Expectation for a skewed distribution Let $X$ be a continuous random variable with distribution $F(x)=P(X\leq x)$ and support $[0,\infty)$. Suppose for a fixed $X_0$ and any $t\geq 0, P(X\geq X_0 +t) \geq P(X\leq X_0-t)$, prove that $E(X)\geq X_0$.
Intuitively, it is cleared to me that the expectation is no less than $X_0$ as the distribution is skewed to the right (the chance of $X$ to rise at least $t$ is no less than the chance of $X$ drops at least $t$). But I couldn't prove it rigorously.
Here is what i tried so far: Let $f(x)=F'(x)$ be the density function. It is easy to deduce that $\int_{X_0}^{\infty} f(x) dx\geq \frac{1}{2}$ and  $\int_0^{X_0} f(x) dx \leq \frac{1}{2}.$
Then $$E(X)=\int_0^{\infty}xf(x)dx=\int_0^{X_0}xdF(x) + \int_{X_0}^{\infty} xf(x) dx$$
$$=[xF(x)]_0^{X_0}-\int_0^{X_0}F(x)dx + \int_{X_0}^{\infty} xf(x) dx$$
$$=X_0F(X_0)-\int_0^{X_0}\int_0^x f(t) dt dx + \int_{X_0}^{\infty} xf(x) dx$$
$$\geq X_0F(X_0)-\int_0^{X_0}\int_0^{X_0} f(t) dt dx + \int_{X_0}^{\infty} X_0f(x) dx$$
$$\geq X_0F(X_0) -\int_0^{X_0} \frac{1}{2} dx + \frac{X_0}{2}$$
$$=X_0F(X_0) -\frac{X_0}{2} + \frac{X_0}{2}=X_0F(X_0)$$
But then $F(X_0) \leq 1$. Please help
 A: \begin{align*}
EX&=\int_0^\infty P(X\geq y)\,dy\\&=\int_0^{X_0}P(X\geq y)\,dy+\int_{X_0}^\infty P(X\geq y)\,dy\\&=X_0-\int_0^{X_0}P(X\leq y)\,dy+\int_{X_0}^\infty P(X\geq y)\,dy\\&\geq X_0+\int_{X_0}^{2X_0}P(X\geq y)\,dy-\int_0^{X_0} P(X\leq y)dy\\&=X_0+\int_0^{X_0}\left(P(X\geq X_0+t)-P(X\leq X_0-t)\right)\,dt\\&\geq X_0.
\end{align*}
A: Integrating the inequality yields $$\begin{align}
\int_0^\infty P(X\geq  x_0+t) dt &\geq \int_0^\infty P(X\leq  x_0+t) dt \\
\iff \int_0^\infty \int_0^\infty 1_{x\geq x_0-t} dP_X(x) dt &\geq \int_0^\infty \int_0^\infty  1_{x\leq x_0-t} dP_X(x) dt \\
\iff \int_{[x_0,\infty)} \int_0^{x-x_0} 1 \;dt dP_X(x) &\geq \int_{[0,x_0]} \int_0^{x_0-x}  1 \;dt dP_X(x) \\
\iff \int_{[x_0,\infty)} (x-x_0) dP_X(x) &\geq \int_{[0,x_0]} (x_0-x) dP_X(x) \tag{1}\\
\iff \int_0^\infty (x-x_0) dP_X(x)&\geq 0 \tag{2}\\
\iff E(X-x_0)&\geq 0
\end{align}$$
Note that going from $(1)$ to $(2)$ requires the continuity of $X$, since we need $$\int_{[0,x_0]} (x-x_0) dP_X(x)  = \int_{[0,x_0)} (x-x_0) dP_X(x) $$
