Understanding the Confidence Interval Problem

The problem asks us to find the relationship between the parameters $\lambda_1$, $\lambda_2$, and $\alpha$ for a specific confidence interval of $\theta$. We are given a random sample $X$ of size 1 drawn from the probability density function (PDF):

$ f(x|\theta) = \begin{cases} \frac{3}{\theta^3} (\theta - x)^2, & \text{if } 0<x<\theta, \\ 0, & \text{otherwise.} \end{cases} $

The proposed confidence interval for $\theta$ is $\left(\frac{X}{1-\lambda_1}, \frac{X}{1-\lambda_2}\right)$, with a confidence coefficient of $1 - \alpha$. We are given that $\lambda_i \in (0,1)$ for $i=1,2$, $\lambda_1 < \lambda_2$, and $\alpha \in (0,1)$.

Deriving the Distribution of a Transformed Variable

To determine the confidence interval, it's often helpful to work with a variable whose distribution is independent of the parameter $\theta$. Let's consider the transformation $Y = \frac{X}{\theta}$.

Since $0 < X < \theta$, the range for $Y$ is $0 < Y < 1$.

We can find the PDF of $Y$ using the transformation method. First, express $X$ in terms of $Y$: $X = Y\theta$. Then, find the Jacobian of the transformation: $\frac{dX}{dY} = \theta$.

The PDF of $Y$, denoted $f_Y(y)$, is given by:

$ f_Y(y) = f_X(x=y\theta|\theta) \left| \frac{dX}{dY} \right| $

Substituting $x = y\theta$ into the PDF of $X$:

$ f_X(y\theta|\theta) = \frac{3}{\theta^3} (\theta - y\theta)^2 = \frac{3}{\theta^3} \theta^2 (1 - y)^2 = \frac{3}{\theta^2} (1 - y)^2 $

Now, incorporating the Jacobian:

$ f_Y(y) = \left( \frac{3}{\theta^2} (1 - y)^2 \right) \cdot \theta = 3(1 - y)^2 $

So, the PDF of $Y$ is $f_Y(y) = 3(1 - y)^2$ for $0 < y < 1$. Importantly, this distribution does not depend on $\theta$.

Expressing the Confidence Interval Condition

A confidence interval $(L(X), U(X))$ for $\theta$ with confidence coefficient $1 - \alpha$ satisfies the condition:

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

In our case, $L(X) = \frac{X}{1-\lambda_1}$ and $U(X) = \frac{X}{1-\lambda_2}$. So we need:

$ P\left(\frac{X}{1-\lambda_1} < \theta < \frac{X}{1-\lambda_2}\right) = 1 - \alpha $

Let's rewrite this inequality in terms of $Y = \frac{X}{\theta}$. Divide the terms by $\theta$:

$ \frac{1}{1-\lambda_1} < \frac{X}{\theta^2} < \frac{1}{1-\lambda_2} $

This is not correct. Let's restart the inequality manipulation.

We have $\frac{X}{1-\lambda_1} < \theta$. Dividing by $\theta$ (which is positive) gives $\frac{X}{\theta(1-\lambda_1)} < 1$. This is still not simplifying well.

Let's invert the terms carefully.

Inequality 1: $ \frac{X}{1-\lambda_1} < \theta $

Multiply by $(1-\lambda_1)$ (which is positive since $\lambda_1 \in (0,1)$): $X < \theta(1-\lambda_1)$.

Divide by $\theta$ (which is positive): $ \frac{X}{\theta} < 1 - \lambda_1 $. This means $ Y < 1 - \lambda_1 $.

Inequality 2: $ \theta < \frac{X}{1-\lambda_2} $

Multiply by $(1-\lambda_2)$ (which is positive since $\lambda_2 \in (0,1)$): $\theta(1-\lambda_2) < X$.

Divide by $\theta$ (which is positive): $ 1 - \lambda_2 < \frac{X}{\theta} $. This means $ 1 - \lambda_2 < Y $.

Combining these, the condition for the confidence interval becomes:

$ P(1 - \lambda_2 < Y < 1 - \lambda_1) = 1 - \alpha $

Calculating the Probability Using the PDF of Y

Now, we calculate the probability $P(1 - \lambda_2 < Y < 1 - \lambda_1)$ using the PDF $f_Y(y) = 3(1 - y)^2$ for $0 < y < 1$.

$ P(1 - \lambda_2 < Y < 1 - \lambda_1) = \int_{1-\lambda_2}^{1-\lambda_1} 3(1 - y)^2 dy $

To evaluate this integral, let $u = 1 - y$. Then $du = -dy$. The limits of integration change as follows:

• When $y = 1 - \lambda_2$, $u = 1 - (1 - \lambda_2) = \lambda_2$.
• When $y = 1 - \lambda_1$, $u = 1 - (1 - \lambda_1) = \lambda_1$.

The integral becomes:

$ \int_{\lambda_2}^{\lambda_1} 3u^2 (-du) = -\int_{\lambda_2}^{\lambda_1} 3u^2 du $

Using the property $\int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx$:

$ = \int_{\lambda_1}^{\lambda_2} 3u^2 du $

Now, we compute the definite integral:

$ \int_{\lambda_1}^{\lambda_2} 3u^2 du = \left[ \frac{3u^3}{3} \right]_{\lambda_1}^{\lambda_2} = [u^3]_{\lambda_1}^{\lambda_2} $

$ = \lambda_2^3 - \lambda_1^3 $

Determining the Relationship

We found that the probability $ P\left(\frac{X}{1-\lambda_1} < \theta < \frac{X}{1-\lambda_2}\right) $ is equal to $ \lambda_2^3 - \lambda_1^3 $.

Since this probability must equal the confidence coefficient $1 - \alpha$, we have:

$ \lambda_2^3 - \lambda_1^3 = 1 - \alpha $

This equation establishes the required relationship between $\lambda_1$, $\lambda_2$, and $\alpha$ for the given interval to be a confidence interval for $\theta$ with coefficient $1 - \alpha$.