Problem Statement

A tester tries to identify a correct brand in a series of trials. Let \(\theta\) be the probability of choosing the correct brand in any trial. Let \(Y_i\) be a Bernoulli random variable taking value 1 for a correct guess in the \(i\)-th trial.

In the first 6 trials, the results are: \(1,1,1,1,1,0\).

We wish to test whether the tester has no discriminatory power against the alternative that she does. So our problem is:

\[H_0: \theta = \frac{1}{2} \quad \text{versus} \quad H_1: \theta > \frac{1}{2}\]

This is a simple versus composite case with:

  • \(\Theta_0 = \{\frac{1}{2}\}\)
  • \(\Theta_1 = (\frac{1}{2}, 1)\)

We assume a uniform prior distribution on \(\theta\) under the alternative:

\[\pi_1(\theta) = 2 \quad \text{if} \quad \frac{1}{2} < \theta < 1\]

For the point null hypothesis, the implied prior distribution is:

\[\pi_0(\theta = \frac{1}{2}) = 1\]

Do not confuse with Geometric Distribution

Some similar problems involve waiting until the first failure, which follows a geometric distribution. However, in this problem:

  • The number of trials (6) is fixed in advance
  • We observe a fixed number of successes (5) and failures (1)
  • This is a Binomial setup, not a Geometric one.
Distribution When Used Our Example
Bernoulli Single trial outcome Each \(Y_i\) individually
Binomial Fixed number of trials, count successes The entire sequence of 6 trials
Geometric Number of trials until first failure Not used here

Step-by-Step Solution

Step 1: Write the Likelihood Function

For independent Bernoulli trials with 5 successes and 1 failure:

\[f(y \mid \theta) = \theta^{\text{successes}} \cdot (1-\theta)^{\text{failures}} = \theta^5 \cdot (1-\theta)^1\]

So:

\[f(y \mid \theta) = \theta^5 (1-\theta)\]

Step 2: Numerator (Under \(H_0: \theta = \frac{1}{2}\))

Since \(H_0\) is a point null:

\[f(y \mid M_0) = f(y \mid \theta_0) = \left(\frac{1}{2}\right)^5 \left(1 - \frac{1}{2}\right) = \left(\frac{1}{2}\right)^5 \cdot \frac{1}{2} = \left(\frac{1}{2}\right)^6\]

\[f(y \mid M_0) = \frac{1}{64} = 0.015625\]

Step 3: Denominator (Under \(H_1: \theta \in (\frac{1}{2}, 1)\))

Under \(H_1\), we must integrate the likelihood over the prior \(\pi_1(\theta) = 2\):

\[f(y \mid M_1) = \int_{1/2}^{1} f(y \mid \theta) \, \pi_1(\theta) \, d\theta = \int_{1/2}^{1} \theta^5 (1-\theta) \cdot 2 \, d\theta\]

\[f(y \mid M_1) = 2 \int_{1/2}^{1} \theta^5 (1-\theta) \, d\theta\]

Step 4: Compute the Integral

\[\int_{1/2}^{1} \theta^5 (1-\theta) \, d\theta = \int_{1/2}^{1} (\theta^5 - \theta^6) \, d\theta\]

\[= \left[ \frac{\theta^6}{6} - \frac{\theta^7}{7} \right]_{1/2}^{1}\]

Evaluate at \(\theta = 1\):

\[\frac{1}{6} - \frac{1}{7} = \frac{7}{42} - \frac{6}{42} = \frac{1}{42}\]

Evaluate at \(\theta = \frac{1}{2}\):

\[\frac{(1/2)^6}{6} - \frac{(1/2)^7}{7} = \frac{1/64}{6} - \frac{1/128}{7} = \frac{1}{384} - \frac{1}{896}\]

Find common denominator: \(384 = 2^7 \cdot 3\), \(896 = 2^7 \cdot 7\), LCM = \(2^7 \cdot 3 \cdot 7 = 2688\)

\[\frac{7}{2688} - \frac{3}{2688} = \frac{4}{2688} = \frac{1}{672}\]

So the definite integral:

\[\int_{1/2}^{1} (\theta^5 - \theta^6) \, d\theta = \frac{1}{42} - \frac{1}{672}\]

Convert \(\frac{1}{42}\) to denominator 672: \(\frac{16}{672}\)

\[\frac{16}{672} - \frac{1}{672} = \frac{15}{672} = \frac{5}{224} \approx 0.02232\]

Step 5: Complete the Denominator

\[f(y \mid M_1) = 2 \times \frac{5}{224} = \frac{10}{224} = \frac{5}{112} \approx 0.04464\]

Step 6: Calculate the Bayes Factor

\[B_{01}(y) = \frac{f(y \mid M_0)}{f(y \mid M_1)} = \frac{1/64}{5/112} = \frac{1}{64} \times \frac{112}{5} = \frac{112}{320}\]

Simplify:

\[\frac{112}{320} = \frac{56}{160} = \frac{28}{80} = \frac{14}{40} = \frac{7}{20} = 0.35\]

The problem states \(B_{01} = \frac{1}{2.86} \approx 0.35\), which matches:

\[2.86 = \frac{1}{0.35} = \frac{20}{7} \approx 2.857\]

R Code Verification

# Compute Bayes factor numerically in R

# Likelihood under H0
lik_H0 <- (1/2)^6

# Prior under H1 (uniform on (0.5, 1))
prior_H1 <- function(theta) 2

# Likelihood function
likelihood <- function(theta) theta^5 * (1 - theta)

# Marginal likelihood under H1 (numerical integration)
marginal_H1 <- integrate(function(theta) likelihood(theta) * prior_H1(theta), 
                         lower = 0.5, upper = 1)$value

# Bayes factor
B01 <- lik_H0 / marginal_H1

# Display results
cat("Likelihood under H0:", lik_H0, "\n")
## Likelihood under H0: 0.015625
cat("Marginal likelihood under H1:", marginal_H1, "\n")
## Marginal likelihood under H1: 0.04464286
cat("Bayes factor B01:", B01, "\n")
## Bayes factor B01: 0.35
cat("B01 as 1/x:", 1/B01, "\n")
## B01 as 1/x: 2.857143

Interpretation

The Bayes factor \(B_{01} = 0.35\) means:

\[B_{01} = \frac{P(\text{data} \mid H_0)}{P(\text{data} \mid H_1)} = 0.35\]

Equivalently:

\[\frac{P(\text{data} \mid H_1)}{P(\text{data} \mid H_0)} = \frac{1}{0.35} \approx 2.86\]

Conclusion: The data are about 2.86 times more likely under \(H_1\) (the tester has some ability, \(\theta > 0.5\)) than under \(H_0\) (pure guessing, \(\theta = 0.5\)). This provides positive evidence against the null hypothesis, though not overwhelming.

Summary

Component Value
Likelihood function \(\theta^5 (1-\theta)\)
\(f(y \mid H_0)\) \(\frac{1}{64} = 0.0156\)
\(f(y \mid H_1)\) \(\frac{5}{112} \approx 0.0446\)
Bayes factor \(B_{01}\) \(\frac{7}{20} = 0.35\)
Evidence for \(H_1\) \(\frac{1}{B_{01}} \approx 2.86\)

Key Takeaway

The geometric distribution does not appear in this example. The calculation uses the Bernoulli/Binomial likelihood because we have a fixed number of trials with a fixed number of successes.