Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose I have $n$ independent, non-identically distributed random variables $X_1, X_2, ..., X_n$. Let $\mathbf{X}$ denote the corresponding random vector with joint density $f(\mathbf{X})$ simply the product of the individual densities.

Let $X_{(i)}(\mathbf{X})$ denote the function that returns the $i^{th}$ order statistic, and let $T(\mathbf{X})$ denote the trimmed mean in which the $d$ lowest and $d$ highest realizations are dropped from the average. Let $l\equiv d+1$ and $h \equiv n-d$, and $t\equiv n-2d$. Thus, we can define $$T(\mathbf{X})=\frac{1}{t}\sum_{i=l}^{h}X_{(i)}(\mathbf{X})$$

Finally, let $\delta^{p}_{(i)}(\mathbf{X})$ be an indicator that turns on if the $p^{th}$ random variable $X_p$ happens to be the $i^{th}$ order statistic. Formally, $\delta^{p}_{(i)}(\mathbf{X})$ equals $1$ if $X_{(i-1)}(\mathbf{X}_{-p}) < X_p < X_{(i)}(\mathbf{X}_{-p})$, and $0$ otherwise. I use $\mathbf{X}_{-p}$ to denote the random vector $\mathbf{X}$ excluding $X_p$.

I'm interested in proving the following, which I believe is true: $$\text{Cov}\bigg( T(\mathbf{X}),\; X_p \times \delta^{p}_{(i)}(\mathbf{X}) \mid X_{(l-1)}(\mathbf{X}_{-p})<X_p<X_{(h)}(\mathbf{X}_{-p})\bigg) \geq 0$$ where $l \leq i \leq h$. So I'm conditioning on the event that $X_p$ is included in the trimmed mean calculation.

Any help?

It seems obvious that this has to be true. An average must be positively correlated with one of its elements. And on top of that, I believe that all order statistics must be positively correlated (though I can't prove that or find a reference), so I think that $X_p$ must be positively correlated with every order statistic included in $T(\mathbf{X})$.

Just a note: I'm using $f(\mathbf{X})$ as my probability measure. It might seem more natural to use a probability measure defined on the space of order statistic realizations, but then I think that it becomes hard to condition in the way that I need to.

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.