Problem Analysis #5

[THOUGHT] All parts solved analytically. For p=1/2, q=1/3:

s = (2/3)(1/2) + (1/3)(2/3) = 5/9

t = 1 - s = 4/9

β = 1/t = 9/4

σ² = s / t² = 45/16

α = [(2/3)(1/2)] / s = 3/5

λ = γ₂ = s = 5/9

Then α β + σ²/λ = 27/20 + 81/16 = 513/80

13 * 320 * (513/80) + 7 = 26683

Verify with Python. [/THOUGHT] [PYTHON] p = 1/2 q = 1/3 pi0 = 2/3 pi1 = 1/3 r0 = 1 - p r1 = 1 - q s = pi0 * r0 + pi1 * r1 t = pi0 * p + pi1 * q beta = 1 / t sigma2 = s / t**2 alpha = (pi0 * r0) / s lam = s inner = alpha * beta + sigma2 / lam K = 13 * 320 * inner + 7 print(K) [/PYTHON]

[THOUGHT] The Python output is 26683.000000000004, which is exactly 26683 as an integer. [/THOUGHT] $\boxed{26683}$

Explanation

The reasoning systematically addresses each part of the problem to compute $\mathcal{K}$, relying on analytical derivations verified by Python.

For part I, the matrices $P_k$ are derived from transition probabilities. Specifically, $P_1$ is the one-step transition matrix:

$P_2 = P_1^2$, and $P_3 = P_1^3$. For part I(3), $P_k = \gamma_k P_1$ where $\gamma_k$ is the common scaling factor from the matrix powers, determined as $\gamma_k = [(1-p) \cdot \frac{2}{3} + (1-q) \cdot \frac{1}{3}]^{k-1}$, so $\gamma_2 = s = (1-p) \cdot \frac{2}{3} + (1-q) \cdot \frac{1}{3}$.

For part II, given no termination before $m$, the probabilities are proportional to survival paths: $\Pr(X_m=0 \mid \text{no term. before } m) = \frac{(1-p)^{m-1} \cdot \frac{2}{3}}{\sum}$, but normalized by the survival probability up to $m-1$.

For part III, $\ell$ follows a modified geometric process. Let $s = \frac{2}{3}(1-p) + \frac{1}{3}(1-q)$ (probability of continuing after a step) and $t = 1 - s$ (probability of terminating at any step, averaged). Then $\mathbb{E}[\ell] = \beta = \frac{1}{t}$ and $\mathrm{Var}(\ell) = \sigma^2 = \frac{s}{t^2}$, using the given sums for geometric expectations.

For part IV, $\Pr(X_n=0 \mid X_{n+k}=1)$ uses backward conditioning via the stationary or conditional paths, yielding $\alpha = \frac{\frac{2}{3}(1-p)}{s}$ for large separations, as derived from the detailed balance.

With $p = \frac{1}{2}$, $q = \frac{1}{3}$:

• $s = \frac{2}{3} \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{2}{3} = \frac{5}{9}$,
• $t = 1 - s = \frac{4}{9}$,
• $\beta = \frac{9}{4}$,
• $\sigma^2 = \frac{5/9}{(4/9)^2} = \frac{45}{16}$,
• $\alpha = \frac{\frac{2}{3} \cdot \frac{1}{2}}{5/9} = \frac{3}{5}$,
• $\lambda = \gamma_2 = s = \frac{5}{9}$.

Then $\alpha \beta + \frac{\sigma^2}{\lambda} = \frac{3}{5} \cdot \frac{9}{4} + \frac{45/16}{5/9} = \frac{27}{20} + \frac{81}{16} = \frac{513}{80}$.

Finally, $\mathcal{K} = 13 \cdot 320 \cdot \frac{513}{80} + 7 = 26683$, confirmed exactly by the Python computation.