题目（高三 · 计数）：设 $x_{1} 、 x_{2} 、 x_{3} 、 x_{4}$ 为互不相等的正实数, 随机变量 $X$ 和 $Y$ 的分布列如下表,若记 $D X, D Y$ 分别为 $X, Y$ 的方差, 则 $D X \ldots D Y$. (填>, <, =)

| $X$ | $x_{1}$ | $x_{2}$ | $x_{3}$ | $x_{4}$ |
| :--- | :--- | :--- | :--- | :--- |


| $Y$ | $\frac{x_{1}+x_{2}}{2}$ | $\frac{x_{2}+x_{3}}{2}$ | $\frac{x_{3}+x_{4}}{2}$ | $\frac{x_{4}+x_{1}}{2}$ |
| :---: | :---: | :---: | :---: | :---: |
| $P$ | $\frac{1}{4}$ | $\frac{1}{4}$ | $\frac{1}{4}$ | $\frac{1}{4}$ |
选项：
答案：答案:>

【解析】 $E X=\frac{1}{4}\left(x_{1}+x_{2}+x_{3}+x_{4}\right)$,

故 $D X=\frac{1}{4}\left[\left(x_{1}-E X\right)^{2}+\left(x_{2}-E X\right)^{2}+\left(x_{3}-E X\right)^{2}+\left(x_{4}-E X\right)^{2}\right]$

$=\frac{1}{4}\left[\sum_{i=1}^{4} x_{i}^{2}-4(E X)^{2}\right]$.
$E Y=\frac{1}{4}\left(\frac{x_{1}+x_{2}}{2}+\frac{x_{2}+x_{3}}{2}+\frac{x_{3}+x_{4}}{2}+\frac{x_{4}+x_{1}}{2}\right)=\frac{1}{4}\left(x_{1}+x_{2}+x_{3}+x_{4}\right)=E X$,

$D Y=\frac{1}{4}\left[\left(\frac{x_{1}+x_{2}}{2}-E X\right)^{2}+\left(\frac{x_{2}+x_{3}}{2}-E X\right)^{2}+\left(\frac{x_{3}+x_{4}}{2}-E X\right)^{2}+\left(\frac{x_{4}+x_{1}}{2}-E X\right)^{2}\right]$

$=\frac{1}{4}\left[\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\left(\frac{x_{2}+x_{3}}{2}\right)^{2}+\left(\frac{x_{3}+x_{4}}{2}\right)^{2}+\left(\frac{x_{4}+x_{1}}{2}\right)^{2}-4(E X)^{2}\right]$.

要比较 $D X, D Y$ 的大小, 只需比较 $\sum_{i=1}^{4} x_{i}^{2}$ 与 $\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\left(\frac{x_{2}+x_{3}}{2}\right)^{2}+\left(\frac{x_{3}+x_{4}}{2}\right)^{2}+\left(\frac{x_{4}+x_{1}}{2}\right)^{2}$,

两者作差并化简得 $\sum_{i=1}^{4} x_{i}^{2}-\left[\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\left(\frac{x_{2}+x_{3}}{2}\right)^{2}+\left(\frac{x_{3}+x_{4}}{2}\right)^{2}+\left(\frac{x_{4}+x_{1}}{2}\right)^{2}\right]$

$=\frac{2 x_{1}^{2}+2 x_{2}^{2}+2 x_{3}^{2}+2 x_{4}^{2}-2\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{1}\right)}{4}$

$=\frac{\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\left(x_{3}-x_{4}\right)^{2}+\left(x_{4}-x_{1}\right)^{2}}{4}$ (1),

由于 $x_{1}, x_{2}, x_{3}, x_{4}$ 为互不相等的正实数, 故(1) $>0$,

也即 $\sum_{i=1}^{4} x_{i}^{2}>\left[\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\left(\frac{x_{2}+x_{3}}{2}\right)^{2}+\left(\frac{x_{3}+x_{4}}{2}\right)^{2}+\left(\frac{x_{4}+x_{1}}{2}\right)^{2}\right]$,

也即 $D X>D Y$, 故答案为>
解析：答案:>

【解析】 $E X=\frac{1}{4}\left(x_{1}+x_{2}+x_{3}+x_{4}\right)$,

故 $D X=\frac{1}{4}\left[\left(x_{1}-E X\right)^{2}+\left(x_{2}-E X\right)^{2}+\left(x_{3}-E X\right)^{2}+\left(x_{4}-E X\right)^{2}\right]$

$=\frac{1}{4}\left[\sum_{i=1}^{4} x_{i}^{2}-4(E X)^{2}\right]$.
$E Y=\frac{1}{4}\left(\frac{x_{1}+x_{2}}{2}+\frac{x_{2}+x_{3}}{2}+\frac{x_{3}+x_{4}}{2}+\frac{x_{4}+x_{1}}{2}\right)=\frac{1}{4}\left(x_{1}+x_{2}+x_{3}+x_{4}\right)=E X$,

$D Y=\frac{1}{4}\left[\left(\frac{x_{1}+x_{2}}{2}-E X\right)^{2}+\left(\frac{x_{2}+x_{3}}{2}-E X\right)^{2}+\left(\frac{x_{3}+x_{4}}{2}-E X\right)^{2}+\left(\frac{x_{4}+x_{1}}{2}-E X\right)^{2}\right]$

$=\frac{1}{4}\left[\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\left(\frac{x_{2}+x_{3}}{2}\right)^{2}+\left(\frac{x_{3}+x_{4}}{2}\right)^{2}+\left(\frac{x_{4}+x_{1}}{2}\right)^{2}-4(E X)^{2}\right]$.

要比较 $D X, D Y$ 的大小, 只需比较 $\sum_{i=1}^{4} x_{i}^{2}$ 与 $\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\left(\frac{x_{2}+x_{3}}{2}\right)^{2}+\left(\frac{x_{3}+x_{4}}{2}\right)^{2}+\left(\frac{x_{4}+x_{1}}{2}\right)^{2}$,

两者作差并化简得 $\sum_{i=1}^{4} x_{i}^{2}-\left[\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\left(\frac{x_{2}+x_{3}}{2}\right)^{2}+\left(\frac{x_{3}+x_{4}}{2}\right)^{2}+\left(\frac{x_{4}+x_{1}}{2}\right)^{2}\right]$

$=\frac{2 x_{1}^{2}+2 x_{2}^{2}+2 x_{3}^{2}+2 x_{4}^{2}-2\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{1}\right)}{4}$

$=\frac{\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\left(x_{3}-x_{4}\right)^{2}+\left(x_{4}-x_{1}\right)^{2}}{4}$ (1),

由于 $x_{1}, x_{2}, x_{3}, x_{4}$ 为互不相等的正实数, 故(1) $>0$,

也即 $\sum_{i=1}^{4} x_{i}^{2}>\left[\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\left(\frac{x_{2}+x_{3}}{2}\right)^{2}+\left(\frac{x_{3}+x_{4}}{2}\right)^{2}+\left(\frac{x_{4}+x_{1}}{2}\right)^{2}\right]$,

也即 $D X>D Y$, 故答案为>