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**Homework Statement::**1) ##X## is number of success out of ##n## trials where ##p## is the probability of success.

1a) Show that ##\mathrm{E}\left[\hat{P}\right]-p=0##, where ##\hat{P}=X/n##.

1b) Find the standard error of ##\hat{P}##, then give calculate the estimated standard error if there are ##5## successes out of ##10## trials.

2) Consider the probability density function ##f(x)=0.5(1+\Theta x)## defined on ##[-1,1]##.

Find the moment estimator for ##\Theta##, then show that ##\hat{\Theta}=3\bar{X}## is an unbiased estimator.

**Relevant Equations::**N/A

1a) ##\mathrm E\left[\hat P\right]-p=\mathrm E\left[X/n\right]-p=\frac1n\mathrm E\left[X\right]-p=\frac1nnp-p=0##.

1b)$$\begin{align*}

\mathrm E\left[{\left(\hat{P}-\mathrm E\left[{\hat{P}}\right]\right)^2}\right]&=\mathrm E\left[{\left(\frac{X}{n}-\mathrm E\left[{\frac Xn}\right]\right)^2}\right]\\

&=\mathrm E\left[{\left(\frac Xn-p\right)^2}\right]\\

&=\mathrm E\left[{\frac{\left(X-np\right)^2}{n^2}}\right]\\

&=\frac{1}{n^2}\mathrm E\left[{\left(X-np\right)^2}\right]\\

&=\frac{1}{n^2}\mathrm E\left[{\left(X-\mathrm E\left[{X}\right]\right)^2}\right]\\

&=\frac{p\left(1-p\right)}{n}\\

\sigma_{\hat{P}}&=\sqrt{\frac{p\left(1-p\right)}{n}}

\end{align*}$$

With ##\hat P=5/10=0.5##, I get ##\hat{\sigma}_{\hat P}=\frac{0.5(1-0.5)}{10}=\frac{\sqrt{10}}{20}##.

2)

Since we only have one parameter to estimate, we use ##\mathrm E\left[{X}\right]##.

$$\begin{align*}

\mathrm E\left[{X}\right]&=\int_{-1}^1x(1+\Theta x)\,dx\\

&=\frac12\int_{-1}^1x\,dx+\frac12\Theta\int_{-1}^1x^2\,dx\\

&=\frac13\Theta

\end{align*}$$The moment estimator is ##\hat{\Theta}=3\bar{X}##.$$\begin{align*}

\mathrm E\left[\hat{\Theta}\right]-\Theta&=3\mathrm E\left[{\bar{X}}\right]-\Theta\\

&=3\times\frac{\Theta}{3}-\Theta\\

&=0

\end{align*}$$