Problem Analysis #5

53.52s

13,647 toks

Problem Statement

Consider the following procedure that generates a sequence of random variables that take the value $0$ or $1$ . For an integer $n \ge 1$ , we denote the $n$ -th random variable of a sequence generated by the procedure as $X_n$ .

$X_1$ becomes $X_1 = 0$ with probability $\dfrac{2}{3}$ and $X_1 = 1$ with probability $\dfrac{1}{3}$ .
For integers $n = 1,2,\ldots$ in order, the following is repeated until the procedure terminates:
The procedure terminates with probability $p$ ( $0 < p < 1$ ) if $X_n = 0$ , and with probability $q$ ( $0 < q < 1$ ) if $X_n = 1$ . Here $p$ and $q$ are fixed constants.
If the procedure does not terminate at step $n$ , then $X_{n+1}$ becomes $0$ with probability $\dfrac{2}{3}$ and $1$ with probability $\dfrac{1}{3}$ .

When the procedure terminates at $n = \ell$ , a sequence of length $\ell$ , composed of random variables $(X_1,\ldots, X_\ell)$ , is generated, and no further random variables are generated.\subsection*{I.}For an integer $k \ge 1$ , consider the matrix $P_k =\begin{pmatrix}\Pr(X_{n+k} = 0 \mid X_n = 0) & \Pr(X_{n+k} = 1 \mid X_n = 0) \\\Pr(X_{n+k} = 0 \mid X_n = 1) & \Pr(X_{n+k} = 1 \mid X_n = 1)\end{pmatrix}.$

[(1)] Express $P_1$ and $P_2$ in terms of $p$ and $q$ .
[(2)] Express $P_3$ using $P_1$ .
[(3)] The matrix $P_k$ can be expressed in the form $P_k = \gamma_k P_1$ for a real number $\gamma_k$ . Find $\gamma_k$ . \subsection*{II.}For an integer $m \ge 2$ , find the respective probabilities that $X_m = 0$ and $X_m = 1$ , given that the procedure does not terminate before $n = m$ .\subsection*{III.}Let $\ell$ be the length of the sequence generated by the procedure. Find the expected value and the variance of $\ell$ . If necessary, you may use $\sum_{m=1}^{\infty} m r^{m-1} = \frac{1}{(1-r)^2},\qquad\sum_{m=1}^{\infty} m^2 r^{m-1} = \frac{1+r}{(1-r)^3}$ for a real number $r$ with $|r| < 1$ .\subsection*{IV.}For an integer $k \ge 1$ , find the probability $\Pr(X_n = 0 \mid X_{n+k} = 1)$ .\subsection*{V. Final quantity}In this part, assume $p = \dfrac{1}{2}$ and $q = \dfrac{1}{3}$ .
Let $\alpha$ be the probability in {\rm IV} for these values of $p$ and $q$ : $\alpha := \Pr(X_n = 0 \mid X_{n+k} = 1).$
Let $\beta$ be the expected value of $\ell$ obtained in {\rm III}, and let $\sigma^2$ be the variance of $\ell$ in {\rm III}.
In {\rm I(3)}, $P_2$ can be written as $P_2 = \gamma_2 P_1$ . Let $\lambda := \gamma_2$ for $p = \dfrac{1}{2}$ and $q = \dfrac{1}{3}$ . Define $\mathcal{K}= 13 \cdot 320\left( \alpha \beta + \frac{\sigma^2}{\lambda} \right) + 7.$ Compute the integer $\mathcal{K}$ .\

Ground Truth

26683

Execution Trace

Reasoning Process

[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]

Tool Output

[PYTHON OUTPUT] 26683.000000000004 [/PYTHON OUTPUT]

Reasoning Process

[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_1 = \begin{pmatrix} (1-p) \cdot \frac{2}{3} & (1-p) \cdot \frac{1}{3} \ (1-q) \cdot \frac{2}{3} & (1-q) \cdot \frac{1}{3} \end{pmatrix}. ] (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.

Final Model Output

26683

Comparative Analysis (P5) (Average across runs)

Avg Time Taken

GPT-5.1 (high)

467.8

Llama-VARCO-8B-Instruct

340.9

Solar-Pro-2 (31B)(high)

321.5

Deepseek-V3.2

272.4

Gemini-3-Pro-Preview

246.2

A.X-4.0 (72B)

191.5

Grok-4.1-fast

53.5

Claude-Opus-4.5

42.6

EXAONE-4.0.1-32B (high)

20.3

HCX-007(high)

18.8

Avg Token Usage

Solar-Pro-2 (31B)(high)

23391.0

A.X-4.0 (72B)

20243.0

GPT-5.1 (high)

19161.0

Deepseek-V3.2

16929.0

Llama-VARCO-8B-Instruct

16067.0

Claude-Opus-4.5

14573.0

Gemini-3-Pro-Preview

14044.0

Grok-4.1-fast

13647.0

EXAONE-4.0.1-32B (high)

6754.0

HCX-007(high)

5076.0