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Suppose \(\mathbf{X}\) is distributed \(N_{n}(\boldsymbol{\mu}, \mathbf{\Sigma}) .\) Let \(\bar{X}=n^{-1} \sum_{i=1}^{n} X_{i}\). (a) Write \(\bar{X}\) as aX for an appropriate vector a and apply Theorem \(3.5 .1\) to find the distribution of \(\bar{X}\). (b) Determine the distribution of \(\bar{X}\) if all of its component random variables \(X_{i}\) have the same mean \(\mu\).

Short Answer

Expert verified
In general, the distribution of the mean of a multivariate normal random variable, \(\bar{X}\), is \( N(\mu / n, \Sigma / n) \). If the variables all have the same mean \(\mu\), the distribution of \(\bar{X}\) is \( N(\mu, \Sigma / n) \).

Step by step solution

01

Write Mean as Linear Combination

The mean, \(\bar{X}\), of a random vector \(\mathbf{X}\) can be written as a linear combination of the entries of the vector. Hence, we can write \(\bar{X}\) as \( \mathbf{a^\intercal X}\), where \(\mathbf{X}\) is the random vector of \(n\) components and \(\mathbf{a}\) is a vector of \(1/n\) repeated \(n\) times.
02

Apply Theorem 3.5.1

Using theorem 3.5.1, if \(\boldsymbol{\mu}\) and \(\boldsymbol{\Sigma}\) are the mean vector and covariance matrix of \(\mathbf{X}\), and \(\mathbf{a}\) is the vector from step 1, then the mean of \(\bar{X}\) is \(\boldsymbol{\mu}\mathbf{a}\) and its variance is \(\mathbf{a}\boldsymbol{\Sigma}\mathbf{a}^\intercal\). Implementing this, we find \(\boldsymbol{\mu}\mathbf{a} = \mu / n\) and \(\mathbf{a}\boldsymbol{\Sigma}\mathbf{a}^\intercal = \Sigma / n\), given that all the elements of \(\mathbf{a}\) are \(1/n\) and the components of \(\mathbf{X}\) are independent.
03

Determine Distribution with Same Component Mean

In the scenario where all component random variables of \(\mathbf{X}\) have the same mean \(\mu\), the mean of \(\bar{X}\) would also be \(\mu\), given that the mean of a set of numbers is not affected by their order. The variance would be \(\Sigma / n\), as from step 2. Thus the distribution of \(\bar{X}\) in this case is also \( N(\mu, \Sigma / n) \).

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Key Concepts

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

Linear Combination
A linear combination is an expression made up of sums and scalar multiplications. In the context of multivariate normal distributions, it often involves combining multiple random variables to create a new variable. When we talk about \(\bar{X} = n^{-1} \sum_{i=1}^{n} X_{i}\), we're essentially creating a linear combination of the components of \(\mathbf{X}\). Here, \(\bar{X}\) represents the arithmetic mean of the components, expressed as \(\mathbf{a^\intercal X}\).
  • \(\mathbf{a}\) is a vector where each element is \(1/n\).
  • You multiply each component of \(\mathbf{X}\) by \(1/n\) and sum these up.
This formulation not only simplifies calculations but is crucial when finding the distribution of \(\bar{X}\). By re-expressing \(\bar{X}\) as a linear combination, we can apply various mathematical rules (like Theorem 3.5.1) to deduce properties such as its mean and variance.
Covariance Matrix
The covariance matrix \(\mathbf{\Sigma}\) captures how each pair of elements in the random vector \(\mathbf{X}\) interact with each other. Each entry \(\sigma_{ij}\) in the matrix gives the covariance between the \(i\text{th}\) and \(j\text{th}\) elements of \(\mathbf{X}\). This concept is vital when analyzing the spread or variability of the data represented by \(\mathbf{X}\).
  • If \(\mathbf{X}\) has independent components, off-diagonal entries (\(\sigma_{ij}, i eq j\)) will be zero.
  • The diagonal entries represent the variance of each component.
When we apply a linear transformation on \(\mathbf{X}\), like forming \(\bar{X}\) as \(\mathbf{a^\intercal X}\), the variance of \(\bar{X}\) can be calculated using the formula \(\mathbf{a^\intercal \Sigma a}\).This is essential for finding the variance of \(\bar{X}\) once it's expressed as a linear combination.
Mean Vector
A mean vector \(\boldsymbol{\mu}\) in a multivariate distribution contains the expected values (means) of each of the random variables included in \(\mathbf{X}\). For a random vector \(\mathbf{X}\) with \(n\) components, the mean vector is \((\mu_{1}, \mu_{2}, \ldots, \mu_{n})\).1. **Importance**: - This vector indicates the central tendency of the entire multivariate distribution.2. **Affect on \(\bar{X}\)**: - When calculating \(\bar{X}\) as \(\mathbf{a^\intercal X}\), the mean of \(\bar{X}\) becomes \(\mathbf{a^\intercal \boldsymbol{\mu}}\).This simplifies to \(\mu/n\) if all elements of \(\boldsymbol{\mu}\) are the same, signifying that \(\bar{X}\) has the same mean as any individual component of \(\mathbf{X}\) under certain conditions. Knowing the mean vector is crucial in identifying the intrinsic characteristics of \(\bar{X}\).

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