如何证明两随机变量线性组合的V标准差有以下不等式成立\(\sigma_v<\sigma_{1 }\)

wufaxian

随机变量1 和随机变量2 的线性组合V 的方差如下图所示

\(\sigma_{V}^{2} = s^{2} \sigma_{1}^{2}+(1-s)^2 \sigma_{2}^{2}+2 s(1-s) c_{12}\)

其中s和（1-s）是线性组合的系数 ，\(c_{12 }\) 代表两个随机变量的协方差！

已知组合V的方差对s 的导数等于0时 s的表达式可以写成！



\(\rho_{12}\) 是随机变量1 和 随机变量2 的线性相关系数。

给定条件：
\(\sigma_{V}^{2} = s_{0}^{2} \sigma_{1}^{2}+(1-s_{0})^2 \sigma_{2}^{2}+2 s_{0}(1-s_{0}) c_{12}\)
\(\sigma_1<\sigma_2\)

求证：在这种情况下\(\sigma_v<\sigma_{1 }\) ！

请忽略下方图片。我删不掉！

wufaxian

补充了一两个条件，修改了一处错误。请各位老师再看看这个问题如何证明。

另外有没有人觉得这两天论坛打开速度特别慢？我想看看是不是只有我一个人感觉慢。

wufaxian

首先将\(\sigma_{v}^{2 }\) 变成标准二次函数的形式：
\begin{aligned}
&\sigma _{v}^{2}=s^{2}\sigma _{1}^{2}+\left( 1-s\right) ^{2}\sigma _{2}^{2}+2s\left( 1-s\right) c_{12}\\
&=s^{2}\sigma _{1}^{2}+\left( 1-2s+s^{2}\right) \sigma _{2}^{2}+2s\rho \sigma _{1}\sigma _{2}-2s^{2}\rho \sigma _{1}\sigma _{2}\\
&=s^{2}\left( \sigma _{1}^{2}+\sigma _{2}^{2}-2\rho \sigma _{1}\sigma _{2}\right) +s\left( 2\rho \sigma _{1}\sigma _{2}-2\sigma _{2}^{2}\right) +\sigma _{2}^{2}
\end{aligned}

将s看作自变量，其他看作系数

\begin{aligned}
&y=ax^{2}+bx+c;a >0\\
&y\geq \dfrac{4ac-b^{2}}{4a}\\
&\sigma _{v}^{2}\geq \dfrac{4\left( \sigma _{1}^{2}+\sigma _{2}^{2}-2\rho \sigma _{1}\sigma _{2}\right) \sigma _{2}^{2}-\left( 2\rho \sigma _{1}\sigma _{2}-2\sigma _{2}^{2}\right) ^{2}}{4\left( \sigma _{1}^{2}+\sigma _{2}^{2}-2\rho \sigma _{1}\sigma _{2}\right) }
\end{aligned}

\begin{aligned}
& \sigma_v^2 \geqslant \frac{4 \sigma_2^2 \sigma_1^2+4 \sigma_2^4-8 \rho \sigma_1 \sigma_2^3-4 \rho^2 \sigma_1^2 \sigma_2^2+8 \rho \sigma_1 \sigma_2^3-4 \sigma_2^4}{4\left(\sigma_1^2+\sigma_2^2-2 \rho \sigma_1 \sigma_2\right)} \\
& \sigma_v^2 \geqslant \frac{\sigma_2^2 \sigma_1^2-\rho^2 \sigma_1^2 \sigma_2^2}{\sigma_1^2+\sigma_2^2-2 \rho \sigma_1 \sigma_2}
\end{aligned}

\begin{aligned}
& \sigma_v^2 \geqslant \frac{\left(1-\rho^2\right) \sigma_1^2 \sigma_2^2}{\sigma_1^2+\sigma_2^2-2 \rho \sigma_1 \sigma_2} \\
& \because \rho<1 \\
& \therefore \sigma_v^2 \leqslant \frac{\left(\rho^2-1\right) \sigma_1^2 \sigma_2^2}{\sigma_1^2+\sigma_2^2-2 \rho \sigma_1 \sigma_2} \\
& \because \sigma_v>0 \\
& \therefore \sigma_v \leqslant \frac{\sqrt{\left(1-\rho^2\right)} \sigma_1 \sigma_2}{\sqrt{\sigma_1^2+\sigma_2^2-2 \rho \sigma_1^2 \sigma_2}} \\
\end{aligned}

\begin{aligned}
& \because \rho<1 \\
& 2 \rho\sigma_1 \sigma_2<2 \sigma_1 \sigma_2 \\
& \\
& \therefore \frac{\sqrt{1-\rho^2} \sigma_1 \sigma_2}{\sqrt{\sigma_1^2+\sigma_2^2-2 \rho \sigma_1 \sigma_2}} < \frac{\sqrt{1-\rho^2} \sigma_1 \sigma_2}{\sqrt{\sigma_1^2+\sigma_2^2-2 \sigma_1 \sigma_2}}=\frac{\sqrt{1-\rho^2} \sigma_1 \sigma_2}{\sigma_1+\sigma_2} \\
& \because 1-\rho^2<1 \\
& \therefore \sqrt{1-\rho^2} \sigma_2<\sigma_2<\sigma_1+\sigma_2 \\
& \therefore \frac{\sqrt{1-\rho^2} \sigma_2}{\sigma_1+\sigma_2}<1 \\
& \therefore \frac{\sqrt{1-\rho^2} \sigma_2 \sigma_1}{\sigma_1+\sigma_2}<\sigma_1 \\
&所以 \ \ \sigma _{v}<\sigma _{1}
\end{aligned}