Question 1207210: If 𝑡1 𝑎𝑛𝑑 𝑡2 are both most efficient estimators with equal variance V
and if 𝑡3 is an average of 𝑡1 𝑎𝑛𝑑 𝑡2. Prove that Var(𝑡3)= 1/2 𝑉(1 + 𝜌)
where 𝜌 is the correlation coefficient between 𝑡1 𝑎𝑛𝑑 𝑡2.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! ## Proving the Variance of the Average Estimator
**Given:**
* t1 and t2 are most efficient estimators with equal variance V.
* t3 = (t1 + t2)/2
**To Prove:**
Var(t3) = (1/2)V(1 + ρ)
**Proof:**
We know that:
* Var(t1) = Var(t2) = V
* Cov(t1, t2) = ρ * σ(t1) * σ(t2) = ρV (since σ(t1) = σ(t2) = √V)
Now, let's calculate the variance of t3:
Var(t3) = Var((t1 + t2)/2)
= (1/4)Var(t1 + t2)
= (1/4)[Var(t1) + Var(t2) + 2Cov(t1, t2)]
= (1/4)[V + V + 2ρV]
= (1/2)V(1 + ρ)
**Therefore, Var(t3) = (1/2)V(1 + ρ).**
This result shows that the variance of the average of two correlated estimators is a function of their individual variances and their correlation coefficient. When the two estimators are positively correlated (ρ > 0), the variance of their average is greater than the individual variances. Conversely, when the two estimators are negatively correlated (ρ < 0), the variance of their average is less than the individual variances.
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