Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid with the distribution \(N\left(\theta, \sigma^{2}\right),-\infty<\theta<\infty\). Prove that a necessary and sufficient condition that the statistics \(Z=\sum_{1}^{n} a_{i} X_{i}\) and \(Y=\sum_{1}^{n} X_{i}\), a complete sufficient statistic for \(\theta\), are independent is that \(\sum_{1}^{n} a_{i}=0 .\)

Short answer

The necessary and sufficient condition for the statistics \(Z=\sum_{1}^{n} a_{i} X_{i}\) and \(Y=\sum_{1}^{n} X_{i}\) to be independent is fulfilled when \(\sum_{1}^{n} a_{i}=0 .\)

Step-by-step solution

Write down the joint pdf of \(X_1, X_2, ..., X_n\)

The joint pdf of \(X_1, X_2, ..., X_n\) is: \[ f(x_1,x_2, ..., x_n) = \frac{1}{(\sqrt{2\pi}\sigma)^n} \exp{-\frac{1}{2\sigma^2}\sum_{i=1}^n (x_i - \theta)^2} \]

Express the joint pdf in terms of Y and Z

The joint pdf can also be expressed in terms of \(Y\) and \(Z\) as: \[ f(y,z) = \exp{-\frac{1}{2\sigma^2}(y - n\theta)^2} \cdot \exp{-\frac{1}{2\sigma^2}(z - ka)^2} \] Here, \(k = \sum_{i=1}^n a_i\) since each \(X_i\) is multiplied by a corresponding \(a_i\) to get \(Z\).

Determine the condition for independence

Now, we want to find when the joint pdf factors into a product of a function of \(Y\) and another function of \(Z\) - this is the condition for \(Y\) and \(Z\) to be independent. For this to happen, we need to remove the second exponential term. This can be achieved by setting \(k = \sum_{i=1}^n a_i = 0\). With this condition, the second exponential term becomes one and the joint pdf factors into a product of a function of \(Y\) and a constant (which is a function of \(Z\)). Thus, \(Y\) and \(Z\) are independent if and only if \(\sum_{i=1}^n a_i = 0\).

Key concepts

Independence of Random Variables

In statistics, two random variables are said to be independent if the occurrence of one does not affect the probability of occurrence of the other. Essentially, knowledge of one variable provides no information about the other. In mathematical terms, random variables \( Y \) and \( Z \) are independent if their joint probability density function (pdf) can be expressed as a product of their individual pdfs.

For example, if \( f(y,z) = g(y) \cdot h(z) \), where \( g(y) \) and \( h(z) \) are the marginal pdfs of \( Y \) and \( Z \) respectively, then \( Y \) and \( Z \) are independent. An important criterion for independence is that any interaction term between \( Y \) and \( Z \) vanishes, leaving a situation where the effect of one variable does not influence the distribution of the other. This was seen clearly in our solution where resolving the second exponential term by setting \( k = \sum_{i=1}^n a_i = 0 \) allowed the joint distribution to break into separate terms for \( Y \) and \( Z \).

Recognizing and proving independence, as in this exercise, can allow statisticians to simplify complex problems significantly, breaking them down into more manageable independent parts.

Normal Distribution

The normal distribution, also known as the Gaussian distribution, is a key concept in statistics. It describes how values of a variable are distributed when influenced by a combination of random, independent effects. Its probability density function (pdf) is characterized by a symmetric, bell-shaped curve centered around the mean \( \theta \), with its spread determined by the standard deviation \( \sigma \).

Mathematically, if \( X \) is a normally distributed random variable with mean \( \theta = 0 \) and variance \( \sigma^2 \), its pdf is given by:
\[f(x) = \frac{1}{\sqrt{2\pi}\sigma} \exp\left(-\frac{(x- \theta)^2}{2\sigma^2}\right)\]
Normal distribution plays a vital role in the field of statistics because of its unique properties, such as the fact that the sum of a large number of independent and identically distributed random variables (by the central limit theorem) tends to be normally distributed, regardless of the original distribution. This makes it incredibly useful for sampling and hypothesis testing.

In our exercise, understanding the properties of normal distribution helped us to express the joint pdf and subsequently analyze the independence of \( Y \) and \( Z \). Whether you're calculating means, variances, or other statistical quantities, grasping normal distribution is crucial.

Joint Probability Density Function

The joint probability density function (pdf) of two or more random variables provides a complete description of their probability structure. It specifies the likelihood that the random variables simultaneously take certain values. For instance, for random variables \( X_1, X_2, ..., X_n \), the joint pdf is formulated by considering the product of the individual pdfs, assuming they are independent or have a known correlation structure.

In our exercise, we looked at the joint pdf of \( X_1, X_2, ..., X_n \), which are normally distributed random variables. The joint pdf was expressed as:
\[f(x_1,x_2,...,x_n) = \frac{1}{(\sqrt{2\pi}\sigma)^n} \exp\left(-\frac{1}{2\sigma^2}\sum_{i=1}^n (x_i - \theta)^2\right)\]
This formula reflects how the values of \( X_1, X_2, ..., X_n \) are jointly distributed. By transforming this joint pdf to describe the distribution of \( Y \) and \( Z \), which are linear combinations of the \( X_i \), a new joint pdf \( f(y,z) \) was derived.

Understanding and working with joint pdfs is crucial when dealing with transformations of normal variables or when investigating dependence structures, as it enables deriving the statistical properties of newly defined variables like \( Y \) and \( Z \). Mastery of joint pdfs allows statisticians to perform analyses accurately and draw insightful conclusions from complex data sets.