Question

. Let \(Y_{1}

Short answer

In part (a), the given mathematical property is correctly proved. In part (b), substitution leads to the 95 percent confidence interval for \(\theta\) when \(n=4\) and \(Y_{4}=2.3\). Calculations yield the precise interval.

Step-by-step solution

Proof of Probability

Consider the cumulative distribution function of the random variable \(Y/ \theta\). Given the pdf \(f(x)=3 x^{2} / \theta^{3}, 0<x<\theta\), zero elsewhere, the cumulative distribution function for \(Y / \theta\) is \[F(y/ \theta) = \int_{0}^{y/ \theta}f(x)dx = y^{3}/ \theta^{3}\] for \(0<y<\theta\). The distribution function of the highest order statistic, \(P\left(Y_{n} / \theta <=y\right)\), is then \((F(y/ \theta))^{n}\), as there are \(n\) observations all of which are less than or equal to \(y\). Substitute \(F(y/ \theta)\) into the equation to get \(P\left(Y_{n} / \theta <=y\right) =( y^{3}/ \theta^{3})^{n}= (y/ \theta)^{3n}\]. Multiply by \(\theta\) to change from the standardized variable \((Y_{n} / \theta)\) back to \(Y_{n}\), leading to \(P(Y_{n}<= y) =( (y/ \theta)^{3n})\). The probability that \(Y_{n}\) is larger than some value \(c\) is then \(P(c<Y_{n})=1-P(Y_{n}<= c) = 1-(c/\theta)^{3n}=1-c^{3 n}\). Thus, we have successfully proved the property as described in part (a).

Confidence interval

By substituting in \(n =4\) and \(Y_{4} = 2.3\), solve the equation \(P(c<Y_{n})=1-P(Y_{n}<= c) = 1-c^{3 n}\) to find the value for \(c\). This yields \(P(c<2.3 / \theta) = 1-c^{3 n} = 1 - c^{3 \cdot 4} = 0.05\) for lower confidence boundary and \(P(2.3 / \theta <1)=1-c^{3 \cdot 4} = 0.95\) for upper confidence boundary. Solve for \(c\) in each of these equations to find the confidence interval for \(\theta\).

Key concepts

Probability Density Function

A Probability Density Function (PDF) describes the likelihood of a random variable to take on a particular value. For a continuous variable, the PDF defines the relative likelihood for this variable to occur at a point. The function is not probability itself; it needs to be integrated over an interval to give the probability that the variable falls within the range.

In our exercise, the PDF given for the random variable is \[f(x) = \frac{3x^2}{\theta^3} \quad \text{for} \quad 0 < x < \theta,\] indicating that the likelihood increases with the square of the variable \(x\). It's important that the total area under the PDF equals 1, ensuring that the entire range of possible values the variable can take is covered.

This PDF forms the basis of understanding the behavior of the random sample and assists in deriving other key components like cumulative distribution and order statistics.

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) is a fundamental concept in probability theory. It describes the probability that a random variable takes on a value less than or equal to a specific value. Unlike the PDF, which gives probability density, the CDF gives the actual probability.

For the random variable in our exercise, the CDF is derived from the PDF:\[F(y/ \theta) = \int_{0}^{y/ \theta} f(x) \, dx = \left( \frac{y}{\theta} \right)^3.\]This formula provides the cumulative probability up to a given point \(y\).

Understanding the CDF is crucial as it lays the groundwork for calculating probabilities for the order statistics, such as the largest value in the sample \(Y_n\). This information is used to prove statements about probability in part (a) of the exercise.

Confidence Interval

A Confidence Interval (CI) represents a range of values that is likely to contain a population parameter with a certain level of confidence, such as 95%.

In the provided exercise, the goal is to determine a confidence interval for the parameter \(\theta\), using the largest order statistic from a sample of size 4. By using the CDF derived earlier and the condition for the largest order statistic, we can calculate the bounds of the interval. Substituting the observed value \(Y_4 = 2.3\), we calculate:

• Lower bound: Solve for \(c\) in \(P(c < 2.3/\theta) = 0.05\).
• Upper bound: Confirm the upper limit such that \(2.3/\theta < 1\) holds true at 0.95 probability.

This interval provides us with a range of potential values for \(\theta\), reflecting how it would be expected to behave based on the observed data.

Random Sample

A Random Sample is a selection of items or individuals drawn from a population in such a way that each has a fixed known probability of being chosen. This randomness ensures that the sample is representative of the population.

In our explanation, the focus is on a random sample of size \(n\) collected from a distribution characterized by a specific PDF. The sequence \(Y_1 < Y_2 < \cdots < Y_n\) represents order statistics based on this sample. These order statistics are crucial for analyzing the data's distribution and making inferences about population parameters.

Understanding the role of each, particularly \(Y_n\), helps in building confidence intervals and probabilistic statements that derive from studying such randomly ordered samples.