Estimator Exercise Homework Solution

  • Thread starter Thread starter archaic
  • Start date Start date
  • Tags Tags
    Exercise
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
11 replies · 2K views
archaic
Messages
688
Reaction score
214
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}##.$$\begin{align*}
\mathrm E\left[\hat{\Theta}\right]-\Theta&=3\mathrm E\left[{\bar{X}}\right]-\Theta\\
&=3\times\frac{\Theta}{3}-\Theta\\
&=0
\end{align*}$$
 
  • Like
Likes   Reactions: etotheipi
on Phys.org
Not a probability guy, but this shouldn't be in precalc
 
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.
 
  • Like
Likes   Reactions: archaic
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?
 
Office_Shredder said:
I think this should have been moved to the calculus homework section.
Done. :smile:
 
  • Like
Likes   Reactions: jim mcnamara
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.
 
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.
 
Now that I look at the variance equation... I could've directly factorized ##\frac1n##. :nb)