题目
设 $X$ 服从区间 $(-\frac{\pi}{2},\frac{\pi}{2})$ 上的均匀分布，$Y=\sin X$, 则 $Cov(X,Y)=$
解析
由题知：
$$f(x)=\left\{\begin{matrix}
\frac{1}{\pi},-\frac{\pi}{2}<x<\frac{\pi}{2}\\
0,Others
\end{matrix}\right.$$
则：
\begin{align}
Cov(X,Y) \\
& =EXY-EXEY \\
& =E(X \sin X)-EXE(\sin X) \\
& =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x \sin x \frac{1}{\pi}dx-\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x\frac{1}{\pi}dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin x \frac{1}{\pi}dx \\
& =2\frac{1}{\pi}\int_{0}^{\frac{\pi}{2}}x \sin x dx-0 \\
& =\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}(-x)d(\cos x) \\
& =\frac{2}{\pi}[(-x)\cos x|_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\cos xd(-x)] \\
& =\frac{2}{\pi}[(-x)\cos x|_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\cos xdx] \\
& =\frac{2}{\pi}(0+\sin x|_{0}^{\frac{\pi}{2}}) \\
& =\frac{2}{\pi}(0+1) \\
& =\frac{2}{\pi}.
\end{align}
EOF