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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}##.$$\begin{align*}
\mathrm E\left[\hat{\Theta}\right]-\Theta&=3\mathrm E\left[{\bar{X}}\right]-\Theta\\
&=3\times\frac{\Theta}{3}-\Theta\\
&=0
\end{align*}$$
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}##.$$\begin{align*}
\mathrm E\left[\hat{\Theta}\right]-\Theta&=3\mathrm E\left[{\bar{X}}\right]-\Theta\\
&=3\times\frac{\Theta}{3}-\Theta\\
&=0
\end{align*}$$