Estimator exercise

[archaic]

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

 

[joshmccraney]

Not a probability guy, but this shouldn't be in precalc

 

[jim mcnamara]

Moved to statistics.

 

[Office_Shredder]

Everything looks fine to me, but I'm not confident the best estimate of the stdev is plugging in your estimate for p to that formula. I don't have any specific reason to think that is the wrong thing to do, it just feels like the type of calculation where there might be some subtly better thing to do.

 

[archaic]

Everything looks fine to me, but I'm not confident the best estimate of the stdev is plugging in your estimate for p to that formula. I don't have any specific reason to think that is the wrong thing to do, it just feels like the type of calculation where there might be some subtly better thing to do.

well, i submitted it anyhow.

Moved to statistics.

hm, so we can post homework in here?

 

[Office_Shredder]

I think this should have been moved to the calculus homework section.

 

[berkeman]

I think this should have been moved to the calculus homework section.

Done.

 

[Stephen Tashi]

$E[X]=\int_{-1}^1x(1+\Theta x)\,dx\\ =\frac12\int_{-1}^1x\,dx+\frac12\Theta\int_{-1}^1x^2\,dx\\$

Are you thinking that $E( g(x) + h(x)) = (1/2) E(g(x)) + (1/2) E(h(x))$?
That isn't true.

 

[WWGD]

Are you thinking that $E( g(x) + h(x)) = (1/2) E(g(x)) + (1/2) E(h(x))$?
That isn't true.

Comes from his definition of the density formula, which contains a 0.5 term.

 

[Office_Shredder]

Yeah there's a 1/2 missing in the first line of his equation series.

 

[archaic]

Are you thinking that $E( g(x) + h(x)) = (1/2) E(g(x)) + (1/2) E(h(x))$?
That isn't true.

Sorry about that! I forgot the $0.5$ of the density function.

 

[archaic]

Now that I look at the variance equation... I could've directly factorized $\frac1n$.