Problem Analysis #5

To solve the problem, I will perform the following steps using Python:

1. Define the Parameters: Set $p = 1/2$ and $q = 1/3$.
2. Verify Matrix Properties:
   • Construct the matrix $P_1$.
   • Calculate $S = \text{trace}(P_1)$.
   • Verify $P_1^2 = S \cdot P_1$.
   • This confirms $\gamma_2 = \lambda = S$.
3. Compute $\alpha$:
   • Use the derived formula $\alpha = \frac{2}{3} \frac{1-p}{S}$.
   • Verify this by computing the ratio of probabilities for specific $n, k$ (e.g., $n=1, k=1$) using exact matrix operations.
4. Compute $\beta$ and $\sigma^2$:
   • Use the formulas for the mean and variance of the geometric-like distribution of $\ell$:
     • $\beta = E[\ell] = \frac{1}{1-S}$.
     • $\sigma^2 = Var(\ell) = \frac{S}{(1-S)^2}$.
5. Calculate $\mathcal{K}$:
   • Compute the expression $\mathcal{K} = 13 \cdot 320 \left( \alpha \beta + \frac{\sigma^2}{\lambda} \right) + 7$.
   • Ensure the result is an integer.

[PYTHON] import numpy as np from fractions import Fraction

def solve(): # Parameters p = Fraction(1, 2) q = Fraction(1, 3)

# Terms for P1
# P1 = [[P(X_{n+1}=0|X_n=0), P(X_{n+1}=1|X_n=0)],
#       [P(X_{n+1}=0|X_n=1), P(X_{n+1}=1|X_n=1)]]

# Row 0 (given X_n=0): continues with 1-p
# Then X_{n+1} is 0 with 2/3, 1 with 1/3
p00 = (1 - p) * Fraction(2, 3)
p01 = (1 - p) * Fraction(1, 3)

# Row 1 (given X_n=1): continues with 1-q
# Then X_{n+1} is 0 with 2/3, 1 with 1/3
p10 = (1 - q) * Fraction(2, 3)
p11 = (1 - q) * Fraction(1, 3)

P1 = np.array([[p00, p01], [p10, p11]])
print("P1:")
print(P1)

# Calculate S (trace)
S = p00 + p11
print(f"S (trace): {S}")

# Verify P1^2 = S * P1
P2 = P1.dot(P1)
S_P1 = float(S) * P1.astype(float) # Cast to float for comparison if needed, or keep object
# Let's do exact check
P2_exact = np.array([[P1[0,0]*P1[0,0] + P1[0,1]*P1[1,0], P1[0,0]*P1[0,1] + P1[0,1]*P1[1,1]],
                     [P1[1,0]*P1[0,0] + P1[1,1]*P1[1,0], P1[1,0]*P1[0,1] + P1[1,1]*P1[1,1]]])

print("P2 exact:")
print(P2_exact)
print("S * P1:")
print(S * P1)

# Lambda is gamma_2 which is S
lam = S

# Alpha
# Formula: alpha = (2/3) * (1-p) / S
alpha = Fraction(2, 3) * (1 - p) / S
print(f"Alpha: {alpha}")

# Verify Alpha numerically for n=1, k=1
# P(X_1=0 | X_2=1) = P(X_1=0, X_2=1) / P(X_2=1)
# P(X_1=0) = 2/3
# P(X_2=1, X_1=0) = P(X_1=0) * P(X_2=1 | X_1=0) = (2/3) * p01
# P(X_2=1) = P(X_1=0)*p01 + P(X_1=1)*p11 = (2/3)*p01 + (1/3)*p11

num = Fraction(2, 3) * p01
den = Fraction(2, 3) * p01 + Fraction(1, 3) * p11
alpha_check = num / den
print(f"Alpha check (n=1, k=1): {alpha_check}")

# Beta (Expected value of l)
# E[l] = 1 / (1 - S)
beta = 1 / (1 - S)
print(f"Beta: {beta}")

# Sigma squared (Variance of l)
# Var(l) = S / (1 - S)^2
sigma2 = S / ((1 - S) ** 2)
print(f"Sigma^2: {sigma2}")

# Calculate K
# K = 13 * 320 * (alpha * beta + sigma2 / lam) + 7
term1 = alpha * beta
term2 = sigma2 / lam
bracket = term1 + term2
K = 13 * 320 * bracket + 7

print(f"Term1 (alpha*beta): {term1}")
print(f"Term2 (sigma2/lam): {term2}")
print(f"Bracket: {bracket}")
print(f"K: {K}")

solve() [/PYTHON]

The calculated values are:

• $p = 1/2$, $q = 1/3$.
• The trace of $P_1$ is $S = 5/9$.
• $\lambda = \gamma_2 = S = 5/9$.
• $\alpha = 3/5$.
• $\beta = 9/4$.
• $\sigma^2 = 45/16$.
• The term inside the parenthesis is $\alpha \beta + \frac{\sigma^2}{\lambda} = \frac{27}{20} + \frac{81}{16} = \frac{513}{80}$.
• $\mathcal{K} = 13 \cdot 320 \cdot \frac{513}{80} + 7 = 13 \cdot 4 \cdot 513 + 7 = 52 \cdot 513 + 7 = 26676 + 7 = 26683$.

\boxed{26683}