I'm in the process of writing my bachelor thesis and while trying to find/understand proof of one theorem I got quite a headache. I would like to get some feedback if the presented proof is correct and if not maybe you can help me.
Theorem:
Suppose we have two functions $f,g : \mathbb{R}^n\to\mathbb{R}$ and for each coordinate $i$ one is non-decreasing while other is non-increasing (note that e.g $f$ doesn't have to be non-increasing/non-decreasing for all coordinates). Let $X=(X_1,...,X_n)$ be a random variable where $X_i$ are independent. Then:
$$\mathrm{Cov}\big(f(X),g(X)\big)\leq0$$
Proof (by induction):
n=1
Let $x,y\in\mathbb{R}$. Consider functions $F,G:\mathbb{R}\to\mathbb{R}$ where without loss of generality $F$ - nonincreasing, $G$ - nondecreasing. Then
$$\big(F(x)-F(y)\big)\big(G(x)-G(y)\big)\leq0\,.$$
Let $Y$ be an independent copy of $X$. Inequality above implies that after conditioning $f,g$ on $X_2=x_2,...,X_n=x_n$ and $Y_2=y_2,...,Y_n=y_n$, we get
$$\mathrm{E}\Big[\big[f(X_1,x_2,...,x_n)-f(Y_1,y_2,...,y_n)\big]\big[g(X_1,x_2,...,x_n)-g(Y_1,y_2,...,y_n)\big]\Big]\leq0\,.$$
Then after multiplying brackets
$$\mathrm{E}\big[f(X_1,x_2,...,x_n)g(X_1,x_2,...,x_n)\big] + \mathrm{E}\big[f(Y_1,y_2,...,y_n)g(Y_1,y_2,...,y_n)\big]\leq \\\mathrm{E}\big[f(X_1,x_2,...,x_n)g(Y_1,y_2,...,y_n)\big] + \mathrm{E}\big[f(Y_1,y_2,...,y_n)g(X_1,x_2,...,x_n)\big]_{\,.}$$
We use the independence of $X,Y$ and assume this holds for $n-1$, meaning
$$\mathrm{E}\big[f(X_1,...,X_{n-1},x_n)g(X_1,...,X_{n-1},x_n)\big] + \mathrm{E}\big[f(Y_1,...,Y_{n-1},y_n)g(Y_1,...,Y_{n-1},y_n)\big]\leq \\\mathrm{E}\big[f(X_1,...,X_{n-1},x_n)\big]\mathrm{E}\big[g(Y_1,...,Y_{n-1},y_n)\big] + \mathrm{E}\big[f(Y_1,...,Y_{n-1},y_n)\big]\mathrm{E}\big[g(X_1,...,X_{n-1},x_n)\big]_{\,.}$$
By taking expected value of both sides $$\mathrm{E}\Big[\mathrm{E}\big[f(X_1,...,X_{n-1},x_n)g(X_1,...,X_{n-1},x_n)\big]\Big] + \mathrm{E}\Big[\mathrm{E}\big[f(Y_1,...,Y_{n-1},y_n)g(Y_1,...,Y_{n-1},y_n)\big]\Big]\leq\\ \mathrm{E}\Big[\mathrm{E}\big[f(X_1,...,X_{n-1},x_n)\big]\Big]\mathrm{E}\Big[\mathrm{E}\big[g(Y_1,...,Y_{n-1},y_n)\big]\Big] + \mathrm{E}\Big[\mathrm{E}\big[f(Y_1,...,Y_{n-1},y_n)\big]\Big]\mathrm{E}\Big[\mathrm{E}\big[g(X_1,...,X_{n-1},x_n)\big]\Big]_{\,,}$$
and applying law of absolute value we get induction result for case $n$: $$\mathrm{E}\big[f(X_1,...,X_n)g(X_1,...,X_n)\big] + \mathrm{E}\big[f(Y_1,...,Y_n)g(Y_1,...,Y_n)\big]\leq \\\mathrm{E}\big[f(X_1,...,X_n)\big]\mathrm{E}\big[g(Y_1,...,Y_n)\big] + \mathrm{E}\big[f(Y_1,...,Y_n)\big]\mathrm{E}\big[g(X_1,...,X_n)\big]_{\,.}$$
Since $X,Y$ are from same distribution this completes the proof as $$0\geq\ 2*\Big[\mathrm{E}\big[f(X)g(X)\big]-\mathrm{E}\big[f(X)\big]\mathrm{E}\big[g(X)\big]\Big]=2\mathrm{Cov}\big(f(X),g(X)\big)\,.$$
This is a general theorem but really I only need a special case where $f$-monotone for each $i$ with random variables $U_i,1-U_i\in\mathcal{U}[0,1]$. Then $\mathrm{Cov}\big(f(U_1,...,U_n),f(1-U_1,...,1-U_n)\big)\leq0\,.$
Not sure if you only want a checkout with the proof, but it may be interesting for you that there is a lot of literature that deals with this problem. It is called the associated random variables theory. Your theorem was proven many times, see e.g. Theorem 2.1. in http://www.stat.cmu.edu/~brian/720/week02/esary-proschan-walkup-1967.pdf (well, you just switch g=-g)