1. ## New estimator

In order to estimate a population mean, $\displaystyle \mu$, 2 surveys were conducted independently and the statistics were noted. ($\displaystyle \overline{X}_1,\overline{X}_2,\sigma _{\overline{X}_1}, \sigma _{\overline{X}_2}$ are obtained). Assume that $\displaystyle \overline{X_1}$ and $\displaystyle \overline{X_2}$ are unbaised. For some $\displaystyle \alpha$ and $\displaystyle \beta$, the two estimates can be combined to give a better estimator:

$\displaystyle X=\alpha \overline{X}_1 + \beta \overline{X}_2$

What choice of $\displaystyle \alpha$ and $\displaystyle \beta$ will minimize the variances, given that $\displaystyle \alpha + \beta =1$?

Since I managed to find one equation which is: $\displaystyle \alpha + \beta =1$. What is another equations which allow me to minimise the variance in order to find suitable $\displaystyle \alpha$ and $\displaystyle \beta$?

2. Originally Posted by noob mathematician
In order to estimate a population mean, $\displaystyle \mu$, 2 surveys were conducted independently and the statistics were noted. ($\displaystyle \overline{X}_1,\overline{X}_2,\sigma _{\overline{X}_1}, \sigma _{\overline{X}_2}$ are obtained). Assume that $\displaystyle \overline{X_1}$ and $\displaystyle \overline{X_2}$ are unbaised. For some $\displaystyle \alpha$ and $\displaystyle \beta$, the two estimates can be combined to give a better estimator:

$\displaystyle X=\alpha \overline{X}_1 + \beta \overline{X}_2$

What choice of $\displaystyle \alpha$ and $\displaystyle \beta$ will minimize the variances, given that $\displaystyle \alpha + \beta =1$?

Since I managed to find one equation which is: $\displaystyle \alpha + \beta =1$. What is another equations which allow me to minimise the variance in order to find suitable $\displaystyle \alpha$ and $\displaystyle \beta$?
Do these surveys have the same sample size (or do we know the sample sizes), and are they otherwise identical?

CB

3. You need a relationship between the variances and are these $\displaystyle \sigma$'s populations st deviations or sample st deviations?
This is not complete.

4. Hi,

There are actually 2 questions on this (However there aren't much more clues to that)

Qn1: Find the conditions on $\displaystyle \alpha$ and $\displaystyle \beta$ that make the combined estimate unbiased. I get $\displaystyle \alpha + \beta =1$

Qn2: What choice of $\displaystyle \alpha$ and $\displaystyle \beta$ minimizes the variances, subject to the condition of unbiasedness above.
Answer I suppose: (it's true that $\displaystyle \alpha + \beta$ is still 1)

$\displaystyle \alpha= \frac{\sigma^2 _{\overline{X}_2}}{\sigma^2 _{\overline{X}_2}+\sigma^2 _{\overline{X}_1}}$, $\displaystyle \beta= \frac{\sigma^2 _{\overline{X}_1}}{\sigma^2 _{\overline{X}_2}+\sigma^2 _{\overline{X}_1}}$

I'm not sure whether is the population needed since we already know their statistics.

$\displaystyle \sigma _{\overline{X}}$ is the standard errors. So $\displaystyle \sigma^2 _{\overline{X}}=Var(\overline{X})$

5. Originally Posted by noob mathematician
Hi,

There are actually 2 questions on this (However there aren't much more clues to that)

Qn1: Find the conditions on $\displaystyle \alpha$ and $\displaystyle \beta$ that make the combined estimate unbiased. I get $\displaystyle \alpha + \beta =1$

Qn2: What choice of $\displaystyle \alpha$ and $\displaystyle \beta$ minimizes the variances, subject to the condition of unbiasedness above.
Ans I suppose: (it's true that $\displaystyle \alpha + \beta$ is still 1)

$\displaystyle \alpha= \frac{\sigma^2 _{\overline{X}_2}}{\sigma^2 _{\overline{X}_2}+\sigma^2 _{\overline{X}_1}}$, $\displaystyle \beta= \frac{\sigma^2 _{\overline{X}_1}}{\sigma^2 _{\overline{X}_2}+\sigma^2 _{\overline{X}_1}}$

I'm not sure whether is the population needed since we already know their statistics.

$\displaystyle \sigma _{\overline{X}}$ is the standard errors. So $\displaystyle \sigma^2 _{\overline{X}}=Var(\overline{X})$
Hi,
With the given answer:$\displaystyle \alpha= \frac{\sigma^2 _{\overline{X}_2}}{\sigma^2 _{\overline{X}_2}+\sigma^2 _{\overline{X}_1}}$, $\displaystyle \beta= \frac{\sigma^2 _{\overline{X}_1}}{\sigma^2 _{\overline{X}_2}+\sigma^2 _{\overline{X}_1}}$
I've tried to work backward in order to attain the 2nd equation:

$\displaystyle X=\alpha\overline{X}_1+\beta\overline{X}_2$
$\displaystyle Var(X)=\alpha^2 Var(\overline{X}_1)+\beta^2Var(\overline{X}_2)=\al pha^2\sigma^2 _{\overline{X}_1}+\beta^2\sigma^2 _{\overline{X}_2}$ (Equation 2)

Let $\displaystyle x=\sigma^2 _{\overline{X}_1}, y=\sigma^2 _{\overline{X}_2}$

Subst the answer into the 1st equation: $\displaystyle (\alpha +\beta=1)$

$\displaystyle \frac{y}{x+y}+\frac{x}{x+y}=1$ so it's valid.

Subst the answer into the 2nd equation:

$\displaystyle Var(X)=(\frac{y}{x+y})^2x+(\frac{x}{x+y})^2y$
$\displaystyle =\frac{xy^2}{(x+y)^2}+\frac{x^2y}{(x+y)^2}$
$\displaystyle =\frac{xy}{x+y}$

However I'm not sure why the new variance estimator becomes $\displaystyle \frac{xy}{x+y}$?