= Solution
Let $Z_m$ be the discount level during policy year $m$, with $Z_0=1$. Under the usual assumption of independent successive annual accidents and repair costs with the same laws each year, the submitted-claim decision depends only on $Z_m$ and that year's cost. Consequently the conditional distribution of $Z_{m+1}$ depends on the past only through $Z_m$: this is the <Markov property>. The <no claims discount system> is therefore a time-homogeneous <Discrete-time Markov chain>. Stationarity of accident probabilities alone, without this independence assumption, would not by itself establish the Markov property.
At level one a claim keeps the policyholder at one, while no submitted claim moves them to two. At level two, a submitted claim moves them to one and no claim to three. At level three, a claim moves them to two and no claim keeps them at three. Thus $p_i$ in the specified <transition matrix> is the annual probability of a submitted claim at current level $i$, not merely of an accident. A small accident paid privately counts as no claim for these transitions.
Let $b>0$ denote the undiscounted annual premium. It is not specified numerically in the source; costs can instead be measured in units of that premium, giving $b=1$. The actual premium payments at the three levels are $b$, $b(1-\alpha)$, and $b(1-\beta)$. Compare the two future paths, assuming no further accidents, to obtain the <two-year premium-loss claim threshold>. From level one, a current claim leads to future levels $(1,2)$, whereas no claim leads to $(2,3)$, giving extra premium
$$
h_1=b+b(1-\alpha)-b(1-\alpha)-b(1-\beta)=b\beta.
$$
From level two the two paths are $(1,2)$ and $(3,3)$, and from level three they are $(2,3)$ and $(3,3)$. Hence
$$
\boxed{h_1=b\beta,\qquad h_2=b(2\beta-\alpha),\qquad h_3=b(\beta-\alpha).}
$$
These are all strictly positive. The PDF's discount fractions are $\alpha$ and $\beta$; the converted TeX has lost these factors in the percentages.
For a repair cost $X$ with the specified <lognormal distribution>, $\log X\sim N(\mu,\sigma^2)$. Put
$$
z_i=\frac{\log h_i-\mu}{\sigma},\qquad\overline\Phi(z)=1-\Phi(z),
$$
where $\Phi$ is the <standard normal distribution function>. Conditional on an accident, the submitted-claim probability is $\mathbb P(X>h_i)=\overline\Phi(z_i)$. Multiplying by the accident probability gives
$$
\boxed{\begin{aligned}
p_1&=p\left[1-\Phi\left(\frac{\log(b\beta)-\mu}{\sigma}\right)\right],\\
p_2&=p\left[1-\Phi\left(\frac{\log(b(2\beta-\alpha))-\mu}{\sigma}\right)\right],\\
p_3&=p\left[1-\Phi\left(\frac{\log(b(\beta-\alpha))-\mu}{\sigma}\right)\right].
\end{aligned}}
$$
Set $b=1$ for the premium-unit convention. Without a monetary premium scale, discount percentages alone cannot determine a numerical threshold for a monetary repair cost.
This nearest-neighbour <Markov chain> is a finite <birth-death chain>. Its <stationary distribution> $\pi$ satisfies the endpoint balance equations
$$
\pi_1(1-p_1)=\pi_2p_2,\qquad
\pi_2(1-p_2)=\pi_3p_3.
$$
These <detailed balance> equations also imply the middle-state balance equation. The <three-level no claims discount equilibrium> is therefore
$$
\boxed{(\pi_1,\pi_2,\pi_3)=\frac1D\bigl(p_2p_3,\ (1-p_1)p_3,\ (1-p_1)(1-p_2)\bigr),}
$$
where
$$
D=p_2p_3+(1-p_1)p_3+(1-p_1)(1-p_2).
$$
Since each $p_i$ lies strictly between zero and one, this is an <irreducible Markov chain>. Positive holding probabilities at the endpoints also make it an <aperiodic Markov chain>. Its unique <stationary distribution> is the limiting distribution even from the initial level one, and gives the steady-state proportions for a homogeneous population of such policyholders.
A submitted claim at level one conditions the repair cost on $X>h_1$. The accident probability cancels in this <conditional expectation>. For $z_1=(\log h_1-\mu)/\sigma$, the <truncated lognormal moment> follows by completing the square:
$$
\begin{aligned}
\mathbb E[X\,1_{\{X>h_1\}}]
&=\int_{z_1}^{\infty}e^{\mu+\sigma z}\frac{e^{-z^2/2}}{\sqrt{2\pi}}\,dz\\
&=e^{\mu+\sigma^2/2}\int_{z_1}^{\infty}\frac{e^{-(z-\sigma)^2/2}}{\sqrt{2\pi}}\,dz\\
&=e^{\mu+\sigma^2/2}\overline\Phi(z_1-\sigma).
\end{aligned}
$$
Dividing by $\mathbb P(X>h_1)=\overline\Phi(z_1)$ yields the expected size of an actual submitted claim:
$$
\boxed{\mathbb E[X\mid\text{claim at level one}]
=e^{\mu+\sigma^2/2}\frac{1-\Phi((\log(b\beta)-\mu-\sigma^2)/\sigma)}{1-\Phi((\log(b\beta)-\mu)/\sigma)}.}
$$
The unconditional expected amount paid per year from level one would instead multiply its tail first moment by $p$; it is not the conditional claim size requested here.
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