Confidence Interval Derivation for Theta

The problem requires finding confidence intervals for the parameter θ given a sample $X$ drawn from a population with a specific probability density function (pdf).

The pdf is provided as:

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

The support of the distribution, $\theta \le x \le 2\theta$, implies constraints on θ relative to $x$. Specifically, $x \ge \theta$ and $x \le 2\theta$, which translates to $\frac{x}{2} \le \theta \le x$.

Pivotal Quantity Identification

To construct confidence intervals, we identify a pivotal quantity—a function of the sample $X$ and parameter θ whose probability distribution is independent of θ. Consider the transformation $Y = \frac{X}{\theta}$.

The range of $Y$ is determined by the support of $X$. Since $1 \le \frac{X}{\theta} \le 2$, the range for $Y$ is $1 \le y \le 2$. By applying the change of variables technique, the pdf of $Y$ is derived:

$f_Y(y) = 2(y-1), \quad \text{for } 1 \le y \le 2$

Since the distribution of $Y$ does not depend on θ, $Y = \frac{X}{\theta}$ serves as a pivotal quantity.

Cumulative Distribution Function (CDF) of Y

The cumulative distribution function (CDF) for the pivotal quantity $Y$ is calculated by integrating its pdf:

$F_Y(y) = \int_1^y f_Y(t) dt = \int_1^y 2(t-1) dt = [t^2 - 2t]_1^y$

$F_Y(y) = (y^2 - 2y) - (1^2 - 2(1)) = y^2 - 2y + 1 = (y-1)^2$

This CDF is valid for $1 \le y \le 2$.

Confidence Interval Construction Principle

A confidence interval for θ, denoted as $[L(X), U(X)]$, with a confidence coefficient of $1 - \alpha$ must satisfy:

$P(L(X) \le \theta \le U(X)) = 1 - \alpha$

This inequality can be transformed into an equivalent statement about the pivotal quantity $Y = \frac{X}{\theta}$:

$P\left(\frac{X}{U(X)} \le Y \le \frac{X}{L(X)}\right) = 1 - \alpha$

Using the CDF $F_Y(y)$, the condition becomes:

$F_Y\left(\frac{X}{L(X)}\right) - F_Y\left(\frac{X}{U(X)}\right) = 1 - \alpha$

Verification of Confidence Intervals

Each provided option is checked against the derived condition $F_Y\left(\frac{X}{L(X)}\right) - F_Y\left(\frac{X}{U(X)}\right) = 1 - \alpha$.

Option A Analysis

The interval is $\left[ \frac{x}{2}, \frac{x}{1+\sqrt{\alpha}} \right]$. Thus, $L(X) = \frac{X}{2}$ and $U(X) = \frac{X}{1+\sqrt{\alpha}}$.

The corresponding interval for $Y$ is $\left[ \frac{X}{U(X)}, \frac{X}{L(X)} \right] = \left[ \frac{X}{X/(1+\sqrt{\alpha})}, \frac{X}{X/2} \right] = \left[ 1+\sqrt{\alpha}, 2 \right]$.

Checking the probability using the CDF:

$F_Y(2) - F_Y(1+\sqrt{\alpha}) = (2-1)^2 - ((1+\sqrt{\alpha})-1)^2 = 1^2 - (\sqrt{\alpha})^2 = 1 - \alpha$

This interval satisfies the condition.

Option B Analysis

The interval is $\left[ \frac{x}{1+\sqrt{1-\frac{\alpha}{2}}}, \frac{x}{1+\sqrt{\frac{\alpha}{2}}} \right]$. Here, $L(X) = \frac{X}{1+\sqrt{1-\frac{\alpha}{2}}}$ and $U(X) = \frac{X}{1+\sqrt{\frac{\alpha}{2}}}$.

The corresponding interval for $Y$ is $\left[ \frac{X}{U(X)}, \frac{X}{L(X)} \right] = \left[ 1+\sqrt{\frac{\alpha}{2}}, 1+\sqrt{1-\frac{\alpha}{2}} \right]$.

Checking the probability:

$F_Y\left(1+\sqrt{1-\frac{\alpha}{2}}\right) - F_Y\left(1+\sqrt{\frac{\alpha}{2}}\right) = \left(\sqrt{1-\frac{\alpha}{2}}\right)^2 - \left(\sqrt{\frac{\alpha}{2}}\right)^2$

$= \left(1-\frac{\alpha}{2}\right) - \frac{\alpha}{2} = 1 - \alpha$

This interval satisfies the condition.

Option C Analysis

The interval is $\left[ \frac{x}{1+\sqrt{1-\alpha}}, x \right]$. Here, $L(X) = \frac{X}{1+\sqrt{1-\alpha}}$ and $U(X) = X$.

The corresponding interval for $Y$ is $\left[ \frac{X}{U(X)}, \frac{X}{L(X)} \right] = \left[ \frac{X}{X}, \frac{X}{X/(1+\sqrt{1-\alpha})} \right] = \left[ 1, 1+\sqrt{1-\alpha} \right]$.

Checking the probability:

$F_Y(1+\sqrt{1-\alpha}) - F_Y(1) = \left(\sqrt{1-\alpha}\right)^2 - (1-1)^2 = 1 - \alpha - 0 = 1 - \alpha$

This interval satisfies the condition.

Option D Analysis

The interval is $\left[ \frac{x}{1+\sqrt{1-\frac{\alpha}{4}}}, \frac{x}{1+\sqrt{\frac{3\alpha}{4}}} \right]$. Thus, $L(X) = \frac{X}{1+\sqrt{1-\frac{\alpha}{4}}}$ and $U(X) = \frac{X}{1+\sqrt{\frac{3\alpha}{4}}}$.

The corresponding interval for $Y$ is $\left[ \frac{X}{U(X)}, \frac{X}{L(X)} \right] = \left[ 1+\sqrt{\frac{3\alpha}{4}}, 1+\sqrt{1-\frac{\alpha}{4}} \right]$.

Checking the probability:

$F_Y\left(1+\sqrt{1-\frac{\alpha}{4}}\right) - F_Y\left(1+\sqrt{\frac{3\alpha}{4}}\right) = \left(\sqrt{1-\frac{\alpha}{4}}\right)^2 - \left(\sqrt{\frac{3\alpha}{4}}\right)^2$

$= \left(1-\frac{\alpha}{4}\right) - \frac{3\alpha}{4} = 1 - \frac{4\alpha}{4} = 1 - \alpha$

This interval satisfies the condition.

All four options represent valid confidence intervals for θ with the specified confidence coefficient $1 - \alpha$, as they all correspond to a probability of $1 - \alpha$ for the pivotal quantity $Y$.