"Explaining Unbiased Expression with Probability

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Homework Statement



In an example my book says that the expression bellow is unbiased.
I can't see why this is exactly...

Homework Equations



<br /> \begin{array}{l}<br /> \hat p = \frac{X}{n} \\\\<br /> E(\hat p) = E\left( {\frac{X}{n}} \right) = \frac{1}{n} \cdot E(X) = \frac{1}{n} \cdot (n \cdot p) = p \\ <br /> \end{array}<br />

The Attempt at a Solution



Could the reason be that the expression comes down to just p, which is simply a probability and we have no better suggestion than to believe that it "hits the target"? (If that didn't make any sense, just ignore it)
 
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A statistic \tau(x_1,x_2,...,x_n) is said to be unbiased for a parameter \theta if E[\tau(x_1,x_2,...,x_n)]=\theta.

It is just a definition.

It is important to know that to say that \hat{p}=\frac{x}{n} is unbiased is WRONG. It is unbiased for a particular PARAMETER.

The expectation of \hat{p} is precisely p. If it so happened that E[\hat{p}]=p-2 then \hat{p} would not be an unbiased estimator for p, it would be an unbiased estimator for p-2.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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