Problem Analysis #36

[PYTHON] from fractions import Fraction

def solve_variance(): # Define probabilities for x = 0, 1, 2, 3 # P(X=x) = |2x - 1| / 12 probs = {} sum_p_first_four = 0 for x in range(4): p = Fraction(abs(2*x - 1), 12) probs[x] = p sum_p_first_four += p

# Calculate a
# Sum of all probs is 1
# sum_p_first_four + a = 1
a = 1 - sum_p_first_four
probs[4] = a

print(f"a = {a}")
print(f"Probabilities: {probs}")

# Calculate E[X]
expected_value = sum(x * probs[x] for x in probs)
print(f"E[X] = {expected_value}")

# Calculate E[X^2]
expected_value_sq = sum((x**2) * probs[x] for x in probs)
print(f"E[X^2] = {expected_value_sq}")

# Calculate V(X)
variance_x = expected_value_sq - expected_value**2
print(f"V(X) = {variance_x}")

# Calculate V((1/a)X) = (1/a)^2 * V(X)
# Since a is a Fraction, we can do this directly
target_variance = (1/a)**2 * variance_x
print(f"V((1/a)X) = {target_variance}")

return target_variance

result = solve_variance() print(f"Result: {result}") [/PYTHON]

The probability mass function is defined as: [P(X=x) = \frac{|2x-1|}{12} \quad \text{for } x \in {0, 1, 2, 3}] [P(X=4) = a]

First, we calculate the probabilities for $x=0, 1, 2, 3$:

• $P(X=0) = \frac{|-1|}{12} = \frac{1}{12}$
• $P(X=1) = \frac{|1|}{12} = \frac{1}{12}$
• $P(X=2) = \frac{|3|}{12} = \frac{3}{12}$
• $P(X=3) = \frac{|5|}{12} = \frac{5}{12}$

The sum of these probabilities is $\frac{1+1+3+5}{12} = \frac{10}{12} = \frac{5}{6}$. Since the total probability must sum to 1, we have: [\frac{5}{6} + a = 1 \implies a = \frac{1}{6}]

Next, we calculate the expected value $E[X]$ and $E[X^2]$: [E[X] = \sum x P(X=x) = 0\cdot\frac{1}{12} + 1\cdot\frac{1}{12} + 2\cdot\frac{3}{12} + 3\cdot\frac{5}{12} + 4\cdot\frac{2}{12} = \frac{0+1+6+15+8}{12} = \frac{30}{12} = \frac{5}{2}] [E[X^2] = \sum x^2 P(X=x) = 0\cdot\frac{1}{12} + 1\cdot\frac{1}{12} + 4\cdot\frac{3}{12} + 9\cdot\frac{5}{12} + 16\cdot\frac{2}{12} = \frac{0+1+12+45+32}{12} = \frac{90}{12} = \frac{15}{2}]

The variance of $X$ is: [V(X) = E[X^2] - (E[X])^2 = \frac{15}{2} - \left(\frac{5}{2}\right)^2 = \frac{30}{4} - \frac{25}{4} = \frac{5}{4}]

We need to compute $V\left(\frac{1}{a}X\right)$. Since $a = \frac{1}{6}$, $\frac{1}{a} = 6$. Using the property $V(cX) = c^2 V(X)$: [V(6X) = 6^2 V(X) = 36 \cdot \frac{5}{4} = 9 \cdot 5 = 45]

\boxed{45}