Estimating Theta from Sample Data

The problem asks us to evaluate several statements about estimators for the unknown parameter $\theta$ based on a random sample $X_1, X_2, \dots, X_n$ from a distribution with the given probability density function (PDF):

$$ f_\theta(x) = \begin{cases} \frac{2(\theta - x)}{\theta^2}, & 0 \le x \le \theta \\ 0, & \text{otherwise} \end{cases} $$

We need to check the unbiasedness and sufficiency of the sample mean $\bar{X}_n = \frac{1}{n} \sum_{i=1}^n X_i$ and the maximum order statistic $X_{(n)} = \max(X_1, X_2, \dots, X_n)$.

Expected Value of a Single Observation

To check for unbiasedness, we first need to calculate the expected value of a single observation $X_i$ from this distribution:

$$ E[X_i] = \int_{-\infty}^{\infty} x f_\theta(x) dx $$

Using the given PDF, the integral is over the range $[0, \theta]$:

$$ E[X_i] = \int_{0}^{\theta} x \frac{2(\theta - x)}{\theta^2} dx $$

We can take the constant term $\frac{2}{\theta^2}$ out of the integral:

$$ E[X_i] = \frac{2}{\theta^2} \int_{0}^{\theta} (x\theta - x^2) dx $$

Now, we integrate term by term:

$$ E[X_i] = \frac{2}{\theta^2} \left[ \frac{x^2 \theta}{2} - \frac{x^3}{3} \right]_{0}^{\theta} $$

Evaluate the expression at the limits of integration:

$$ E[X_i] = \frac{2}{\theta^2} \left( \left( \frac{\theta^2 \theta}{2} - \frac{\theta^3}{3} \right) - \left( \frac{0^2 \theta}{2} - \frac{0^3}{3} \right) \right) $$

$$ E[X_i] = \frac{2}{\theta^2} \left( \frac{\theta^3}{2} - \frac{\theta^3}{3} \right) $$

Combine the terms within the parenthesis:

$$ E[X_i] = \frac{2}{\theta^2} \left( \frac{3\theta^3 - 2\theta^3}{6} \right) = \frac{2}{\theta^2} \left( \frac{\theta^3}{6} \right) $$

Simplify the expression:

$$ E[X_i] = \frac{2 \theta^3}{6 \theta^2} = \frac{\theta}{3} $$

So, the expected value of a single observation is $\frac{\theta}{3}$.

Expected Value of the Sample Mean

The sample mean is $\bar{X}_n = \frac{1}{n} \sum_{i=1}^n X_i$. Using the linearity of expectation, the expected value of the sample mean is:

$$ E[\bar{X}_n] = E\left[\frac{1}{n} \sum_{i=1}^n X_i\right] = \frac{1}{n} \sum_{i=1}^n E[X_i] $$

Since each $X_i$ comes from the same distribution, $E[X_i]$ is the same for all $i$, which we found to be $\frac{\theta}{3}$.

$$ E[\bar{X}_n] = \frac{1}{n} \sum_{i=1}^n \frac{\theta}{3} = \frac{1}{n} \left( n \cdot \frac{\theta}{3} \right) = \frac{\theta}{3} $$

Checking Unbiasedness of 3 times the Sample Mean

An estimator $T$ for a parameter $\psi$ is unbiased if $E[T] = \psi$. We want to check if $3\bar{X}_n$ is an unbiased estimator for $\theta$. We calculate its expected value:

$$ E[3\bar{X}_n] = 3 E[\bar{X}_n] $$

Using the expected value of the sample mean we just calculated:

$$ E[3\bar{X}_n] = 3 \left( \frac{\theta}{3} \right) = \theta $$

Since $E[3\bar{X}_n] = \theta$, the statistic $3\bar{X}_n$ is an unbiased estimator for $\theta$. This confirms that statement 3 is correct.

Sufficiency of Statistics

A statistic $T(X_1, \dots, X_n)$ is sufficient for a parameter $\theta$ if the conditional distribution of the sample $X_1, \dots, X_n$ given $T(X_1, \dots, X_n) = t$ does not depend on $\theta$. The Factorization Theorem states that $T$ is sufficient for $\theta$ if and only if the likelihood function $L(\theta | x_1, \dots, x_n) = \prod_{i=1}^n f_\theta(x_i)$ can be factored into two non-negative functions, $g$ and $h$, such that $L(\theta | x_1, \dots, x_n) = g(T(x_1, \dots, x_n) | \theta) h(x_1, \dots, x_n)$, where $h(x_1, \dots, x_n)$ does not depend on $\theta$.

The joint density (likelihood function) is:

$$ L(\theta | x_1, \dots, x_n) = \prod_{i=1}^n \frac{2(\theta - x_i)}{\theta^2} \cdot I(0 \le x_i \le \theta \text{ for all } i) $$

The indicator function $I(0 \le x_i \le \theta \text{ for all } i)$ is equivalent to $I(0 \le \min(x_i) \text{ and } \max(x_i) \le \theta)$, which is $I(x_{(1)} \ge 0) \cdot I(x_{(n)} \le \theta)$. Since $x_i \ge 0$ is given by the support, we focus on $I(x_{(n)} \le \theta)$.

$$ L(\theta | x_1, \dots, x_n) = \frac{2^n}{\theta^{2n}} \prod_{i=1}^n (\theta - x_i) \cdot I(x_{(n)} \le \theta) $$

The term $\prod_{i=1}^n (\theta - x_i)$ involves all the individual $x_i$ values. It is a polynomial in $\theta$ whose coefficients depend on elementary symmetric polynomials of $x_1, \dots, x_n$, such as $\sum x_i$, $\sum_{i<j} x_i x_j$, etc. The presence of these terms means the likelihood cannot generally be factored into a part depending only on $x_{(n)}$ and $\theta$, and another part independent of $\theta$. Thus, $X_{(n)}$ is not sufficient.

Similarly, the presence of symmetric polynomials of order higher than 1 (like $\sum_{i<j} x_i x_j$) in the expansion of $\prod_{i=1}^n (\theta - x_i)$ implies that the likelihood cannot be factored into a part depending only on $\sum X_i$ (or $\bar{X}_n$) and $\theta$, and another part independent of $\theta$. Thus, $3\bar{X}_n$ (or $\bar{X}_n$) is not sufficient.

Checking Unbiasedness of the Maximum Order Statistic

To check if $X_{(n)}$ is unbiased for $\theta$, we would need to calculate $E[X_{(n)}]$. This involves finding the PDF of $X_{(n)}$ and integrating $y$ multiplied by this PDF. The PDF of $X_{(n)}$ is $f_{X_{(n)}}(y) = n [F_\theta(y)]^{n-1} f_\theta(y)$, where $F_\theta(y)$ is the CDF of $X$. We found $F_\theta(y) = \frac{2y}{\theta} - \frac{y^2}{\theta^2} = \frac{y(2\theta - y)}{\theta^2}$ for $0 \le y \le \theta$.

$$ f_{X_{(n)}}(y) = n \left(\frac{y(2\theta - y)}{\theta^2}\right)^{n-1} \frac{2(\theta - y)}{\theta^2} \quad \text{for } 0 \le y \le \theta $$

Calculating $E[X_{(n)}] = \int_0^\theta y f_{X_{(n)}}(y) dy$ is complex. As shown in the thought process (or can be shown with detailed calculation for $n > 1$), $E[X_{(n)}]$ does not equal $\theta$. For instance, for $n=1$, $X_{(1)}=X_1$, and $E[X_{(1)}] = E[X_1] = \theta/3 \ne \theta$. For $n=2$, $E[X_{(2)}] = 7\theta/15 \ne \theta$. Thus, $X_{(n)}$ is not an unbiased estimator for $\theta$ for $n \ge 2$.

Conclusion

Based on our analysis:

• Statement 1: $X_{(n)}$ is sufficient for $\theta$. This is incorrect.
• Statement 2: $X_{(n)}$ is unbiased for $\theta$. This is incorrect.
• Statement 3: $3 \bar{X}_n$ is unbiased for $\theta$. This is correct, as $E[3\bar{X}_n] = \theta$.
• Statement 4: $3 \bar{X}_n$ is sufficient for $\theta$. This is incorrect.

Therefore, the correct statement is that $3 \bar{X}_n$ is unbiased for $\theta$.