# 5.1 - Distribution of Sample Mean Vector

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As noted previously $$\bar{x}$$ is a function the random data, and hence $$\bar{x}$$ is also a random vector with a mean, a variance-covariance matrix and a distribution. We have already seen that the mean of the sample mean vector is equal to the population mean vector μ.

#### Variance

Before considering the sample variance-covariance matrix for the mean vector $$\bar{x}$$, let us revisit the univariate setting.

Univariate Setting: You should recall from introductory statistics that the population variance of the sample mean, generated from independent samples of size n, is equal to the population variance, σ2 divided by n.

$\text{var}(\bar{x}) = \frac{\sigma^2}{n}$

This, of course, is a function of the unknown population variance σ2. We can estimate this by simply substituting s2 in the sample variance σ2 yielding our estimate for the variance of the population mean as shown below:

$\widehat{\text{var}}(\bar{x}) = \frac{s^2}{n}$

If we were to take the square root of this quantity we would obtain the standard error of the mean. The standard error of the mean is a measure of the uncertainty of our estimate of the population mean. If the standard error is large, then we are less confident of our estimate of the mean. Conversely, if the standard error is small, then we are more confident of our estimate. What is meant by large or small depends on the application at hand. But in any case, since the standard error is a decreasing function of sample size, the larger our sample the more confident we can be of our estimate of the population mean.

Multivariate Setting: The population variance-covariance matrix replaces the variance of the $$\bar{x}$$’s generated from independent samples of size n, taking a form similar to the univariate setting as shown below:

$\text{var}(\bar{\textbf{x}}) = \frac{1}{n}\Sigma$

Again, this is a function of the unknown population variance-covariance matrix Σ. An estimate of the variance-covariance matrix of $$\bar{x}$$ can be obtain by substituting the sample variance-covariance matrix S for the population variance-covariance matrix Σ, yielding the estimate as shown below:

$\widehat{\text{var}}(\bar{\textbf{x}}) = \frac{1}{n}\textbf{S}$

### Distribution

Let's consider the distribution of the sample mean vector, first looking at the univariate setting and comparing this to the multivariate setting.

Univariate Setting: Here we are going to make the additional assumption that X1, X2,...,Xn are independently sampled from a normal distribution with mean μ and variance σ2. In this case, $$\bar{x}$$ is normally distributed as

$\bar{x} \sim N\left(\mu, \frac{\sigma^2}{n}\right)$

Multivariate Setting: Similarly, for the multivariate setting, we are going to assume that the data vectors X1, X2,...,Xn are independently sampled from a multivariate normal distribution with mean vector μ and variance-covariance matrix Σ. Then, in this case, the sample mean vector, $$\bar{x}$$, is distributed as multivariate normal with mean vector μ and variance-covariance matrix $$\frac{1}{n}\Sigma$$, the variance-covariance matrix for $$\bar{x}$$. In statistical notation we write:

$\bar{\textbf{x}} \sim N \left(\mu, \frac{1}{n}\Sigma \right)$

### Law of Large Numbers

At this point we will drop the assumption that the individual observations are sampled from a normal distribution and look at laws of large numbers. These will hold regardless of the distribution of the individual observations.

Univariate Setting: In the univariate setting, we see that if the data are independently sampled, then the sample mean, $$\bar{x}$$, is going to converge (in probability) to the population mean μ. What does this mean exactly? It means that as the sample size gets larger and larger the sample mean will tend to approach the true value for a population μ.

Multivariate Setting: A similar result is involved in the multivariate setting , the sample mean vector, $$\bar{x}$$, will also converge (in probability) to the mean vector μ;. As our sample size gets larger and larger, each of the individual components of that vector, $$\bar{x}_{j}$$, will converge to the corresponding mean, μj.

$\bar{x}_j \stackrel{p}\rightarrow \mu_j$

### Central Limit Theorem

Just as in the univariate setting we also have a multivariate Central Limit Theorem. But first, let's review the univariate Central Limit Theorem.

Univariate Setting. If all of our individual observations, X1, X2,...,Xn, are independently sampled from a population with mean μ and variance σ2, then, the sample mean, $$\bar{x}$$, is approximately normally distributed with mean μ and variance σ2/n. Note that in the distribution property described above, normality was a requirement. Under normality, even for small samples, the data are normally distributed. The Central Limit Theorem is a more general result which holds regardless of the distribution of the original data. Significance of CLT lies in the fact that the sample mean is approximately normally distributed for large samples whatever be the distribution of the individual observations.

Multivariate Setting. A similar result is available in the multivariate setting. If our data vectors X1, X2,...,Xn, are independently sampled from a population with mean vector μ and variance-covariance matrix Σ, then the sample mean vector, $$\bar{x}$$, is going to be approximately normally distributed with mean vector μ and variance-covariance matrix  $$\frac{1}{n}\Sigma$$.

This Central Limit Theorem is a key result that we will take advantage of later on in this course when we talk about hypothesis tests about individual mean vectors or collections of mean vectors under different treatment regimens.