A triathlon consists of $3$ disciplines: swimming, cycling and running. A triathlete does a training session every day. However he doesn’t want to pay for professional coaching advice so instead his strategy for training is to choose the next day’s training discipline uniformly likely from the two disciplines that he didn’t do on the current day.
1)Supposing that on day zero the training session undertaken is swimming; what is the probability that the first day’s session is swimming?
2)Supposing that on day zero the training session undertaken is swimming; what is the probability that the $n$th day’s session is swimming?
So I constructed a stochastic matrix $P$ in the following way:
$$P=\begin{pmatrix} 0 & 1\over 2 & 1\over 2 \\ 1\over 2 & 0 & 1\over 2 \\ 1\over 2 & 1\over 2 & 0 \\ \end{pmatrix}$$ since going from swimming or any other sport to a different sport has probability $1\over 2$.
So the answer to the first question is $0$ since if we start at swimming then the next day is supposed to be a different sport day.
and then for the second question I tried finding the $p^n_{s_0s_n}$ -th value- where $p_{i_0,j_n}$ means probability of starting at day $i$ sat day $0$ and finishing at day $j$ on day $n$ , and $p^n$ is the value of the specific entry on with the matrix to the $n$-th power. Also $s$ is denoted as swimming day.
But in pursuit of finding this value I got stuck. Any help would be appreciated
The given matrix $P$ is diagonalizable, and we can write $$ P = \underbrace{ \begin{bmatrix} 1&1&0\\1&0&1\\1&-1&-1 \end{bmatrix}} _T \underbrace{ \begin{bmatrix} 1\\&-1/2\\&&-1/2 \end{bmatrix}} _D \underbrace{ \frac 13 \begin{bmatrix} 1&1&1\\2&-1&-1\\-1&2&-1 \end{bmatrix}} _{S=T^{-1}}\ . $$ Then $$ P^n=\underbrace{(TDS)(TDS)\dots(TDS)}_{n\text{ times}} =TD^nS\ , $$ since touching $ST$ matrices are collapsing to the identity matrix.
We need to isolate the $(1,1)$ entry in the product, the "swimming-swimming" entry, so we compute: $$ \begin{aligned} &[1\ 0\ 0]\cdot P^n\cdot\begin{bmatrix}1\\0\\0\end{bmatrix} \\ &\qquad= [1\ 0\ 0]T\cdot D^n\cdot S\begin{bmatrix}1\\0\\0\end{bmatrix} \\ &\qquad= [1\ 1\ 0] \cdot \begin{bmatrix} 1^n\\&(-1/2)^n\\&&(-1/2)^n \end{bmatrix} \cdot \frac 13\begin{bmatrix}1\\2\\-1\end{bmatrix} \\ &\qquad= [1\ (-1/2)^n\ 0] \cdot \frac 13\begin{bmatrix}1\\2\\-1\end{bmatrix} \\ &\qquad= \frac 13\left(1+2\left(-\frac 12\right)^n\right)\ . \end{aligned} $$