Density on the square, expected value

Question

Let $f: [0,1]^2 \rightarrow \Bbb R^{+}$ a density function on the square.

I suppose that the random variable $X=(X_1,X_2)$ has the density f with respect to the lebesgue measure.

I denote $\bar{X}=(\bar{X_1},\bar{X_2})$ the average point in the square.

$E(\bar{X_1})= \int_0^1 \int_0^1 x f(x,y) dx dy$

$E(\bar{X_2})= \int_0^1 \int_0^1 y f(x,y) dx dy$

Is it true, or not, that any line on the square which countains $\bar{X}$ separates the square into two areas which has the same expected value?

Thx a lot

Answer 1

No:

Consider $f(x,y) = g(x)$ where $g$ is a distribution for which the mean is different from the median, and the line $$\left\{ x = EX_1 = \int_0^1 xg(x)dx \right\}$$

Then both areas $$ \int_{x>EX_1} f(x,y) dx dy = P(x>EX_1) \\ \int_{x<EX_1} f(x,y) dx dy = P(x<EX_1) $$ are different.