什么时候$p$增加，这意味着它的累积分布函数

$$P(x) = \int_0^x p(x)\,\mathrm{d}x$$

是凸的。自从$P(0)=0$和$P(1)=1,$凸性意味着$P$位于连接线段之上或之下$(0,0)$和$(1,1),$一个覆盖三角形区域的段$1/2.$ （图形为图形曲线，线段为图中的虚线。）因此，图形下的面积$P$不能超过$1/2:$

$$\frac{1}{2} \ge \int_0^1 P(x)\,\mathrm{d}x.$$

该区域是图中阴影区域的区域：

自从

$$E_p(X) = \int_0^1 (1-P(x))\,\mathrm{d}x = 1 - \int_0^1 P(x)\,\mathrm{d}x \ge 1 - \frac{1}{2},$$ 它遵循$E_p(X) \ge 1/2.$ This minimum value is the best possible because it can be attained by the uniform distribution where $p(x)=1$ for $0\le x \le 1.$

First, change the problem to say that the density is non-decreasing and piecewise constant. Solve that problem first and then return to this problem.

Suppose there is a density $f(x)$ that minimizes $EX$ where $n$ is a positive integer and for the integers $1 \le i \le n$ the density is constant and equal to $a_i$ on each of the the intervals $((i-1)/n,i/n)$. The constants are non-decreasing and $a_1 \le a_2 \le ... \le a_n$. If there is any $j$ where the inequality is strict $a_j<a_{j+1}$, then you could define a new density equal to $f(x)$ outside of the interval $((j-1)/n,(j+1)/n)$ but constant and equal to the average $\frac{a_j+a_{j+1}}{2}$ in the interval $((j-1)/n,(j+1)/n)$ and then $EX$ would be strictly smaller. Thus, all $a_i$ must be equal to $1/n$. The uniform density has that property and is the only density that has that property for all $n$. If X is uniform, $EX=\frac{1}{2}$.

That proof only works for piecewise constant densities, but for an arbitrary density, the integral is defined as the limit of these piecewise constant approximations. Thus, the minimum over arbitrary non-decreasing densities is also $\frac{1}2$.

For the original problem, where you want $f(x)$ to be increasing, for any small $\epsilon>0$, you can have $EX=\frac{1}{2}+\epsilon$ by taking $f(X)=1+6\epsilon(2 x-1)$. But you can never have $EX=\frac{1}{2}$ with an increasing density because of the argument above.