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})
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.