ &= P^2_{2,5} + \mathbb P(X_2=2\mid X_0=1)\mathbb P(X_0=1) + \mathbb P(X_2=2\mid X_0=5)\mathbb P(X_0=5)\\ &= P_{2,5}^2 + P_{1,2}^2\cdot\frac 12 + P^2_{2,5}\cdot\frac12. \end{align} Computing $P^2$, we have
$$ P^2 = \left( \begin{array}{ccccc} \frac{29}{100} & \frac{3}{25} & \frac{7}{50} & \frac{3}{10} & \frac{3}{20} \\ 0 & \frac{13}{25} & 0 & \frac{11}{25} & \frac{1}{25} \\ \frac{13}{50} & \frac{4}{25} & \frac{4}{25} & \frac{3}{10} & \frac{3}{25} \\ \frac{3}{50} & \frac{11}{25} & \frac{3}{100} & \frac{21}{50} & \frac{1}{20} \\ \frac{3}{25} & \frac{1}{25} & \frac{3}{25} & \frac{7}{20} & \frac{37}{100} \\ \end{array} \right) $$ and thus $$ \mathbb P(X_2=2,X_4=5) = \frac1{25} + \frac3{50} + \frac1{50} =\frac3{25}. $$