Estimator Exercise Homework Solution

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Homework Help Overview

The discussion revolves around statistical estimation, specifically focusing on the properties of estimators related to success probabilities in trials and moment estimators derived from probability density functions.

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Approaches and Questions Raised

  • Participants explore the unbiasedness of estimators and the calculation of standard errors. There are attempts to clarify the application of probability density functions and moment estimators. Some participants question the appropriateness of certain calculations and the placement of the thread within the forum.

Discussion Status

There are ongoing discussions about the validity of the calculations presented, particularly regarding the standard deviation estimates and the integration of the probability density function. Some participants express uncertainty about the methods used, while others provide feedback on the clarity of the mathematical expressions.

Contextual Notes

There are indications of confusion regarding the categorization of the homework topic, with suggestions that it may belong in different sections of the forum. Additionally, some participants note potential issues with the definitions and assumptions used in the calculations.

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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}##.$$\begin{align*}
\mathrm E\left[\hat{\Theta}\right]-\Theta&=3\mathrm E\left[{\bar{X}}\right]-\Theta\\
&=3\times\frac{\Theta}{3}-\Theta\\
&=0
\end{align*}$$
 
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Not a probability guy, but this shouldn't be in precalc
 
Moved to statistics.
 
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.
 
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Office_Shredder said:
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. :oldshy:
jim mcnamara said:
Moved to statistics.
hm, so we can post homework in here?
 
I think this should have been moved to the calculus homework section.
 
Office_Shredder said:
I think this should have been moved to the calculus homework section.
Done. :smile:
 
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archaic said:
## 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.
 
Stephen Tashi said:
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.
 
  • #10
Yeah there's a 1/2 missing in the first line of his equation series.
 
  • #11
Stephen Tashi said:
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.
 
  • #12
Now that I look at the variance equation... I could've directly factorized ##\frac1n##. :nb)
 

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