Problem Analysis #5

321.54s

23,391 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

To solve the given problem, we will break it down into manageable parts and use the provided information to find the final answer.

Part I: Transition Matrices

Matrix ( P_1 ): The transition matrix ( P_1 ) is given by: [ P_1 = \begin{pmatrix} \Pr(X_{n+1} = 0 \mid X_n = 0) & \Pr(X_{n+1} = 1 \mid X_n = 0) \ \Pr(X_{n+1} = 0 \mid X_n = 1) & \Pr(X_{n+1} = 1 \mid X_n = 1) \end{pmatrix} ] Given the transition probabilities: [ \Pr(X_{n+1} = 0 \mid X_n = 0) = \frac{2}{3}(1 - p), \quad \Pr(X_{n+1} = 1 \mid X_n = 0) = \frac{1}{3}(1 - p) ] [ \Pr(X_{n+1} = 0 \mid X_n = 1) = \frac{2}{3}(1 - q), \quad \Pr(X_{n+1} = 1 \mid X_n = 1) = \frac{1}{3}(1 - q) ] Therefore, [ P_1 = \begin{pmatrix} \frac{2}{3}(1 - p) & \frac{1}{3}(1 - p) \ \frac{2}{3}(1 - q) & \frac{1}{3}(1 - q) \end{pmatrix} ]
Matrix ( P_2 ): The matrix ( P_2 ) is given by: [ P_2 = P_1 \cdot P_1 ] Calculating ( P_2 ): [ P_2 = \begin{pmatrix} \frac{2}{3}(1 - p) & \frac{1}{3}(1 - p) \ \frac{2}{3}(1 - q) & \frac{1}{3}(1 - q) \end{pmatrix} \cdot \begin{pmatrix} \frac{2}{3}(1 - p) & \frac{1}{3}(1 - p) \ \frac{2}{3}(1 - q) & \frac{1}{3}(1 - q) \end{pmatrix} ] [ P_2 = \begin{pmatrix} \frac{2}{3}(1 - p) \cdot \frac{2}{3}(1 - p) + \frac{1}{3}(1 - p) \cdot \frac{2}{3}(1 - q) & \frac{2}{3}(1 - p) \cdot \frac{1}{3}(1 - p) + \frac{1}{3}(1 - p) \cdot \frac{1}{3}(1 - q) \ \frac{2}{3}(1 - q) \cdot \frac{2}{3}(1 - p) + \frac{1}{3}(1 - q) \cdot \frac{2}{3}(1 - q) & \frac{2}{3}(1 - q) \cdot \frac{1}{3}(1 - p) + \frac{1}{3}(1 - q) \cdot \frac{1}{3}(1 - q) \end{pmatrix} ] Simplifying: [ P_2 = \begin{pmatrix} \frac{4}{9}(1 - p)^2 + \frac{2}{9}(1 - p)(1 - q) & \frac{2}{9}(1 - p)^2 + \frac{1}{9}(1 - p)(1 - q) \ \frac{4}{9}(1 - q)(1 - p) + \frac{2}{9}(1 - q)^2 & \frac{2}{9}(1 - q)(1 - p) + \frac{1}{9}(1 - q)^2 \end{pmatrix} ] [ P_2 = \begin{pmatrix} \frac{2(1 - p) + (1 - q)}{3} \cdot \frac{2}{3}(1 - p) & \frac{2(1 - p) + (1 - q)}{3} \cdot \frac{1}{3}(1 - p) \ \frac{2(1 - p) + (1 - q)}{3} \cdot \frac{2}{3}(1 - q) & \frac{2(1 - p) + (1 - q)}{3} \cdot \frac{1}{3}(1 - q) \end{pmatrix} ] [ P_2 = \gamma_2 P_1 \quad \text{where} \quad \gamma_2 = \frac{2(1 - p) + (1 - q)}{3} ]
General Form ( P_k ): By induction, we can show that: [ P_k = \gamma_k P_1 \quad \text{where} \quad \gamma_k = \left( \frac{2(1 - p) + (1 - q)}{3} \right)^{k-1} ]

Part II: Probabilities Given Survival

Given that the procedure does not terminate before ( n = m ), the distribution of ( X_m ) is: [ \Pr(X_m = 0 \mid \text{survived up to } m-1) = \frac{2}{3}, \quad \Pr(X_m = 1 \mid \text{survived up to } m-1) = \frac{1}{3} ]

Part III: Expected Value and Variance of ( \ell )

Expected Value ( \beta ): The expected value ( \beta ) is given by: [ \beta = \frac{3}{q + 2p} ] For ( p = \frac{1}{2} ) and ( q = \frac{1}{3} ): [ \beta = \frac{3}{\frac{1}{3} + 2 \cdot \frac{1}{2}} = \frac{3}{\frac{1}{3} + 1} = \frac{3}{\frac{4}{3}} = \frac{9}{4} ]
Variance ( \sigma^2 ): The variance ( \sigma^2 ) is given by: [ \sigma^2 = \frac{3(3 + q + 2p)}{(q + 2p)^3} ] For ( p = \frac{1}{2} ) and ( q = \frac{1}{3} ): [ \sigma^2 = \frac{3(3 + \frac{1}{3} + 2 \cdot \frac{1}{2})}{(\frac{1}{3} + 1)^3} = \frac{3(3 + \frac{1}{3} + 1)}{(\frac{4}{3})^3} = \frac{3 \cdot \frac{13}{3}}{\frac{64}{27}} = \frac{13}{\frac{64}{27}} = \frac{351}{64} ]

Part IV: Conditional Probability ( \alpha )

The conditional probability ( \alpha ) is given by: [ \alpha = \frac{2(1 - p)}{2(1 - p) + (1 - q)} ] For ( p = \frac{1}{2} ) and ( q = \frac{1}{3} ): [ \alpha = \frac{2(1 - \frac{1}{2})}{2(1 - \frac{1}{2}) + (1 - \frac{1}{3})} = \frac{2 \cdot \frac{1}{2}}{2 \cdot \frac{1}{2} + \frac{2}{3}} = \frac{1}{1 + \frac{2}{3}} = \frac{1}{\frac{5}{3}} = \frac{3}{5} ]

Part V: Final Calculation

Scalar ( \lambda ): [ \lambda = \gamma_2 = \frac{2(1 - p) + (1 - q)}{3} ] For ( p = \frac{1}{2} ) and ( q = \frac{1}{3} ): [ \lambda = \frac{2(1 - \frac{1}{2}) + (1 - \frac{1}{3})}{3} = \frac{2 \cdot \frac{1}{2} + \frac{2}{3}}{3} = \frac{1 + \frac{2}{3}}{3} = \frac{\frac{5}{3}}{3} = \frac{5}{9} ]
Final Quantity ( \mathcal{K} ): [ \mathcal{K} = 13 \cdot 320 \left( \alpha \beta + \frac{\sigma^2}{\lambda} \right) + 7 ] Substituting the values: [ \alpha \beta = \frac{3}{5} \cdot \frac{9}{4} = \frac{27}{20} ] [ \frac{\sigma^2}{\lambda} = \frac{\frac{351}{64}}{\frac{5}{9}} = \frac{351}{64} \cdot \frac{9}{5} = \frac{3159}{320} ] [ \alpha \beta + \frac{\sigma^2}{\lambda} = \frac{27}{20} + \frac{3159}{320} = \frac{432}{320} + \frac{3159}{320} = \frac{3591}{320} ] [ \mathcal{K} = 13 \cdot 320 \cdot \frac{3591}{320} + 7 = 13 \cdot 3591 + 7 = 46683 + 7 = 46690 ]

Thus, the final answer is: [ \boxed{46690} ]

Final Model Output

46690

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