Problem Analysis #5

1012.14s
72,753 toks
Problem Statement

Consider the following procedure that generates a sequence of random variables that take the value 00 or 11. For an integer n1n \ge 1, we denote the nn-th random variable of a sequence generated by the procedure as XnX_n.

  • X1X_1 becomes X1=0X_1 = 0 with probability 23\dfrac{2}{3} and X1=1X_1 = 1 with probability 13\dfrac{1}{3}.

  • For integers n=1,2,n = 1,2,\ldots in order, the following is repeated until the procedure terminates:

  • The procedure terminates with probability pp (0<p<10 < p < 1) if Xn=0X_n = 0, and with probability qq (0<q<10 < q < 1) if Xn=1X_n = 1. Here pp and qq are fixed constants.

  • If the procedure does not terminate at step nn, then Xn+1X_{n+1} becomes 00 with probability 23\dfrac{2}{3} and 11 with probability 13\dfrac{1}{3}.

When the procedure terminates at n=n = \ell, a sequence of length \ell, composed of random variables (X1,,X)(X_1,\ldots, X_\ell), is generated, and no further random variables are generated.\subsection*{I.}For an integer k1k \ge 1, consider the matrixPk=(Pr(Xn+k=0Xn=0)Pr(Xn+k=1Xn=0)Pr(Xn+k=0Xn=1)Pr(Xn+k=1Xn=1)).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 P1P_1 and P2P_2 in terms of pp and qq.

  • [(2)] Express P3P_3 using P1P_1.

  • [(3)] The matrix PkP_k can be expressed in the form Pk=γkP1P_k = \gamma_k P_1 for a real number γk\gamma_k. Find γk\gamma_k. \subsection*{II.}For an integer m2m \ge 2, find the respective probabilities that Xm=0X_m = 0 and Xm=1X_m = 1, given that the procedure does not terminate before n=mn = 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 usem=1mrm1=1(1r)2,m=1m2rm1=1+r(1r)3\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 rr with r<1|r| < 1.\subsection*{IV.}For an integer k1k \ge 1, find the probability Pr(Xn=0Xn+k=1)\Pr(X_n = 0 \mid X_{n+k} = 1).\subsection*{V. Final quantity}In this part, assume p=12p = \dfrac{1}{2} and q=13q = \dfrac{1}{3}.

  • Let α\alpha be the probability in {\rm IV} for these values of pp and qq: α:=Pr(Xn=0Xn+k=1). \alpha := \Pr(X_n = 0 \mid X_{n+k} = 1).

  • Let β\beta be the expected value of \ell obtained in {\rm III}, and let σ2\sigma^2 be the variance of \ell in {\rm III}.

  • In {\rm I(3)}, P2P_2 can be written as P2=γ2P1P_2 = \gamma_2 P_1. Let λ:=γ2 \lambda := \gamma_2 for p=12p = \dfrac{1}{2} and q=13q = \dfrac{1}{3}. DefineK=13320(αβ+σ2λ)+7.\mathcal{K}= 13 \cdot 320\left( \alpha \beta + \frac{\sigma^2}{\lambda} \right) + 7.Compute the integer K\mathcal{K}.\

Ground Truth
26683
Execution Trace
Reasoning Process

The solution requires computing a complex expression involving parameters derived from the described stochastic process. After careful analysis:

  • Part I identifies (P_1 = \begin{pmatrix} (1-p)\cdot 2/3 & (1-p)\cdot 1/3 \ (1-q)\cdot 2/3 & (1-q)\cdot 1/3 \end{pmatrix}), (P_2 = M P_1), and (\gamma_k = M^{k-1}) with (M = \frac{3-2p-q}{3}).

  • Part II gives (\Pr(X_m = 0 \mid \text{no termination before } m) = \frac{2}{3}), (\Pr(X_m = 1 \mid \text{no termination before } m) = \frac{1}{3}).

  • Part III yields (\mathbb{E}[\ell] = \frac{3}{2p+q}) and (\operatorname{Var}(\ell) = \frac{3(3-2p-q)}{(2p+q)^2}).

  • Part IV yields (\Pr(X_n = 0 \mid X_{n+k} = 1) = \frac{2(1-p)}{2(1-p) + (1-q)}).

For the specific values (p = \frac12), (q = \frac13):

  • (\alpha = \frac{3}{5})
  • (\beta = \frac{9}{4})
  • (\sigma^2 = \frac{45}{16})
  • (\lambda = \frac{5}{9})

Then

[ \alpha\beta + \frac{\sigma^2}{\lambda} = \frac{27}{20} + \frac{81}{16} = \frac{ tích 513}{80}, ]

[ 13 \cdot 320 \cdot \frac{ needs to compute precisely. [ 13 \cdot 320 \cdot \frac{513}{80} = 13 \cdot \frac{320}{80} \cdot 513 = 13 \cdot 4 \cdot 513 = 13 \cdot 2052 = 26676, ]

[ K = 26676 + 7 = 26683. ]

Thus the integer (\mathcal{K}) is (\boxed{26683}).

Reasoning Process

[THOUGHT] We need to compute K = 13 * 320 * (αβ + σ^2/λ) + 7 with p=1/2, q=1/3. We have derived: α = 3/5, β = 9/4, σ^2 = 45/16, λ = 5/9. Compute αβ = (3/5)(9/4) = 27/20 = 1.35. σ^2/λ = (45/16) / (5/9) = (45/16)(9/5) = (459)/(165) = 405/80 = 81/16 = 5.0625. Sum = 27/20 + 81/16 = (108/80 + 405/80) = 513/80. Then 13 * 320 * (513/80) + 7 = (13 * 320 * charg on the fly.

We'll compute exactly with Python.

[PYTHON]

Compute K step by step

p = 1/2 q = 1/3

α = Pr(X_n = 0 | X_{n+k} = 1) for general k, using initial distribution of X_1

Using formula α = (π0 * b) / (π0 * b + π1 * d) where π0=2/3, π1=1/3, b=(1-p)/3, d=(1-q)/3.

π0 = 2/3 π1 = 1/3 b = (1-p)/3 d = (1-q)/3 α = (π0 * b) / (π0 * b + π1 * d)

β = E[ℓ] = 1 / ((2/3)p + (1/3)q)

r = (2/3)*p + (1/3)*q β = 1 / r

σ^2 = Var(ℓ) = (1 - r) / r**2

σ2 = (1 - r) / (r**2)

λ = γ_2 = (3 - 2p - q)/3

λ = (3 - 2*p - q) / 3

Now compute expression

import sympy as sp K = 13 * 320 * (α * β + σ2 / λ) + 7 K, float(K) [/PYTHON]

Tool Output
[PYTHON OUTPUT] (26683.000000000004, 26683.000000000004) [/PYTHON OUTPUT]
Reasoning Process

[TH0N] import sympy as sp

Given values

p = sp.Rational(1, 2) q = sp.Rational(1, 3)

Compute α = Pr(X_n = 0 | X_{n+k} = 1)

π0 = sp.Rational(2, 3) π1 = sp.Rational(1, 3) b = (1 - p) / 3 # = (1/2)/3 = 1/6 d = (1 - q) / 3 # = (2/3)/3 = 2/9 α = (π0 * b) / (π0 * b + π1 * d)

Compute β = E[ℓ] = 1 / r, where r = (2/3)p + (1/3)q

r = (sp.Rational(2, 3)) * p + (sp.Rational(1, 3)) * q β = 1 / r

Compute σ² = Var(ℓ) = (1 - r)/r²

σ2 = (1 - r) / (r ** 2)

Compute λ = γ_2 = (3 - 2p - q)/3

λ = (3 - 2*p - q) / 3

Compute the expression: 13 * 320 * (αβ + σ²/λ) + 7

K = 13 * 320 * (α * β + σ2 / λ) + 7 K, K.evalf() </PYTHON>

Reasoning Process

[TH0N] 26683 [/THOUGHT]

\boxed{26683}

Final Model Output
26683

Comparative Analysis (P5) (Average across runs)

Avg Time Taken

K-EXAONE-236B-A23B
802.4
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
Kanana-2-30B-Thinking
234.2
K-EXAONE-236B-A23B
201.8
A.X-4.0 (72B)
191.5
Kanana-2-30B-Thinking-2601
171.1
Solar-Open-100B
133.9
GPT-5.2 (high)
78.5
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
axk1
13.1

Avg Token Usage

K-EXAONE-236B-A23B
71914.3
Kanana-2-30B-Thinking-2601
36339.0
Kanana-2-30B-Thinking
33424.0
K-EXAONE-236B-A23B
26257.0
Solar-Pro-2 (31B)(high)
23391.0
Solar-Open-100B
23022.0
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
GPT-5.2 (high)
8615.0
EXAONE-4.0.1-32B (high)
6754.0
HCX-007(high)
5076.0
axk1
4852.0