## 7.7 Sampling distributions of $$\hat{\mu}$$ and $$\hat{\Sigma}$$

Consider the GWN model in matrix form (7.1). The estimator of the $$N \times 1$$ matrix $$\mu$$ is the sample mean vector (7.16) and the estimator of the $$N \times N$$ matrix $$\Sigma$$ is the sample covariance matrix (7.17). In this section we give the sampling distributions for the mean vector $$\hat{\mu}$$ and for the elements of the covariance matrix $$\hat{\Sigma}$$.

### 7.7.1 Sampling distribution of $$\hat{\mu}$$

The following proposition gives the sampling distribution of the $$N \times 1$$ vector $$\hat{\mu}$$.

Proposition 7.8 (Sampling distribution of $$\hat{\mu}$$) Under the GWN model and for large enough T, the sampling distribution for $$\hat{\mu}$$ is multivariate normal:

$\begin{equation} \hat{\mu} \sim N\left(\mu, ~ \frac{1}{T}\hat{\Sigma}\right). \end{equation}$

This result shows that the elements of the vector $$\hat{\mu}$$ are correlated, with the correlations being the same as the asset return correlations.

Example 3.9 (Joint distribution of $$\hat{\mu}_1$$ and $$\hat{\mu}_2$$)

Consider the bivariate case ($$N=2$$). Proposition 7.8 tells us the joint distribution between $$\hat{\mu}_1$$ and $$\hat{\mu}_2$$ is bivariate normal:

$\begin{equation} \left( \begin{array}{c} \hat{\mu}_1 \\ \hat{\mu}_2 \end{array} \right) \sim N\left( \left(\begin{array}{c} \mu_1 \\ \mu_2 \end{array} \right), ~ \frac{1}{T} \left( \begin{array}{cc} \hat{\sigma}_1^2 & \hat{\sigma}_{12} \\ \hat{\sigma}_{12} & \hat{\sigma}_{2}^{2} \end{array} \right) \right). \end{equation}$

Hence, $$\widehat{\mathrm{cov}}(\hat{\mu}_1, \hat{\mu}_2) = \frac{1}{T}\hat{\sigma}_{12}$$ and $$\widehat{\mathrm{cor}}(\hat{\mu}_1, \hat{\mu}_2)=\hat{\rho}_{12}$$.

The fact that the elements of $$\hat{\mu}$$ are correlated and the correlations are equal to the correlations between the corresponding assets is a useful result. For example, if assets 1 and 2 are strongly positively (negatively) correlated, then $$\hat{\mu}_1$$ and $$\hat{\mu}_2$$ are strongly positively (negatively) correlated. As a result, a positive (negative) estimate of $$\hat{\mu}_1$$ will tend to be associated with a positive (negative) estimate of $$\hat{\mu}_2$$.

$$\blacksquare$$

### 7.7.2 Sampling distribution of the elements of $$\hat{\Sigma}$$

We can also determine the sampling distribution for the vector of unique elements of $$\hat{\Sigma}$$. This requires a bit of new matrix notation because the joint distribution of a $$N \times N$$ symmetric matrix of random variables is defined as the joint distribution of the $$N(N+1)/2$$ elements of the matrix stacked into a $$N(N+1)/2 \times 1$$ vector.

Define $$\hat{\theta}$$ as the $$N(N+1)/2 \times 1$$ vector of unique elements of $$\Sigma$$. These elements can be extracted using the $$\mathrm{vech}(\cdot)$$ operator, which extracts the elements on and below the main diagonal of $$\hat{\Sigma}$$ column-wise and stacks them into a $$N(N+1)/2$$ vector. For example, if $$N=3$$ then $$\hat{\theta}$$ is the $$6 \times 1$$ vector

$\begin{equation} \hat{\theta} = \mathrm{vech}(\hat{\Sigma}) = \mathrm{vech} \left(\begin{array}{ccc} \hat{\sigma}_1^2 & \hat{\sigma}_{12} & \hat{\sigma}_{13} \\ \hat{\sigma}_{12} & \hat{\sigma}_2^2 & \hat{\sigma}_{23} \\ \hat{\sigma}_{13} & \hat{\sigma}_{23} & \hat{\sigma}_{3}^2 \end{array} \right) = \left(\begin{array}{c} \hat{\sigma}_1^2 \\ \hat{\sigma}_{12}\\ \hat{\sigma}_{13}\\ \hat{\sigma}_2^2\\ \hat{\sigma}_{23}\\ \hat{\sigma}_{3}^2 \end{array} \right). \end{equation}$

Proposition 7.9 (Sampling distribution of the unique elements of $$\hat{\Sigma}$$) Under the GWN model and for large enough T, the sampling distribution for the $$N(N+1)/2 \times 1$$ vector $$\hat{\theta} = \mathrm{vech}(\hat{\Sigma})$$ is multivariate normal:

$\begin{equation} \hat{\theta} \sim N\left(\theta, ~ \frac{1}{T}\hat{\Omega}\right). \end{equation}$

where $$\frac{1}{T}\hat{\Omega}$$ is an $$N(N+1)/2 \times N(N+1)/2$$ matrix giving the estimated variances and covariances of the unique elements of $$\hat{\Sigma}$$. The element of $$\frac{1}{T}\hat{\Omega}$$ corresponding to $$\widehat{\mathrm{cov}}(\hat{\sigma}_{ij},\hat{\sigma}_{lm})$$ is given by $$T^{-1}(\hat{\sigma}_{il}\hat{\sigma}_{jm} + \hat{\sigma}_{im}\hat{\sigma}_{jl})$$ for all $$i,j,l,m = 1,2,\ldots,N$$, including $$i=j=l=m.$$ If $$i=j$$ and $$l=m$$ then

\begin{align*} \widehat{\mathrm{cov}}(\hat{\sigma}_{ij},\hat{\sigma}_{lm}) & = \widehat{\mathrm{cov}}(\hat{\sigma}_{ii},\hat{\sigma}_{ll}) = \widehat{\mathrm{cov}}(\hat{\sigma}_i^2,\hat{\sigma}_l^2) \\ & =\frac{\hat{\sigma}_{il}\hat{\sigma}_{il} + \hat{\sigma}_{il}\hat{\sigma}_{il}}{T} = \frac{\hat{\sigma}_{il}^2 + \hat{\sigma}_{il}^2}{T} = \frac{2\hat{\sigma}_{il}^2}{T} \end{align*}.

If $$i=j=l=m$$ then

\begin{align*} \widehat{\mathrm{cov}}(\hat{\sigma}_{ij},\hat{\sigma}_{lm}) & = \widehat{\mathrm{cov}}(\hat{\sigma}_i^2,\hat{\sigma}_i^2) = \widehat{\mathrm{var}}(\hat{\sigma}_i^2) \\ & =\frac{\hat{\sigma}_i^2\hat{\sigma}_i^2 + \hat{\sigma}_i^2\hat{\sigma}_i^2}{T} = \frac{2\hat{\sigma}_i^4}{T} \end{align*}.

In Proposition 7.9, it is not easy to see what are the elements of $$\hat{\Omega}$$. A simple example can make things more clear. Consider the bivariate case $$N=2$$. Then the result in Proposition 7.9 gives:

\begin{align*} \hat{\theta} = \left(\begin{array}{c} \hat{\sigma}_1^2 \\ \hat{\sigma}_{12} \\ \hat{\sigma}_2^2 \end{array} \right) & \sim N\left( \left(\begin{array}{c} \sigma_1^2\\ \sigma_{12} \\ \sigma_2^2 \end{array} \right),~ \frac{1}{T}\left( \begin{array}{ccc} \hat{\omega}_{11} & \hat{\omega}_{12} & \hat{\omega}_{13} \\ \hat{\omega}_{21} & \hat{\omega}_{22} & \hat{\omega}_{23} \\ \hat{\omega}_{31} & \hat{\omega}_{32} & \hat{\omega}_{33} \end{array} \right) \right) \\ & \sim N\left( \left(\begin{array}{c} \sigma_1^2\\ \sigma_{12} \\ \sigma_2^2 \end{array} \right),~ \left( \begin{array}{ccc} \widehat{\mathrm{var}}(\hat{\sigma}_1^2) & \widehat{\mathrm{cov}}(\hat{\sigma}_1^2, \hat{\sigma}_{12}) & \widehat{\mathrm{cov}}(\hat{\sigma}_1^2, \hat{\sigma}_2^2) \\ \widehat{\mathrm{cov}}(\hat{\sigma}_{12}, \hat{\sigma}_1^2) & \widehat{\mathrm{var}}(\hat{\sigma}_{12}) & \widehat{\mathrm{cov}}(\hat{\sigma}_{12}, \hat{\sigma}_2^2) \\ \widehat{\mathrm{cov}}(\hat{\sigma}_1^2, \hat{\sigma}_2^2) & \widehat{\mathrm{cov}}(\hat{\sigma}_2^2, \hat{\sigma}_{12}) & \widehat{\mathrm{var}}(\hat{\sigma}_2^2) \end{array} \right) \right) \\ & \sim N\left( \left(\begin{array}{c} \sigma_1^2\\ \sigma_{12} \\ \sigma_2^2 \end{array} \right),~ \frac{1}{T}\left( \begin{array}{ccc} 2\hat{\sigma}_1^4 & 2\hat{\sigma}_1^2\hat{\sigma}_{12} & 2\hat{\sigma}_{12}^2 \\ 2\hat{\sigma}_1^2 \hat{\sigma}_{12} & \hat{\sigma}_1^2 \hat{\sigma}_2^2 + \hat{\sigma}_{12}^2 & 2\hat{\sigma}_{12}\hat{\sigma}_2^2 \\ 2\hat{\sigma}_{12} & 2\hat{\sigma}_{12} \hat{\sigma}_2^2 & 2\hat{\sigma}_2^4 \end{array} \right) \right). \end{align*}

Notice that the results for the diagonal elements of $$\frac{1}{T}\hat{\Omega}$$ match those given earlier (see equations (7.20) and (7.22)). What is new here are formulas for the off-diagonal elements.

### 7.7.3 Joint sampling distribution between $$\hat{\mu}$$ and $$\mathrm{vech}(\hat{\Sigma})$$

When we study hypothesis testing in Chapter 9 and the statistical analysis of portfolios in Chapter 15 we will need to use results on the joint sampling distribution between $$\hat{\mu}$$ and $$\mathrm{vech}(\hat{\Sigma})$$. The follow proposition states the result:

Proposition 7.10 (Joint sampling distribution between $$\hat{\mu}$$ and $$\mathrm{vech}(\hat{\Sigma})$$) Under the GWN model and for large enough T, the joint sampling distribution for the $$N+N(N+1)/2 \times 1$$ vector $$(\hat{\mu}', \hat{\theta}')'$$, where $$\hat{\theta} = \mathrm{vech}(\hat{\Sigma})$$ is multivariate normal $\begin{equation} \left(\begin{array}{c} \hat{\mu} \\ \hat{\theta} \end{array} \right) \sim N \left( \left( \begin{array}{c} \mu \\ \theta \end{array} \right), ~ \frac{1}{T} \left( \begin{array}{cc} \hat{\Sigma} & 0 \\ 0 & \hat{\Omega} \end{array} \right) \right). \end{equation}$

Proposition 7.10 tells us that the vectors $$\hat{\mu}$$ and $$\hat{\theta}$$ are jointly multivariate normally distributed and that all elements of $$\hat{\mu}$$ are independent of all the elements of $$\hat{\theta}$$.

Example 2.35 (Joint distribution of $$(\mu,\sigma)'$$)

Consider the GWN model for a single asset and suppose we are interested in the joint distribution of the vector $$(\hat{\mu}, \hat{\sigma}^2)'$$. Then Proposition 7.10 gives us the result:

$\begin{equation} \left( \begin{array}{c} \hat{\mu} \\ \hat{\sigma}^2 \end{array} \right) \sim N \left( \left( \begin{array}{c} \mu \\ \sigma^2 \end{array} \right), ~ \left( \begin{array}{cc} \frac{\hat{\sigma}^2}{T} & 0 \\ 0 & \frac{2\hat{\sigma}^4}{T} \end{array} \right) \right). \tag{7.44} \end{equation}$

Here, we see that $$\mathrm{cov}(\hat{\mu}, \hat{\sigma}^2)=0$$, which implies that $$\hat{\mu}$$ and $$\hat{\sigma}^2$$ are independent since they are jointly normally distributed. Furthermore, since $$\hat{\mu}$$ and $$\hat{\sigma}^2$$ are independent and function of $$\hat{\mu}$$ is independent of any function of $$\hat{\sigma}^2$$. For example, $$\hat{\sigma} = \sqrt{\hat{\sigma}^2}$$ is independent of $$\hat{\mu}$$. Using (7.34) we can then deduce the joint distribution of $$(\hat{\mu}, \hat{\sigma}^2)'$$:

$\begin{equation} \left( \begin{array}{c} \hat{\mu} \\ \hat{\sigma} \end{array} \right) \sim N \left( \left( \begin{array}{c} \mu \\ \sigma \end{array} \right), ~ \left( \begin{array}{cc} \frac{\hat{\sigma}^2}{T} & 0 \\ 0 & \frac{\hat{\sigma}^2}{2T} \end{array} \right) \right). \tag{7.45} \end{equation}$

$$\blacksquare$$