Question

Let \( x_1 = 0, x_2 = 1, x_3 = 1, x_4 = 1, x_5 = 0 \) be observed values of a random sample of size 5 from \( {Bin}(1, \theta) \) distribution, where \( \theta \in (0, 0.7] \). Then the maximum likelihood estimate of \( \theta \) based on the above sample is ________ (rounded off to two decimal places).

Hint:

For binomial distributions, the maximum likelihood estimate (MLE) of \( \theta \) is simply the sample mean \( \frac{\sum_{i=1}^{n} x_i}{n} \), since the likelihood function is maximized when \( \theta \) equals the proportion of successes in the sample.

Answer 1

For a binomial distribution \( {Bin}(1, \theta) \), the likelihood function is given by: \[ L(\theta) = \prod_{i=1}^{n} \theta^{x_i} (1 - \theta)^{1 - x_i}. \] For the given sample, the likelihood function becomes: \[ L(\theta) = \theta^{x_1} (1 - \theta)^{1 - x_1} \times \theta^{x_2} (1 - \theta)^{1 - x_2} \times \cdots \times \theta^{x_5} (1 - \theta)^{1 - x_5}. \] Thus, the log-likelihood is: \[ \ln L(\theta) = \sum_{i=1}^{5} x_i \ln \theta + (1 - x_i) \ln (1 - \theta). \] Substituting the observed values \( x_1 = 0, x_2 = 1, x_3 = 1, x_4 = 1, x_5 = 0 \): \[ \ln L(\theta) = (0 + 1 + 1 + 1 + 0) \ln \theta + (1 + 0 + 0 + 0 + 1) \ln (1 - \theta). \] Simplifying: \[ \ln L(\theta) = 3 \ln \theta + 2 \ln (1 - \theta). \] To find the maximum likelihood estimate, we take the derivative of \( \ln L(\theta) \) with respect to \( \theta \) and set it equal to 0: \[ \frac{d}{d\theta} \left( 3 \ln \theta + 2 \ln (1 - \theta) \right) = \frac{3}{\theta} - \frac{2}{1 - \theta} = 0. \] Solving for \( \theta \): \[ \frac{3}{\theta} = \frac{2}{1 - \theta} \quad \Rightarrow \quad 3(1 - \theta) = 2\theta \quad \Rightarrow \quad 3 - 3\theta = 2\theta \quad \Rightarrow \quad 3 = 5\theta \quad \Rightarrow \quad \theta = \frac{3}{5} = 0.6. \] Thus, the maximum likelihood estimate of \( \theta \) is \( \boxed{0.60} \).