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Let $X_1,X_2, \dots,X_n$ be a sample from a population with distribution function $F(x-\theta)$, where $F$ is symmetric around $0$. The $\alpha$ trimmed mean $T_n(\alpha)=\dfrac{1}{n-2\lfloor n\alpha \rfloor }\displaystyle\sum_{i=\lfloor n \alpha \rfloor+1}^{n-\lfloor n \alpha\rfloor}X_{(i)}$ is asymptotically normally distributed, i.e. $$\sqrt n(T_n (\alpha) -\theta) \sim \mathcal N(0, \sigma^2(\alpha,F))$$ where $$\sigma^2(\alpha,F)=\dfrac{2\displaystyle\int_0^{F^{-1}(1-\alpha)}t^2~dF(t)+2\alpha(F^{-1}(1-\alpha))^2}{(1-2\alpha)^2}$$ Propose a method of choosing $\alpha$ from the data.compare the difference between your proposed adaptive choice of $\alpha$ and its theoretical counterpart. Investigate your proposed methodology when $F$ is standard Cauchy. i.e. $f(x)=\dfrac{1}{\pi(1+x^2)},x\in \mathbb{R}.$(Density function).

From this paper and this I understand that how I can find the value of $\alpha$. But then how I can investigate it for Cauchy Distribution. Please help. Thanks in advance.

EDIT: According to whuber's comment here I need to show $\sigma^2(\alpha,F)=\frac{1}{2\alpha-1}+\frac{2(\pi \alpha+1)\cot(\pi\alpha)}{\pi(2\alpha-1)^2}$. I understand that here I need to use standard cauchy. I want to know how to find value of $F^{-1}(1-\alpha)$ and $\int_0^{F^{-1}(1-\alpha)}t^2~dF(t)$. Thanks in advance.

Argha
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    In your case (if I have done the calculations correctly!), $\sigma^2(\alpha,F)=\frac{1}{2\alpha-1}+\frac{2(\pi \alpha+1)\cot(\pi\alpha)}{\pi(2\alpha-1)^2}.$ There's no closed form expression for its minimum value on $(0, 1/2)$ but it is straightforward to minimize numerically in the vicinity of $\alpha=0.178691.$ – whuber Feb 20 '14 at 21:49
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    Is this work for some subject? – Glen_b Feb 20 '14 at 22:23
  • One approach to investigate its behavior would be to do some kind of simulation study, but I don't know if that's what's being asked for here. – Glen_b Feb 20 '14 at 22:25
  • @whuber: I initially try to minimize $\sigma^2(\alpha,F)$ by differentiate it w.r.t $\alpha$ and equate it to zero for finding optimum value of $\alpha$. But I am unable get any rigid expression. So, it is clear that there's no closed form expression for its minimum value on $(0,1/2)$. But I don't understand how you get $\alpha=0.178691$. Please explain. – Argha Feb 21 '14 at 04:18
  • @Glen_b : I am a student. I am doing a course on Robust Statistics. This problem is under that course. And simulation study is permissible. But how I use that. Please help. – Argha Feb 21 '14 at 04:23
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    Thanks for being completely clear. You can ask "homework-related" questions, but you should add the `self-study` tag to your post and read its [tag wiki info](http://stats.stackexchange.com/tags/self-study/info). You may need to add to the information in your question and modify the kind of question you ask a little. – Glen_b Feb 21 '14 at 05:51
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    Argha, your experience echoes mine: I think it is not possible to find a closed form expression and therefore I sought a numerical solution. Most automatic "black-box" solvers will have no trouble with this, because the function is convex on $(0,1/2)$. For instance, the Newton-Raphson method ought to work well. Plotting $\sigma^2(\alpha,F)$ against $\alpha$ is a good idea because it reveals much. For instance, $\sigma^2$ increases less than $10\%$ above its minimum throughout the (wide) interval $(0.12,0.26)$. This gives you wide latitude to select a good value of $\alpha$. – whuber Feb 21 '14 at 15:41
  • @whuber : Thank you. But please help me to find value of $F^{−1}(1−α)$ and $\int_0^{{F^{−1}}(1−α)}t^2 dF(t)$. And what I take as initial value for newton raphson, – Argha Feb 21 '14 at 15:48
  • @whuber Is $F^{−1}(1−α)=\cot(\pi\alpha)$? – Argha Feb 21 '14 at 15:49
  • Any initial value close to the minimum will work just fine. Finding $F$ and $F^{-1}$ is a matter of elementary integration: $F(t)=\frac{\tan ^{-1}(t)}{\pi }+\frac{1}{2},$ whose inverse (on the interval $[0,1]$) is $-\cot (\pi x).$ – whuber Feb 21 '14 at 15:52
  • Related: https://stats.stackexchange.com/q/373526/119261. – StubbornAtom May 31 '20 at 14:54

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