乘积大于二分之一

在单位正方形上积分。条件 $XY>1/2$ 等价于 $y>\\frac{1}{2x}$ 且 $x\\ge 1/2$。\n\n面积为\n\n$$\n\\int_{1/2}^{1}\\left(1-\\frac{1}{2x}\\right)dx=\\left[x-\\frac{1}{2}\\ln x\\right]_{1/2}^{1}=\\frac{1}{2}(1-\\ln 2).\n$$\n\n因此\n\n$$\n\\boxed{\\mathbb{P}(XY>1/2)=\\frac{1}{2}(1-\\ln 2)\\approx 0.1534}.\n$$

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英文解析

Imagine the surface $f(x,y) = x\cdot y$ plotted above the unit square. We need to find the area of that part of the domain where $f(x,y) = x\cdot y > 1 / 2$ . If we project this down onto the $x - y$ space, we need only find the area within the unit square above the isovalue curve $x\cdot y = 1 / 2$ . That is, we need the area within the unit square that is above $y = 1 / (2x)$

A quick sketch shows that this area is only in the right half of the unit square. So, our answer is one half less the area below $y = 1 / (2x)$ for $x > = 1 / 2$ . The answer is given as follows.

$$\begin{array}{l}{{P=\frac{1}{2}-\int_{x=1/2}^{x=1}\frac{1}{2\nu}d\nu}}\\ {{\quad=\frac{1}{2}-\int_{x=1/2}^{x=1}\frac{1}{2}\nu^{-1}d\nu}}\\ {{\quad=\frac{1}{2}(1-\ln(x))_{x=1/2}^{x=1})}}\\ {{\quad=\frac{1}{2}(1+\ln(\frac{1}{2}))\approx0.1534}}\end{array}$$