Estimation with Equicorrelated Normal Samples

Worked Solution

How to Think About It: Equicorrelated samples come up constantly in finance -- think of returns from stocks in the same sector, or repeated measurements on the same day from a mean-reverting process. The key question is: does correlation destroy the unbiasedness of the sample mean, or just inflate its variance? The answer to (a) is encouraging: $\bar{X}$ is still unbiased. But the effective sample size shrinks dramatically when $r$ is large -- 1000 highly correlated observations can be worth less than 10 independent ones. The test statistic in (b) must account for this or you will massively overstate your significance.

Quick Estimate for (b): Suppose $n = 100$, $r = 0.5$. Effective sample size: $n_{\text{eff}} = 100/(1 + 99 \times 0.5) \approx 100/50.5 \approx 2$. If you naively use $\sqrt{n} = 10$ in your z-statistic, you overstate it by a factor of $\sqrt{50} \approx 7$. This is why ignoring correlation in clustered data leads to spurious significance.

Approach: Use linearity of expectation for unbiasedness, compute $\text{Var}(\bar{X})$ directly from the covariance structure, then standardize.

Formal Solution:

(a) Estimating $\mu$:

Unbiasedness of $\bar{X}$: $$E[\bar{X}] = \frac{1}{n}\sum_{i=1}^n E[X_i] = \mu$$ Correlation does not affect the mean, only the variance. So $\bar{X}$ is unbiased regardless of $r$.

Variance of $\bar{X}$: $$\text{Var}(\bar{X}) = \frac{1}{n^2} \sum_{i,j} \text{Cov}(X_i, X_j) = \frac{1}{n^2}\left[n \cdot 1 + n(n-1)r\right] = \frac{1 + (n-1)r}{n}$$

Is $\bar{X}$ the best linear unbiased estimator (BLUE)? By symmetry of the equicorrelated structure, the Gauss-Markov theorem applied to $\Sigma$ yields that the GLS estimator is $\hat{\mu}_{\text{GLS}} = (\mathbf{1}^\top \Sigma^{-1} \mathbf{1})^{-1} \mathbf{1}^\top \Sigma^{-1} X$. For the equicorrelated structure, $\Sigma^{-1} \mathbf{1} \propto \mathbf{1}$, so GLS reduces to OLS, meaning $\bar{X}$ is in fact the BLUE.

(b) Test statistic for $H_0: \mu = \mu_0$:

Since $\bar{X}$ is a linear combination of jointly normal variables, $\bar{X} \sim N(\mu, (1+(n-1)r)/n)$ exactly. Under $H_0$:

$$T = \frac{\bar{X} - \mu_0}{\sqrt{(1+(n-1)r)/n}} \sim N(0,1)$$

The effective sample size is: $$n_{\text{eff}} = \frac{n}{1 + (n-1)r}$$

When $r > 0$: $n_{\text{eff}} < n$, so you have less information than the raw count suggests. As $r \to 1$: $n_{\text{eff}} \to 1$ -- all observations collapse to a single draw.

When $r < 0$: $n_{\text{eff}} > n$ -- negative correlation adds information. This is the benefit of diversification.

(c) Valid range of $r$:

The matrix $\Sigma = (1-r)I + r\mathbf{1}\mathbf{1}^\top$ must be positive definite (PD). Its eigenvalues are: - $1 - r$ with multiplicity $n - 1$ (eigenvectors orthogonal to $\mathbf{1}$) - $1 + (n-1)r$ with multiplicity $1$ (eigenvector $\mathbf{1}/\sqrt{n}$)

For PD, both must be positive: $$1 - r > 0 \implies r < 1$$ $$1 + (n-1)r > 0 \implies r > -\frac{1}{n-1}$$

So the valid range is: $$r \in \left(-\frac{1}{n-1},\; 1\right)$$

At $r = -1/(n-1)$, the matrix is singular -- all observations sum to a constant almost surely, so you learn nothing by taking more draws.

Answer: - (a) $\bar{X}$ is the BLUE for $\mu$ with variance $(1+(n-1)r)/n$. - (b) $T = (\bar{X} - \mu_0)/\sqrt{(1+(n-1)r)/n} \sim N(0,1)$ under $H_0$. - (c) $r \in (-1/(n-1),\; 1)$.