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Chapter 10: Problem 37

Let \(Y_{1}

Short Answer

Expert verified
The ratio of the joint p.d.f of the random sample to that of the order statistics is \(1 / n!\), irrespective of the underlying p.d.f. This suggests that the order statistics are joint sufficient statistics for the unknown parameter 'f'.

Step by step solution

01

Joint p.d.f of Original Sample

The original sample \(X_1, X_2, ..., X_n\) has a joint probability density function. Since the sample variables are independent and identically distributed (i.i.d), this is a product of the individual probability densities: \[ f(x_1, x_2, ..., x_n) = f(x_1) f(x_2) \cdots f(x_n) \]
02

Joint p.d.f of Order Statistics

The joint p.d.f. for the order statistics \(Y_1 < Y_2 < ... < Y_n\) is given by the formula: \[ f(y_1, y_2, ..., y_n) = n! f(y_1) f(y_2) \cdots f(y_n) \] The \(n!\) factor comes from the number of permutations of \(n\) objects, which represents all possible ways the order statistics can be arranged. All \(n!\) configuration(s) are equally likely.
03

Calculation of the ratio

The ratio of the joint p.d.f of the original sample and the joint p.d.f of the order statistics is given by: \[ \frac {f(x_1, x_2, ..., x_n)} {f(y_1, y_2, ..., y_n)} = \frac {f(x_1) f(x_2) \cdots f(x_n)} {n! f(y_1) f(y_2) \cdots f(y_n)} = \frac {1} {n!} \] This concludes that the ratio is indeed \(1 / n!\) which is independent of the underlying p.d.f 'f'.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Joint Probability Density Function
The joint probability density function is a way of expressing the probability that a set of random variables falls within a particular range of values. Imagine you're looking at not just one, but several variables and you want to comprehend how they might "behave" together. For a setup where we have random variables ordered as \(Y_1 < Y_2 < \, ... < Y_n\), finding their joint probability density function helps us understand how likely particular values are when these variables are considered together.

In the context of the exercise, the joint p.d.f. of the order statistics \(Y_1 < Y_2 < ... < Y_n\) from a sample \(X_1, X_2, ..., X_n\) is given by \(n! f(y_1) f(y_2) \cdots f(y_n)\). The \(n!\) arises due to permutations, representing all possible ways these variables can be rearranged; hence, all are equally probable. Understanding this joint p.d.f. is crucial when assessing the distribution's behavior and the data's overall shape.
Sufficient Statistics
Sufficient statistics are values that summarize all the necessary information about a data set with respect to parameter estimation. Essentially, if you have these statistics, you don鈥檛 need to know anything else about your sample to make inferences about the parameter.

In the exercise, the order statistics \(Y_1
Independent and Identically Distributed (i.i.d.)
Independent and identically distributed (i.i.d.) describes a collection of random variables that share the same probability distribution and are mutually independent, meaning that the outcome of one variable does not influence the others. This concept is foundational in probability and statistics because it allows for simplifications in calculations and proofs.

In our exercise, it鈥檚 assumed that the sample \(X_1, X_2, ..., X_n\) is i.i.d. This property is crucial because it implies that the joint probability density function of the sample can be computed as the product of individual probabilities \[f(x_1) f(x_2) \cdots f(x_n)\]. This independence among variables simplifies our understanding of the sample's behavior, as it eliminates additional complexities that might arise if the variables were correlated.

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Most popular questions from this chapter

Let \(Y_{1}

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample of size \(n>2\) from a distribution with p.d.f. \(f(x ; \theta)=\theta e^{-\theta x}, 00 .\) Then \(Y=\sum_{1}^{n} X_{1}\) is a sufficient statistic for \(\theta .\) Prove that \((n-1) / Y\) is the best estimator of \(\theta\).

Let the p.d.f. \(f\left(x ; \theta_{1}, \theta_{2}\right)\) be of the form $$ \exp \left[p_{1}\left(\theta_{1}, \theta_{2}\right) K_{1}(x)+p_{2}\left(\theta_{1}, \theta_{2}\right) K_{2}(x)+S(x)+q\left(\theta_{1}, \theta_{2}\right)\right], \quad a

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample of size \(n\) from a beta distribution with parameters \(\alpha=\theta>0\) and \(\beta=2\). Show that the product \(X_{1} X_{2} \cdots X_{n}\) is a sufficient statistic for \(\theta\).

Let \(X\) be a random variable with a p.d.f. of a regular case of the exponential class. Show that \(E[K(X)]=-q^{\prime}(\theta) / p^{\prime}(\theta)\), provided these derivatives exist, by differentiating both members of the equality $$ \int_{a}^{b} \exp [p(\theta) K(x)+S(x)+q(\theta)] d x=1 $$ with respect to \(\theta .\) By a second differentiation, find the variance of \(K(X)\).
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