0
$\begingroup$

Hi all I have a question about a proof that I don't understand, enter image description here

My question is about the line after "We also have that....", I don't understand how $P(\hat{\theta_n} \geq \theta -\frac{x}{n})$ becomes $1-(1-\frac{x}{\theta*n})^n$

Could someone kindly explains?

$\endgroup$
  • 2
    $\begingroup$ Isn’t that simply $P(\hat{\theta}_n \geq ...) = 1 - P(\hat{\theta}_n \leq ...)$ and for the latter expression one used the equation above...? $\endgroup$ – Fabian Werner Dec 14 '19 at 5:22
  • 1
    $\begingroup$ (and then pull out the $\theta$ in $\theta-x/n$...) $\endgroup$ – Fabian Werner Dec 14 '19 at 5:24
  • 1
    $\begingroup$ please add the self-study tag and provide more details $\endgroup$ – Xi'an Dec 14 '19 at 9:33
0
$\begingroup$

The result follows since $$\begin{align*}P(\hat\theta_n\geq\theta-x/n)&=1-P(\hat\theta_n<\theta-x/n)\\&=1-P(\hat\theta_n\leq\theta-x/n)\\&=1-\Big(\frac{\theta-x/n}{\theta}\Big)^n\\&=1-(1-x/(\theta n))^n,\end{align*}$$ where the second equality follows by continuity of $\theta_n$ (it is continuous because its cdf is continuous), and where the third equality follows from the fact that since $x\in[0,\theta]$, $(\theta-x/n)/\theta\in[1-1/n,1]$, assuming $\theta\geq1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.