If $X$, $Y$ are i.i.d. with mean $m$ then $E((X-Y)\mathrm{sign}(X-m))>0$

Question

It is obvious that if $X$ and $Y$ are independent and uniformly distributed on $[a,b]$ random variables, and $E[X]=E[Y]=m\equiv\frac{a+b}{2}$, then the random variable $Z=(X-Y)\mathrm{sign}(X-m)$ has the positive expectation, that is, $E[Z]>0$.

Indeed, $$ E[Z]=\frac{1}{(b-a)^2}\int_{a}^{b}\int_{a}^{b}(x-y)\mathrm{sign}(x-m)dxdy= $$ $$ =\frac{1}{b-a}\int_{a}^{b}x\mathrm{sign}(x-m)dx-\frac{1}{(b-a)^2}\int_{a}^{b}\mathrm{sign}(x-m)dx\int_{a}^{b}ydy= $$ $$ =\frac{1}{b-a}\Big(\int_{m}^{b}xdx-\int_{a}^{m}xdx\Big)-\frac{m}{b-a}\int_{a}^{b}\mathrm{sign}(x-m)dx= $$ $$ =\frac{1}{b-a}\Big(\int_{m}^{b}xdx-\int_{a}^{m}xdx\Big)-\frac{m}{b-a}\Big(\int_{m}^{b}dx-\int_{a}^{m}dx\Big)=\frac{b-a}{4}>0. $$

Now, let $X$ and $Y$ be independent and identically distributed random variables with the probability density function $f(x)$, $E[X]=E[Y]=m$, and let $Z=(X-Y)\mathrm{sign}(X-m)$. Is it true that $E[Z]>0$?

My calculations: $$ E[Z]=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x-y)\mathrm{sign}(x-m)f(x)f(y)dxdy= $$ $$ =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x\mathrm{sign}(x-m)f(x)f(y)dxdy-\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}y\mathrm{sign}(x-m)f(x)f(y)dxdy= $$ $$ =\int_{-\infty}^{\infty}xf(x)\mathrm{sign}(x-m)dx-m\int_{-\infty}^{\infty}f(x)\mathrm{sign}(x-m)dx= $$ $$ =\int_{m}^{\infty}xf(x)dx-\int_{-\infty}^{m}xf(x)dx-m\Big(\int_{m}^{\infty}f(x)dx-\int_{-\infty}^{m}f(x)dx\Big). $$

And what will be if $X$ and $Y$ are not independent but identically distributed random variables with the joint probability density function $f(x,y)$, $E[X]=E[Y]=m$; $Z=(X-Y)\mathrm{sign}(X-m)$? Is it true that $E[Z]>0$ in this case too?

In this case we have $$ E[Z]=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x-y)\mathrm{sign}(x-m)f(x,y)dxdy= $$ $$ =\int_{m}^{\infty}\int_{-\infty}^{\infty}(x-y)f(x,y)dxdy-\int_{-\infty}^{m}\int_{-\infty}^{\infty}(x-y)f(x,y)dxdy. $$

Answer 1

Hint: show that $\mathbb{E}[Z] = 2m\mathbb{E}[\mathbb{1}_{X < m}(1 - \frac{X}{m})]$, therefore yielding $\mathbb{E}[Z] > 0$.

Detailed hint: We have the following identities: $\mathbb{E}[Z] =\int_{m}^{\infty}xf(x)dx-\int_{-\infty}^{m}xf(x)dx-m\Big(\int_{m}^{\infty}f(x)dx-\int_{-\infty}^{m}f(x)dx\Big).$

(1) $\int_{-\infty}^{\infty}xf(x)dx = m = \int_{-\infty}^{m}xf(x)dx + \int_{m}^{\infty}xf(x)dx $, hence $\int_{m}^{\infty}xf(x)dx = m -\int_{-\infty}^{m}xf(x)dx$

(2) $\int_{-\infty}^{\infty}f(x)dx= 1 =\int_{-\infty}^{m}f(x)dx + \int_{m}^{\infty}f(x)dx$ hence $\int_{m}^{\infty}f(x)dx = 1 - \int_{-\infty}^{m}f(x)dx$

Thus $\mathbb{E}[Z] = 2m\big(\int_{-\infty}^{m} f(x)(1-\frac{x}{m})dx\big) $

Answer 2

Regarding the independent case, observe that \begin{align} \mathbb{E}[Z]\ =&\ \int_{m}^{\infty}\color{red}{xf(x)}dx-\int_{-\infty}^{m}\color{blue}{xf(x)}dx-m\Big(\int_{m}^{\infty}\color{red}{f(x)}dx-\int_{-\infty}^{m}\color{blue}{f(x)}dx\Big) \\ =&\ \int_m^\infty\color{red}{(x - m)f(x)}dx - \int_{-\infty}^m\color{blue}{(x - m)f(x)}dx \\ =&\ \int_m^\infty \color{red}{(x - m)f(x)}dx + \int_{-\infty}^m\color{blue}{(m - x)f(x)}dx \\ =&\ \int_{-\infty}^{\infty} |x - m|f(x)dx \\ =&\ \mathbb{E}[|X - m|] \end{align} $\mathbb{E}[Z]$ equals the expectation of a positive random variable (i.e. $|X - m|$), thus it is positive. Regarding the dependent case, consider the counterexample $X = Y$.