1

For DTMC with $S=\{1,2\}$ and transition probabilities

$$P = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$

How do we see that $(P_{00})^{(n)} = 1$ if $n$ is even or $0$ if $n$ is odd ??

(Where $P_{ij}$ is a one-step transition probability).

Note as John proved below $(P_{00})^{(n)} = (P_{11})^{(n)}$

I'd like to know how $(P_{00})^{(n)}$ can equal 1 or 0 ??

is that because $(P_{00})$ refers us to look at the top left value in the matrix? (where $P_{11}$ would refer us to look at the 2nd row and 2nd column of the matrix)

The reason I am confused is I understand $(P_{00})^{(n)}= P(X_{n} =0| X_{0} = 0)$ and $(P_{11})^{(n)} = P(X_{n} =1| X_{0}=1)$ and don't see how these two say the same thing.

Max
  • 25
Max
  • 11

1 Answers1

1

Since $$P^2 = \begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix},$$ it is clear that, $$P^{2n} = (P^2)^n = \begin{bmatrix}1&0\\0&1\end{bmatrix}^n = \begin{bmatrix}1&0\\0&1\end{bmatrix}$$ and hence $$ P^{2n+1} = P^{2n}P=\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}= \begin{bmatrix}0&1\\1&0\end{bmatrix}.$$

Math1000
  • 36,983