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I calculated that $$ Var(\theta_1)=\frac{2\sigma^2}{(x_n-x_1)^2},~~~~~Var(\theta_2)=\frac{\sigma^2}{\sum_{i=1}^{n}(x_i-\overline{x})^2}. $$ Now I have to show that for $\sigma^2 >0$ and $\overline{x}\neq\frac{(x_1+x_n)}{2}$ it is $$ Var(\theta_1) > Var(\theta_2). $$

I have to show that it is $$ \frac{2\sigma^2}{(x_n-x_1)^2} > \frac{\sigma^2}{\sum_{i=1}^{n}(x_i-\overline{x})^2}. $$

Do you have an idea how to show that?

Unfortunately, I have not.

Edit:

I've already tried to show that $$ Var(\theta_2)/Var(\theta_1)<1, $$ i.e. that $$ \frac{(x_n-x_1)^2}{\sum_{i=1}^{n}(x_i-\overline{x})^2}=\frac{x_1^2+x_n^2-2x_1x_n}{\sum_{i=1}^{n}x_i^2-n\overline{x}^2}<1 $$ but with no success.

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  • $\begingroup$ You could look at the ratio and show that it is $ >1$. $\endgroup$ – Hans Engler May 14 '14 at 22:17
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We wish to show that

$$ \frac{2\sigma^2}{(x_1 - x_n)^2} \geq \frac{\sigma^2}{\sum_{i=1}^n (x_i - \bar{x})^2} $$

Cancelling the sigmas and inverting the fractions shows that the desired result is

$$ \sum_{i=1}^n (x_i - \bar{x})^2 \geq \frac{1}{2}(x_1 - x_n)^2.$$ Now, in order to avoid typing $\bar{x}$ all over the place, let's define $y_i = x_i - \bar{x}$. We then have $y_1 - y_n = x_1 - x_n$, and the desired result is \begin{align*} \sum_{i=1}^n y_i^2 &\geq \frac{1}{2}(y_1 - y_n)^2\\ &= \frac{1}{2}(y_1^2 + y_n^2 - 2y_1y_n). \end{align*} Subtracting $y_1^2$ and $y_n^2$ from both sides yields that we want to show \begin{align*} \sum_{i=2}^{n-1} y_i^2 &\geq -\frac{1}{2}(y_1^2 + y_n^2 + 2y_1y_n)\\ &= -\frac{(y_1+y_n)^2}{2}, \end{align*} which is trivially true as the left hand side is a sum of non-negative terms, and must therefore be greater than zero.

For $\bar{x} \neq \frac{1}{2}(x_1 + x_n)$ we see that the right hand side is negative, and the inequality is strict.

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  • 1
    $\begingroup$ Nice, +1. $ $ $ $ $\endgroup$ – Did May 19 '14 at 15:42

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