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!


2 Answers 2


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 $$\mathrm{cov}[T,T']=\mathrm{var}[T]$$

  • $\begingroup$ Thank you so much, that's quite straightforward and quite a clever way to solve this! $\endgroup$ 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 $\sum_{i=1}^n \alpha_i^2 = \sum_{i=1}^n \alpha_i \beta_i$.

  • $\begingroup$ Two quite different proofs with the same basic idea.Nice. $\endgroup$ Feb 14 at 4:23
  • $\begingroup$ Clever, is $\alpha_i = \frac{1}{n}$the unique solution for $T$ to be minimized? $\endgroup$ Feb 15 at 1:59
  • $\begingroup$ 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 $\endgroup$
    – angryavian
    Feb 15 at 2:29

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