Problem Analysis #5

467.85s

19,161 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] I will symbolically derive all needed expressions using Python (sympy) and then compute the final integer value of K for p = 1/2 and q = 1/3. [/THOUGHT] [PYTHON] from sympy import symbols, Matrix, Rational, factor, simplify

Define symbols

p, q, k = symbols('p q k', positive=True)

Survival factors

a = 1 - p b = 1 - q

Base transition (given survival): rows identical [2/3, 1/3]

B = Matrix([[Rational(2,3), Rational(1,3)], [Rational(2,3), Rational(1,3)]])

T = sub-stochastic transition including survival

T = Matrix([[aRational(2,3), aRational(1,3)], [bRational(2,3), bRational(1,3)]])

P1 = T print("P1 =") print(P1)

Compute P2, P3

P2 = T2 P3 = T3 print("\nP2 =") print(P2) print("\nP3 =") print(P3)

Check rank-1 structure: find lambda0 such that T^2 = lambda0*T

Since T is rank 1, eigenvalue should be v^T u

lambda0 = Rational(2,3)*a + Rational(1,3)*b print("\nlambda0 =", lambda0)

Verify T**2 = lambda0 * T

print("\nT**2 - lambda0T =") print(simplify(P2 - lambda0T))

General gamma_k: Tk = lambda0(k-1) * T

So gamma_k = lambda0**(k-1)

Part II: distribution of X_m given not terminated before m (m >= 2)

Initial distribution of X1

pi1 = Matrix([[Rational(2,3), Rational(1,3)]]) # row vector

Compute pi1*T and normalize to check distribution for m=2

pi2_unnorm = pi1 * T # this is unnormalized wrt survival

Condition on survival before time 2: divide by sum of entries

total2 = (pi2_unnorm * Matrix([[1],[1]]))[0] pi2_cond = pi2_unnorm / total2 print("\nConditional distribution at m=2 (should be 2/3,1/3):") print(pi2_cond.simplify())

Show that for general m>=2, conditional distribution remains [2/3,1/3]

pi_m_unnorm = pi1 * T(m-1) = lambda0(m-2) * pi1*T

After normalization, lambda0 cancels and we get same as for m=2 implicitly.

Part III: length ℓ distribution

r = termination prob per step given survival distribution [2/3,1/3]

r = Rational(2,3)*p + Rational(1,3)*q lambda0_check = 1 - r print("\nr (termination prob per step) =", r) print("1 - r =", simplify(1 - r), "should equal lambda0:", simplify(lambda0 - (1-r)))

P(ℓ >= m) = lambda0**(m-1)

So ℓ ~ geometric with parameter r, support {1,2,...}: P(ℓ=m) = r * (1-r)**(m-1)

E[ℓ] and Var[ℓ] for this geometric:

E_l = 1/r Var_l = (1-r)/r**2 print("\nE[l] =", E_l) print("Var[l] =", Var_l)

Part IV: α = P(X_n=0 | X_{n+k}=1)

Use n=1 (stationary-like), Bayes with P_k = gamma_k P1

gamma_k = lambda0**(k-1)

P_k elements:

P_k_01 = gamma_k * P1[0,1] # (0,1) P_k_11 = gamma_k * P1[1,1] # (1,1)

P(X1=0) and P(X1=1)

P_X1_0 = Rational(2,3) P_X1_1 = Rational(1,3)

P(X_{1+k}=1)

P_Xkp1_1 = P_X1_0P_k_01 + P_X1_1P_k_11 P_Xkp1_1_simplified = simplify(P_Xkp1_1) print("\nP(X_{1+k}=1) general =", P_Xkp1_1_simplified)

Bayes for alpha:

alpha_general = simplify(P_k_01 * P_X1_0 / P_Xkp1_1_simplified) print("\nAlpha general =", alpha_general)

Simplify alpha more:

alpha_simplified = simplify(alpha_general) print("Alpha simplified =", alpha_simplified)

Now plug in p = 1/2, q = 1/3 for final quantities in part V

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

Numeric r, lambda0, alpha, beta, sigma^2, lambda (gamma_2)

r_val = r.subs({p: p_val, q: q_val}) lambda0_val = lambda0.subs({p: p_val, q: q_val}) alpha_val = alpha_simplified.subs({p: p_val, q: q_val}) E_l_val = simplify(E_l.subs({p: p_val, q: q_val})) Var_l_val = simplify(Var_l.subs({p: p_val, q: q_val}))

gamma2_val = lambda0_val # since gamma_2 = lambda0^(2-1) = lambda0

print("\nFor p=1/2, q=1/3:") print("r =", r_val) print("lambda0 =", lambda0_val) print("alpha =", alpha_val) print("E[l] (beta) =", E_l_val) print("Var[l] (sigma^2) =", Var_l_val) print("lambda (gamma_2) =", gamma2_val)

Compute K = 13320(alpha*beta + sigma^2 / lambda) + 7

beta_val = E_l_val sigma2_val = Var_l_val lam_val = gamma2_val

K = 13320(alpha_val*beta_val + sigma2_val/lam_val) + 7 K_simplified = simplify(K) print("\nK simplified =", K_simplified) print("K as integer =", int(K_simplified)) [/PYTHON]

Tool Output

[PYTHON OUTPUT] P1 = Matrix([[2/3 - 2*p/3, 1/3 - p/3], [2/3 - 2*q/3, 1/3 - q/3]]) P2 = Matrix([[4*p**2/9 + 2*p*q/9 - 10*p/9 - 2*q/9 + 2/3, 2*p**2/9 + p*q/9 - 5*p/9 - q/9 + 1/3], [4*p*q/9 - 4*p/9 + 2*q**2/9 - 8*q/9 + 2/3, 2*p*q/9 - 2*p/9 + q**2/9 - 4*q/9 + 1/3]]) P3 = Matrix([[-8*p**3/27 - 8*p**2*q/27 + 32*p**2/27 - 2*p*q**2/27 + 20*p*q/27 - 14*p/9 + 2*q**2/27 - 4*q/9 + 2/3, -4*p**3/27 - 4*p**2*q/27 + 16*p**2/27 - p*q**2/27 + 10*p*q/27 - 7*p/9 + q**2/27 - 2*q/9 + 1/3], [-8*p**2*q/27 + 8*p**2/27 - 8*p*q**2/27 + 32*p*q/27 - 8*p/9 - 2*q**3/27 + 14*q**2/27 - 10*q/9 + 2/3, -4*p**2*q/27 + 4*p**2/27 - 4*p*q**2/27 + 16*p*q/27 - 4*p/9 - q**3/27 + 7*q**2/27 - 5*q/9 + 1/3]]) lambda0 = -2*p/3 - q/3 + 1 T**2 - lambda0*T = Matrix([[0, 0], [0, 0]]) Conditional distribution at m=2 (should be 2/3,1/3): r (termination prob per step) = 2*p/3 + q/3 1 - r = -2*p/3 - q/3 + 1 should equal lambda0: 0 E[l] = 1/(2*p/3 + q/3) Var[l] = (-2*p/3 - q/3 + 1)/(2*p/3 + q/3)**2 P(X_{1+k}=1) general = 3**(-k - 1)*(-2*p - q + 3)**k Alpha general = 2*(1 - p)/(-2*p - q + 3) Alpha simplified = 2*(p - 1)/(2*p + q - 3) For p=1/2, q=1/3: r = 4/9 lambda0 = 5/9 alpha = 3/5 E[l] (beta) = 9/4 Var[l] (sigma^2) = 45/16 lambda (gamma_2) = 5/9 K simplified = 26683 K as integer = 26683 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] Using the validated Python results, the computed value of (\mathcal{K}) is 26683. [ \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