Problem Analysis #5

42.57s

14,573 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] Let me work through this problem step by step.

Part I: Finding P_1, P_2, and γ_k

For P_1: Given X_n = 0 or 1, we need to find the probability of X_{n+1} = 0 or 1, given that the procedure doesn't terminate at step n.

If X_n = 0: The procedure continues with probability (1-p), and then X_{n+1} = 0 with prob 2/3, X_{n+1} = 1 with prob 1/3. If X_n = 1: The procedure continues with probability (1-q), and then X_{n+1} = 0 with prob 2/3, X_{n+1} = 1 with prob 1/3.

So P_1 = [[2(1-p)/3, (1-p)/3], [2(1-q)/3, (1-q)/3]]

For P_2, we need P_2 = P_1 * P_1 (matrix multiplication).

For γ_k: We need to find γ_k such that P_k = γ_k * P_1.

Part II: Probabilities of X_m = 0 and X_m = 1

Given the procedure doesn't terminate before n = m, we need conditional probabilities.

Part III: Expected value and variance of ℓ

Part IV: Pr(X_n = 0 | X_{n+k} = 1)

Part V: Calculate K with p = 1/2, q = 1/3

Let me compute all of this using Python. [/THOUGHT] [PYTHON] import sympy as sp from sympy import Rational, simplify, symbols, Matrix

Define symbols

p, q = symbols('p q')

Part I(1): P_1

P_1[i,j] = Pr(X_{n+1} = j | X_n = i) where the procedure continues

If X_n = 0: continues with prob (1-p), then X_{n+1} = 0 with 2/3, = 1 with 1/3

If X_n = 1: continues with prob (1-q), then X_{n+1} = 0 with 2/3, = 1 with 1/3

P1 = Matrix([ [Rational(2,3)(1-p), Rational(1,3)(1-p)], [Rational(2,3)(1-q), Rational(1,3)(1-q)] ])

print("P_1 =") print(P1)

Part I(1): P_2 = P_1 * P_1

P2 = P1 * P1 P2 = simplify(P2) print("\nP_2 = P_1 * P_1 =") print(P2)

Part I(2): Express P_3 using P_1

P3 = P1 * P1 * P1 P3 = simplify(P3) print("\nP_3 =") print(P3)

Part I(3): Find gamma_k such that P_k = gamma_k * P_1

Let's check if P_2 = gamma_2 * P_1

P_2[0,0] / P_1[0,0] should equal gamma_2

gamma_2 = simplify(P2[0,0] / P1[0,0]) print("\ngamma_2 =", gamma_2)

Verify P_2 = gamma_2 * P_1

print("Verification P_2 = gamma_2 * P_1:", simplify(P2 - gamma_2 * P1))

General pattern: gamma_k = ((1-p)2/3 + (1-q)1/3)^(k-1)

Let's verify with gamma_3

gamma_3 = simplify(P3[0,0] / P1[0,0]) print("gamma_3 =", gamma_3)

The trace of P_1 is (1-p)2/3 + (1-q)1/3

trace_P1 = simplify(P1[0,0] + P1[1,1]) print("trace(P_1) =", trace_P1)

So gamma_k = (trace(P_1))^(k-1) = ((2(1-p) + (1-q))/3)^(k-1)

[/PYTHON]

Tool Output

[PYTHON OUTPUT] P_1 = Matrix([[2/3 - 2*p/3, 1/3 - p/3], [2/3 - 2*q/3, 1/3 - q/3]]) P_2 = P_1 * P_1 = Matrix([[2*(p - 1)*(2*p + q - 3)/9, (p - 1)*(2*p + q - 3)/9], [2*(q - 1)*(2*p + q - 3)/9, (q - 1)*(2*p + q - 3)/9]]) P_3 = Matrix([[-2*(p - 1)*(2*p + q - 3)**2/27, -(p - 1)*(2*p + q - 3)**2/27], [-2*(q - 1)*(2*p + q - 3)**2/27, -(q - 1)*(2*p + q - 3)**2/27]]) gamma_2 = -2*p/3 - q/3 + 1 Verification P_2 = gamma_2 * P_1: Matrix([[0, 0], [0, 0]]) gamma_3 = (2*p + q - 3)**2/9 trace(P_1) = -2*p/3 - q/3 + 1 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] Let me continue the calculations to find all the required values. [/THOUGHT] [PYTHON] import sympy as sp from sympy import Rational, simplify, symbols, Matrix, sqrt

Define symbols

p, q = symbols('p q')

gamma_k = ((2(1-p) + (1-q))/3)^(k-1) = (1 - (2p+q)/3)^(k-1)

Let r = (2(1-p) + (1-q))/3 = 1 - (2p+q)/3

r = Rational(2,3)(1-p) + Rational(1,3)(1-q) r = simplify(r) print("r = (2(1-p) + (1-q))/3 =", r)

Part II: Pr(X_m = 0) and Pr(X_m = 1) given procedure doesn't terminate before m

Given the procedure doesn't terminate before m, X_m is generated

The key insight: regardless of the history, X_m = 0 with prob 2/3 and X_m = 1 with prob 1/3

This is because each new X is generated with these fixed probabilities

print("\nPart II: Pr(X_m = 0 | no termination before m) = 2/3") print("Part II: Pr(X_m = 1 | no termination before m) = 1/3")

Part III: Expected value and variance of ℓ

Let's compute the probability that the sequence has length exactly m

P(ℓ = m) = P(doesn't terminate at 1, ..., m-1, terminates at m)

Let's define:

At each step n, if X_n = 0, terminate with prob p; if X_n = 1, terminate with prob q

X_n = 0 with prob 2/3, X_n = 1 with prob 1/3 (for n >= 1)

Probability of termination at step n given we reach step n:

P(terminate | reach n) = (2/3)p + (1/3)q

Probability of NOT terminating at step n given we reach step n:

P(continue | reach n) = (2/3)(1-p) + (1/3)(1-q) = 1 - (2p+q)/3 = r

So P(ℓ = m) = r^(m-1) * (1-r) for m >= 1

This is a geometric distribution with parameter (1-r)

E[ℓ] = 1/(1-r)

Var[ℓ] = r/(1-r)^2

one_minus_r = simplify(1 - r) print("\n1 - r =", one_minus_r)

E_ell = simplify(1 / one_minus_r) print("E[ℓ] = 1/(1-r) =", E_ell)

Var_ell = simplify(r / one_minus_r**2) print("Var[ℓ] = r/(1-r)^2 =", Var_ell)

Part IV: Pr(X_n = 0 | X_{n+k} = 1)

Using Bayes' theorem:

P(X_n = 0 | X_{n+k} = 1) = P(X_{n+k} = 1 | X_n = 0) * P(X_n = 0) / P(X_{n+k} = 1)

P(X_{n+k} = 1 | X_n = 0) = P_k[0,1] = gamma_k * P_1[0,1] = gamma_k * (1-p)/3

P(X_{n+k} = 1 | X_n = 1) = P_k[1,1] = gamma_k * P_1[1,1] = gamma_k * (1-q)/3

P(X_n = 0) = 2/3, P(X_n = 1) = 1/3

P(X_{n+k} = 1) = P(X_{n+k} = 1 | X_n = 0)P(X_n = 0) + P(X_{n+k} = 1 | X_n = 1)P(X_n = 1)

= gamma_k * (1-p)/3 * 2/3 + gamma_k * (1-q)/3 * 1/3

= gamma_k/9 * (2(1-p) + (1-q))

= gamma_k/9 * 3r = gamma_k * r / 3

P(X_n = 0 | X_{n+k} = 1) = [gamma_k * (1-p)/3 * 2/3] / [gamma_k * r / 3]

= 2(1-p)/3 / r

= 2(1-p) / (2(1-p) + (1-q))

alpha = simplify(Rational(2,3)*(1-p) / r) print("\nalpha = P(X_n = 0 | X_{n+k} = 1) =", alpha)

Simplify alpha

alpha_simplified = simplify(2*(1-p) / (2*(1-p) + (1-q))) print("alpha simplified =", alpha_simplified)

Part V: With p = 1/2, q = 1/3

p_val = Rational(1, 2) q_val = Rational(1, 3)

Calculate r with these values

r_val = Rational(2,3)(1-p_val) + Rational(1,3)(1-q_val) print("\nWith p=1/2, q=1/3:") print("r =", r_val)

gamma_2 = r (since gamma_k = r^(k-1), so gamma_2 = r)

lambda_val = r_val print("lambda = gamma_2 =", lambda_val)

E[ℓ] = 1/(1-r)

one_minus_r_val = 1 - r_val beta_val = 1 / one_minus_r_val print("beta = E[ℓ] =", beta_val)

Var[ℓ] = r/(1-r)^2

sigma2_val = r_val / one_minus_r_val**2 print("sigma^2 = Var[ℓ] =", sigma2_val)

alpha = 2(1-p) / (2(1-p) + (1-q))

alpha_val = 2*(1-p_val) / (2*(1-p_val) + (1-q_val)) print("alpha =", alpha_val)

Calculate K

K = 13 * 320 * (alpha * beta + sigma^2 / lambda) + 7

K = 13 * 320 * (alpha_val * beta_val + sigma2_val / lambda_val) + 7 print("\nK =", K) print("K simplified =", simplify(K)) [/PYTHON]

Tool Output

[PYTHON OUTPUT] r = (2(1-p) + (1-q))/3 = -2*p/3 - q/3 + 1 Part II: Pr(X_m = 0 | no termination before m) = 2/3 Part II: Pr(X_m = 1 | no termination before m) = 1/3 1 - r = 2*p/3 + q/3 E[ℓ] = 1/(1-r) = 3/(2*p + q) Var[ℓ] = r/(1-r)^2 = 3*(-2*p - q + 3)/(2*p + q)**2 alpha = P(X_n = 0 | X_{n+k} = 1) = 2*(p - 1)/(2*p + q - 3) alpha simplified = 2*(p - 1)/(2*p + q - 3) With p=1/2, q=1/3: r = 5/9 lambda = gamma_2 = 5/9 beta = E[ℓ] = 9/4 sigma^2 = Var[ℓ] = 45/16 alpha = 3/5 K = 26683 K simplified = 26683 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] The Python calculation gives us the final answer. Let me verify the key values:

With p = 1/2 and q = 1/3:

r = 5/9
λ = γ₂ = 5/9
β = E[ℓ] = 9/4
σ² = Var[ℓ] = 45/16
α = 3/5

Then: K = 13 × 320 × (α × β + σ²/λ) + 7 K = 13 × 320 × (3/5 × 9/4 + (45/16)/(5/9)) + 7 K = 13 × 320 × (27/20 + 81/16) + 7 K = 26683

The answer is 26683. [/THOUGHT]

\boxed{26683}

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

Problem Analysis #5

Define symbols

Part I(1): P_1

P_1[i,j] = Pr(X_{n+1} = j | X_n = i) where the procedure continues

If X_n = 0: continues with prob (1-p), then X_{n+1} = 0 with 2/3, = 1 with 1/3

If X_n = 1: continues with prob (1-q), then X_{n+1} = 0 with 2/3, = 1 with 1/3

Part I(1): P_2 = P_1 * P_1

Part I(2): Express P_3 using P_1

Part I(3): Find gamma_k such that P_k = gamma_k * P_1

Let's check if P_2 = gamma_2 * P_1

P_2[0,0] / P_1[0,0] should equal gamma_2

Verify P_2 = gamma_2 * P_1

General pattern: gamma_k = ((1-p)*2/3 + (1-q)*1/3)^(k-1)

Let's verify with gamma_3

The trace of P_1 is (1-p)*2/3 + (1-q)*1/3

So gamma_k = (trace(P_1))^(k-1) = ((2(1-p) + (1-q))/3)^(k-1)

Define symbols

gamma_k = ((2(1-p) + (1-q))/3)^(k-1) = (1 - (2p+q)/3)^(k-1)

Let r = (2(1-p) + (1-q))/3 = 1 - (2p+q)/3

Part II: Pr(X_m = 0) and Pr(X_m = 1) given procedure doesn't terminate before m

Given the procedure doesn't terminate before m, X_m is generated

The key insight: regardless of the history, X_m = 0 with prob 2/3 and X_m = 1 with prob 1/3

This is because each new X is generated with these fixed probabilities

Part III: Expected value and variance of ℓ

Let's compute the probability that the sequence has length exactly m

P(ℓ = m) = P(doesn't terminate at 1, ..., m-1, terminates at m)

Let's define:

At each step n, if X_n = 0, terminate with prob p; if X_n = 1, terminate with prob q

X_n = 0 with prob 2/3, X_n = 1 with prob 1/3 (for n >= 1)

Probability of termination at step n given we reach step n:

P(terminate | reach n) = (2/3)*p + (1/3)*q

Probability of NOT terminating at step n given we reach step n:

P(continue | reach n) = (2/3)(1-p) + (1/3)(1-q) = 1 - (2p+q)/3 = r

So P(ℓ = m) = r^(m-1) * (1-r) for m >= 1

This is a geometric distribution with parameter (1-r)

E[ℓ] = 1/(1-r)

Var[ℓ] = r/(1-r)^2

Part IV: Pr(X_n = 0 | X_{n+k} = 1)

Using Bayes' theorem:

P(X_n = 0 | X_{n+k} = 1) = P(X_{n+k} = 1 | X_n = 0) * P(X_n = 0) / P(X_{n+k} = 1)

P(X_{n+k} = 1 | X_n = 0) = P_k[0,1] = gamma_k * P_1[0,1] = gamma_k * (1-p)/3

P(X_{n+k} = 1 | X_n = 1) = P_k[1,1] = gamma_k * P_1[1,1] = gamma_k * (1-q)/3

P(X_n = 0) = 2/3, P(X_n = 1) = 1/3

P(X_{n+k} = 1) = P(X_{n+k} = 1 | X_n = 0)*P(X_n = 0) + P(X_{n+k} = 1 | X_n = 1)*P(X_n = 1)

= gamma_k * (1-p)/3 * 2/3 + gamma_k * (1-q)/3 * 1/3

= gamma_k/9 * (2(1-p) + (1-q))

= gamma_k/9 * 3r = gamma_k * r / 3

P(X_n = 0 | X_{n+k} = 1) = [gamma_k * (1-p)/3 * 2/3] / [gamma_k * r / 3]

= 2(1-p)/3 / r

= 2(1-p) / (2(1-p) + (1-q))

Simplify alpha

Part V: With p = 1/2, q = 1/3

Calculate r with these values

gamma_2 = r (since gamma_k = r^(k-1), so gamma_2 = r)

E[ℓ] = 1/(1-r)

Var[ℓ] = r/(1-r)^2

alpha = 2(1-p) / (2(1-p) + (1-q))

Calculate K

K = 13 * 320 * (alpha * beta + sigma^2 / lambda) + 7

Comparative Analysis (P5) (Average across runs)

Avg Time Taken

Avg Token Usage

General pattern: gamma_k = ((1-p)2/3 + (1-q)1/3)^(k-1)

The trace of P_1 is (1-p)2/3 + (1-q)1/3

P(terminate | reach n) = (2/3)p + (1/3)q

P(X_{n+k} = 1) = P(X_{n+k} = 1 | X_n = 0)P(X_n = 0) + P(X_{n+k} = 1 | X_n = 1)P(X_n = 1)