Problem 1651

1. Define the problem and the Markov Chain:
- The particle performs a random walk on the integer points of the semi-axis $x \ge 0$.
- The transition probabilities are:
- Moves to the right (from $x$ to $x+1$) with probability $a$.
- Moves to the left (from $x$ to $x-1$) with probability $b$, but if $x = 0$, it stands still.
- Stands still with the remaining probability $1 - a - b$.

2. Steady-state probability distribution:
- Let $\pi(k)$ be the steady-state probability of being at position $k$.
- For $k \ge 1$, the balance equations are:
$$\pi(k) a = b \pi(k+1)$$
- For $k = 0$, the balance equation is:
$$\pi(0) = \pi(0) (1 - a) + b \pi(1)$$
- From the first balance equation, we can express $\pi(k+1)$ in terms of $\pi(k)$:
$$\pi(k+1) = \frac{a}{b} \pi(k)$$
- By iterating this relationship, we get:
$$\pi(k) = \left(\frac{a}{b}\right)^k \pi(0)$$
- To find $\pi(0)$, we use the normalization condition:
$$\sum_{k=0}^{\infty} \pi(k) = 1$$
$$\pi(0) \sum_{k=0}^{\infty} \left(\frac{a}{b}\right)^k = 1$$
- The series $\sum_{k=0}^{\infty} \left(\frac{a}{b}\right)^k$ converges if $\frac{a}{b} < 1$, which implies $a < b$:
$$\sum_{k=0}^{\infty} \left(\frac{a}{b}\right)^k = \frac{1}{1 - \frac{a}{b}} = \frac{b}{b - a}$$
- Therefore:
$$\pi(0) \cdot \frac{b}{b - a} = 1 \implies \pi(0) = \frac{b - a}{b}$$
- The steady-state distribution is:
$$\pi(k) = \left(\frac{a}{b}\right)^k \frac{b - a}{b}$$

3. **Expectation of $x$:**
- The expectation $E[x]$ is given by:
$$E[x] = \sum_{k=0}^{\infty} k \pi(k)$$
$$E[x] = \sum_{k=0}^{\infty} k \left(\frac{a}{b}\right)^k \frac{b - a}{b}$$
- Using the formula for the sum of an infinite series:
$$\sum_{k=0}^{\infty} k r^k = \frac{r}{(1 - r)^2} \quad \text{for} \quad |r| < 1$$
- Here, $r = \frac{a}{b}$:
$$E[x] = \frac{b - a}{b} \cdot \frac{\frac{a}{b}}{\left(1 - \frac{a}{b}\right)^2} = \frac{b - a}{b} \cdot \frac{\frac{a}{b}}{\left(\frac{b - a}{b}\right)^2} = \frac{a}{b - a}$$

4. **Expectation of $x^2$:**
- The expectation $E[x^2]$ is given by:
$$E[x^2] = \sum_{k=0}^{\infty} k^2 \pi(k)$$
$$E[x^2] = \sum_{k=0}^{\infty} k^2 \left(\frac{a}{b}\right)^k \frac{b - a}{b}$$
- Using the formula for the sum of an infinite series:
$$\sum_{k=0}^{\infty} k^2 r^k = \frac{r(1 + r)}{(1 - r)^3} \quad \text{for} \quad |r| < 1$$
- Here, $r = \frac{a}{b}$:
$$E[x^2] = \frac{b - a}{b} \cdot \frac{\frac{a}{b} \left(1 + \frac{a}{b}\right)}{\left(1 - \frac{a}{b}\right)^3} = \frac{b - a}{b} \cdot \frac{\frac{a}{b} \left(\frac{b + a}{b}\right)}{\left(\frac{b - a}{b}\right)^3}$$
$$E[x^2] = \frac{a(b + a)}{(b - a)^2}$$

The final answer is: