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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).

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.

further notice that:

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 $$ P^{2n+1} = P^{2n}P=\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}= \begin{bmatrix}0&1\\1&0\end{bmatrix}.$$

Max
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  • The matrix $P$ should be indexed by $S$. If $S={1,2}$, there is no such thing as $P_{00}$ or $(P^n){00}$. The question probably asks to show that $(P^n){11}$ is $0$ when $n$ is odd and $1$ when $n$ is even (and this is obvious by the computations in the post, right?) – Did Apr 11 '15 at 18:04

1 Answers1

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A matrix $A$ is called symmetric if $A^T = A$, where $A^T$ denotes the transpose of $A$. For a 2 by 2 symmetric matrix $A$, we have $(A_{10}) = (A_{01})$

$P$ is symmetric.

Therefore $P^n$ is also symmetric because $(P^n)^T = (P \cdots P)^T = P^T \cdots P^T = (P^T)^n = P^n$.

So we have $(P_{10})^{(n)} = (P_{01})^{(n)}$.

But $P^n$ must also be a transition matrix because $P$ is a transition matrix. So each column must sum to 1.

Therefore $(P_{00})^{(n)}+(P_{10})^{(n)} = 1$ and $(P_{01})^{(n)}+(P_{11})^{(n)}=1$.

Using that $(P_{10})^{(n)} = (P_{01})^{(n)}$, we get $(P_{00})^{(n)}+(P_{10})^{(n)} = 1$ and $(P_{10})^{(n)}+(P_{11})^{(n)}=1$.

Therefore $(P_{00})^{(n)} = 1 - (P_{10})^{(n)}$ and $(P_{11})^{(n)} = 1 - (P_{10})^{(n)}$.

Hence $(P_{00})^{(n)} = (P_{11})^{(n)}$.

For an arbitrary 2 by 2 transition matrix $P$, $(P_{00})^{(n)}$ and $(P_{11})^{(n)}$ will not be the same. But the symmetric nature of your particular $P$ makes them the same here.

LucasSilva
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  • This does not seem to answer the OP's request, which is to understand how $P$ indexed by $S={1,2}$ may yield an entry $P_{00}$ (it may not). – Did Apr 11 '15 at 18:05