A surveyor wishes to lay out a square region with each side having length $L$. However, because of measurement error, he instead lays out a rectangle in which the north-south sides both have length $X$ and the east-west sides both have length $Y$. Suppose that $X$ and $Y$ are independent and the each is uniformly distributed on the interval $[L-A,L+A]$ (where $0 < A < L$). What is the expected area of the resulting rectangle?
1
$X$ 和 $Y$ 独立, 且都服从 $[L-A,L+A]$ 上的均匀分布. 因此,
\[
f(x,y)=f_X(x)\cdot f_Y(y),
\]
并且 $f_X(x)=f_Y(y)=\frac{1}{2A}$, 故 $f(x,y)=\frac{1}{2A}\cdot\frac{1}{2A}=\frac{1}{4A^2}$.
矩形的面积为 $S=XY$, 是一个二维随机变量. 其期望值为
\[
\begin{split}
E(S)=E(XY)&=\int_{L-A}^{L+A}\int_{L-A}^{L+A}xy\cdot f(x,y)dxdy\\
&=\int_{L-A}^{L+A}\int_{L-A}^{L+A}xy\cdot\frac{1}{4A^2}dxdy\\
&=\frac{1}{4A^2}\cdot\int_{L-A}^{L+A}\int_{L-A}^{L+A}xydxdy\\
&=\frac{1}{4A^2}\cdot\biggl(\int_{L-A}^{L+A}xdx\biggr)\cdot\biggl(\int_{L-A}^{L+A}ydy\biggr)\\
&=\frac{1}{4A^2}\cdot\biggl(\int_{L-A}^{L+A}xdx\biggr)^2\\
&=\frac{1}{4A^2}\cdot\biggl(\frac{1}{2}x^2\biggr|_{L-A}^{L+A}\biggr)^2\\
&=\frac{1}{4A^2}\cdot\biggl(\frac{1}{2}\bigl((L+A)^2-(L-A)^2\bigr)\biggr)^2\\
&=\frac{1}{4A^2}\cdot\biggl(\frac{1}{2}\cdot 4LA\biggr)^2\\
&=\frac{1}{4A^2}\cdot 4L^2 A^2\\
&=L^2
\end{split}
\]