I established the reccurence equation for a Markov Chain but are not able to finde the base cases.
We are interested in whether the sum of $t$ throws of a fair die is divisible by $k$ for some $k \geq 6$.
We then model this as a Markov Chain with states $S=\{0, \dotsc, k-1\}$, where, if $X_i$ is the thrown number in throw $i$, then $Y_t:=\sum_{i=1}^t X_i \mod k$ is the state of the Markov chain that tells us whether the sum is divisible by $k$.
Let $I_i:=\{i +l \mod k \mid l \in \{1, \dotsc, 6\}\}$. We have the transition probabilities $p_{ij} = \frac{1}{6}$ if $j \in I_i$ and 0 otherwise, where $p_{ij}$ is the probability to go from state $i$ to state $j$ with one die throw.
Now I would like to find a stationary Distribution. In order to do that I want to find the hitting times $h_{ii}$ defined as the expected number of steps to get from $i$ to $i$ again for every $i \in S$. (and $h_{ji}$ to get from $j$ to $i$)
For that, I have established the following reccurence equation:
$$h_{ii} = 1 + \frac{1}{6} \sum_{j \in I_i} h_{ji}.$$
But I have no idea how to get here to the base case. Can you help me? Thank you for any hint.
My idea was to maybe argue about symmetry, but I am not sure.
The stationary distribution of a Markov chain is found by solving the matrix equation
$$v(T-I)=0$$ Where $T$ is the transition matrix and $I$ is the identity matrix with 1 on the diagonal and 0 elsewhere. The equation can be solved by Gaussian elimination. You will get an infinite family of solutions, and you want to normalize it so the sum of the components of the vector is 1.
To calculate the mean time to get from $i$ to $i$ again, you compute first the mean time to get from any state to $i$. You do this by solving
$$k_i=0$$ $$k_j=1+\frac{1}{6}\sum{k_p}$$ Where $j$ is a non stopping state.
Then the probability of getting from $i$ to $i$ again is $1+ \frac{1}{6}\sum{k_p}$