Unbiased family of estimators and variance.

Question

Given $X_1, \dots, X_n$ simple random sample with distribution $F_X$ -unknown-, i have to estimate $\mu = \mathrm{E}(X)$.

Now, given the famility of estimators $\tilde{T} = \bigg\{\displaystyle\sum_{i=1}^n \alpha_iX_i : \ \displaystyle\sum_{i=1}^n \alpha_i = 1\bigg\}$.

I have to (1) Prove that if $\hat{\mu} \in \tilde{T},$ then $\hat{\mu}$ it's unbiased; and (2) Show that for every estimator $\beta \in \tilde{T}$, $\mathrm{Var}(\bar{X_n})<\mathrm{Var}\{\beta\}$.

Here's my attempt:

(1) $\hat{\mu} \in \tilde{T} \implies \hat{\mu} = \displaystyle\sum_{i=1}^n \alpha_iX_i \implies \mathrm{E}(\hat{\mu}) = \mathrm{E}\bigg(\displaystyle\sum_{i=1}^n \alpha_iX_i\bigg) = \displaystyle\sum_{i=1}^n \alpha_i\mathrm{E}(X_i)$.

Now, since $X_1,\dots, X_n$ it's a sample mean, then all of them have the same distribution, let's say $X$. Follows $\mathrm{E}(\hat{\mu}) = \displaystyle\sum_{i=1}^n \alpha_i\mathrm{E}(X_i) = \displaystyle\sum_{i=1}^n \alpha_i\mathrm{E}(X) = \mathrm{E}(X)\displaystyle\sum_{i=1}^n \alpha_i = \mathrm{E}(X)$ and then follows $ \hat{\mu}$ unbiased ?

(2) I know that $\mathrm{V}(\bar{X}) = \displaystyle\frac{\sigma^2}{n}$, if $\mathrm{V}(\beta) < \displaystyle\frac{\sigma^2}{n}$ and since $\beta \in \tilde{T}$ we have $\beta = \displaystyle\sum_{i=1}^n \beta_iX_i$ for $\beta_i$ scalars, then $\mathrm{V}(\beta) =\mathrm{V}\bigg(\displaystyle\sum_{i=1}^n \beta_iX_i\bigg) = \displaystyle\sum_{i=1}^n \beta_i^2\mathrm{V}(X_i) > \displaystyle\sum_{i=1}^n \mathrm{V}(X_i) = n\sigma ^2$ and should be $n\sigma ^2 < \displaystyle\frac{\sigma^2}{n}$.

And follows that the assumption was wrong.

Well, I see no problems with the first part. Speaking about the second part, do you assume $\alpha_i$ to be positive? — Caran-d'Ache, Jul 08 '13 at 04:52
By the way is there any need to name the same coefficients $\alpha_i$ and $\beta_i$ at the same time? — Caran-d'Ache, Jul 08 '13 at 04:59
Does this answer your question? Finding the MVUE out of all unbiased estimators $\sum_{i=1}^ka_iW_i$ using Lagrange multiplier — StubbornAtom, Oct 06 '21 at 18:18

score 0 · Answer 1 · answered Jul 08 '13 at 06:36

Correct me if I did not understand you properly, but you have not set any restrictions on the coefficients.
So assuming that $\mathrm{V}(\bar{X}) = \displaystyle\frac{\sigma^2}{n}$ one can say that if $\beta \in \tilde{T}$ then $\mathrm{V}(\beta) = \displaystyle\sum_{i=1}^n \alpha_i^2\mathrm{V}(X_i) = \sigma ^2\displaystyle\sum_{i=1}^n \alpha_i^2 $. And $\mathrm{V}(\beta) = \mathrm{V}(\bar{X}) \ n \ \displaystyle\sum_{i=1}^n \alpha_i^2$. So if $n \ \displaystyle\sum_{i=1}^n \alpha_i^2>1 \ \text{or} \ n>\frac{1}{\sum_{i=1}^n \alpha_i^2}$ then $\mathrm{V}(\beta)>\mathrm{V}(\bar{X})$. And the first part, as I have said looks correct.

Actually, you're right. I should look for another way to prove it because there was no restriction to the coefficients aside from being real numbers. Thanks! — balaya, Jul 08 '13 at 14:17

Unbiased family of estimators and variance.

1 Answers1