1. **Stating the problem:** We have a generating function $$G_X(s) = \sum_{k=0}^{\infty} P(X=k)s^{k}$$ with $$G(s) = \frac{1}{3} + \frac{1}{3}s + \frac{1}{3}s^{2}$$. We want to find the expectation $$E[X]$$ and variance $$Var(X)$$ using the formulas: $$E[X] = G'_X(1)$$ $$Var(X) = G''_X(1) + G'_X(1) - (G'_X(1))^{2}$$ 2. **Find the first derivative $$G'(s)$$:** $$G(s) = \frac{1}{3} + \frac{1}{3}s + \frac{1}{3}s^{2}$$ Taking derivative term-by-term, $$G'(s) = 0 + \frac{1}{3} + \frac{2}{3}s = \frac{1}{3} + \frac{2}{3}s$$ 3. **Calculate $$G'(1)$$:** $$G'(1) = \frac{1}{3} + \frac{2}{3} \times 1 = \frac{1}{3} + \frac{2}{3} = 1$$ So, $$E[X] = 1$$. 4. **Find the second derivative $$G''(s)$$:** $$G''(s) = 0 + 0 + \frac{2}{3} = \frac{2}{3}$$ 5. **Calculate $$G''(1)$$:** $$G''(1) = \frac{2}{3}$$ 6. **Calculate variance $$Var(X)$$:** By the formula, $$Var(X) = G''(1) + G'(1) - (G'(1))^{2} = \frac{2}{3} + 1 - 1^{2} = \frac{2}{3} + 1 -1 = \frac{2}{3}$$ So, $$Var(X) = \frac{2}{3}$$. --- 7. **Regarding matrix $$P$$:** Given $$P = \begin{pmatrix} 0.5 & 0.3 & 0.2 \\ 0.1 & 0.6 & 0.3 \\ 0.4 & 0.0 & 0.6 \end{pmatrix}$$ Calculate $$P^{(3)} = P^{(2)} P$$ as: - First calculate $$P^{(2)} = P \times P$$. - Then $$P^{(3)} = P^{(2)} \times P$$. 8. **Stationary distribution $$\pi = (\pi_1, \pi_2, \pi_3)$$:** Solve $$\pi P = \pi$$ with $$\pi_1 + \pi_2 + \pi_3 = 1$$. This involves solving the system: $$\pi_1 = 0.5\pi_1 + 0.1\pi_2 + 0.4\pi_3$$ $$\pi_2 = 0.3\pi_1 + 0.6\pi_2 + 0\pi_3$$ $$\pi_3 = 0.2\pi_1 + 0.3\pi_2 + 0.6\pi_3$$ Together with $$\pi_1 + \pi_2 + \pi_3 = 1$$. 9. **Fundamental matrix $$N = (I - Q)^{-1}$$:** Where $$Q$$ is a submatrix of $$P$$ representing transient states if given in a Markov chain context. --- ### Final answers: **Expectation:** $$E[X] = 1$$ **Variance:** $$Var(X) = \frac{2}{3}$$