# Covariance of Best Linear Unbiased Estimators and arbitrary LUE

I'm working on a problem involving two linear unbiased estimators $$T$$ and $$T'$$ of a parameter $$\theta$$, defined from a sample $$\{X_1, \dots, X_n\}$$ with mean $$\theta$$ and finite variance. I aim to prove that $$\text{Cov}_\theta(T, T') = \text{Var}_\theta(T)$$, given that:

• $$T = \sum_{i=1}^{n} \alpha_i X_i$$ and $$T'$$ is another linear unbiased estimator of $$\theta$$, with $$T' = \sum_{i=1}^{n} \beta_i X_i$$.
• Both $$T$$ and $$T'$$ are unbiased ($$\mathbb{E}[T] = \mathbb{E}[T'] = \theta$$).
• $$T$$ has minimum variance among all such estimators.

Through my derivations, I reached that $$\text{Cov}_\theta(T, T') = \sigma^2 \sum_{i=1}^{n} \alpha_i \beta_i$$ and $$\text{Var}_\theta(T) = \sigma^2 \sum_{i=1}^{n} \alpha_i^2$$. From these, I inferred that if $$\text{Cov}_\theta(T, T') = \text{Var}_\theta(T)$$, it should imply $$\sum_{i=1}^{n} \alpha_i \beta_i = \sum_{i=1}^{n} \alpha_i^2$$, but it seems hard to prove, and I am honestly not even sure if this is a solvable equation. I would need to prove this in order to conclude my proof.

I'm unsure if what I got to is universally valid or if I've overlooked critical assumptions about the relationship between $$T$$ and $$T'$$. Here's a summary of my steps:

1. Established the unbiasedness of $$T$$ and $$T'$$.
2. Calculated $$\text{Cov}_\theta(T, T')$$ using the expectation of their product minus the product of their expectations.
3. Arrived at the conclusion based on the expressions for covariance and variance.

Is there a flaw in my reasoning, or are there specific conditions under which this relationship holds true? I'm particularly interested in understanding the assumptions required for $$\text{Cov}_\theta(T, T') = \text{Var}_\theta(T)$$ to be valid. Any insights or references to similar proofs would be greatly appreciated!

Consider the variance of $$T_a=aT'+(1-a)T$$, which (for any $$a$$) is another linear unbiased estimator

$$\mathrm{var}[T_a]=a^2\mathrm{var}[T']+(1-a)^2\mathrm{var}[T]+2a(1-a)\mathrm{cov}[T,T']$$ or rearranged $$\mathrm{var}[T_a]=a^2\left(\mathrm{var}[T']+\mathrm{var}[T]-2\mathrm{cov}[T,T']\right)+2a(\mathrm{cov}[T,T']-\mathrm{var}[T])+\left(\mathrm{var}[T]\right)$$

By assumption this must be minimised at $$a=0$$, because $$T$$ is the best linear unbiased estimator. It's a standard high-school fact about quadratics that the minimum is at $$a_{\text{opt}}=\frac{-2(\mathrm{cov}[T,T']-\mathrm{var}[T])}{2\left(\mathrm{var}[T']+\mathrm{var}[T]-2\mathrm{cov}[T,T']\right)}$$ and for this to be 0, we need $$\mathrm{cov}[T,T']=\mathrm{var}[T]$$

• Thank you so much, that's quite straightforward and quite a clever way to solve this! Feb 15 at 1:58

Unbiasedness of $$T$$ and $$T'$$ imply $$\sum_{i=1}^n \alpha_i = \sum_{i=1}^n \beta_i = 1$$.

$$T$$ having minimium variance means $$\alpha_1, \ldots, \alpha_n$$ minimize $$\sum_{i=1}^n \alpha_i^2$$ subject to $$\sum_{i=1}^n \alpha_i = 1$$. By Cauchy-Schwarz, we have $$\left(\sum_{i=1}^n \alpha_i^2 \right)\left(\sum_{i=1}^n 1\right) \ge \left(\sum_{i=1}^n (\alpha_i \cdot 1)\right)^2 = 1,$$ for any $$\alpha_1, \ldots, \alpha_n$$. For $$\sum_{i=1}^n \alpha_i^2$$ to be minimized, we need $$\alpha_i = 1/n$$.

Combining this with the fact that $$\sum_{i=1}^n \beta_i = 1$$ should allow you to prove $$\sum_{i=1}^n \alpha_i^2 = \sum_{i=1}^n \alpha_i \beta_i$$.

• Two quite different proofs with the same basic idea.Nice. Feb 14 at 4:23
• Clever, is $\alpha_i = \frac{1}{n}$the unique solution for $T$ to be minimized? Feb 15 at 1:59
• Yes. For Cauchy-Schwarz to attain equality in this case, there exists some $c$ such that $\alpha_i = c \cdot 1$ for all $i$. The constraint on $\sum_{i=1}^n \alpha_i$ forces $c=1/n$. @TahaRhaouti Feb 15 at 2:29