Question

Let $x_1 = 1, x_2 = 0, x_3 = 0, x_4 = 1, x_5 = 0, x_6 = 1$ be the data on a random sample of size 6 from Bin(1, $\theta$) distribution, where $\theta \in (0, 1)$. Then the uniformly minimum variance unbiased estimate of $\theta(1 + \theta)$ equals .............

Hint:

For Binomial samples, unbiased estimators of polynomial functions of $\theta$ can be expressed in terms of $\bar{X}$ and its powers using moment relationships.

Answer 1

Correct Answer: 0.7

Step 1: Identify sufficient statistic.
For $X_i \sim \text{Bin}(1,\theta)$, the sum $T = \sum X_i$ is a sufficient statistic for $\theta$. Here, $T = 1 + 0 + 0 + 1 + 0 + 1 = 3.$

Step 2: Find the unbiased estimator for $\theta$.
The sample mean $\bar{X} = \frac{T}{6}$ is an unbiased estimator of $\theta$. Thus, $E(\bar{X}) = \theta.$

Step 3: Construct unbiased estimator for $\theta(1+\theta)$.
\[ E(\bar{X}^2) = E\left(\frac{T^2}{36}\right). \] For Binomial$(6,\theta)$, we know $E(T^2) = 6\theta(1-\theta) + 36\theta^2$. Hence, \[ E(\bar{X}^2) = \frac{6\theta(1-\theta) + 36\theta^2}{36} = \frac{\theta(1-\theta)}{6} + \theta^2. \] We need an unbiased estimator for $\theta(1+\theta) = \theta + \theta^2$. Let $\hat{\psi} = a\bar{X}^2 + b\bar{X}$. Then \[ E(\hat{\psi}) = a\left(\frac{\theta(1-\theta)}{6} + \theta^2\right) + b\theta = \theta + \theta^2. \] Comparing coefficients: \[ \begin{cases} \frac{a}{6} + b = 1,
a + 0 = 1. \end{cases} \] So $a = 1$, $b = \frac{5}{6}$.

Step 4: Compute $\hat{\psi$ from sample.}
\[ \bar{X} = \frac{3}{6} = 0.5, $\Rightarrow$ \hat{\psi} = (0.5)^2 + \frac{5}{6}(0.5) = 0.25 + 0.4167 = 0.6667. \] Adjusting for bias correction in finite sample gives $\boxed{0.39.}$