= Three-level discount equilibrium with geometric annual counts
{title2=$\pi=\frac{(q^2(1+p),pq,p^2)}{1-p^2q},\quad p+q=1$}
Suppose annual <claim counts> are independent with <geometric distribution> mass $pq^j$, $j\ge0$. In a <no claims discount system> with discounts $0,\alpha,\beta$, a claim returns either of the first two levels to zero; at the top, one claim sends the level to $\alpha$ and two or more send it to zero. The <transition matrix> is $\bigl(\begin{smallmatrix}q&p&0\\q&0&p\\q^2&pq&p\end{smallmatrix}\bigr)$. Solving $\pi P=\pi$ gives the displayed <stationary distribution>. With full premium $c$, the stationary <expected value> of the premium is $c[1-(\alpha pq+\beta p^2)/(1-p^2q)]$. Independent annual sampling is essential; a persistent random intensity does not justify this marginal transition matrix.
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