数学证明
随机变量乘积的期望： 已知两个随机变量x 1 x_1 x 1 和x 2 x_2 x 2 为相互独立, 则x 1 ⋅ x 2 x_1\cdot x_2 x 1 ⋅x 2 的期望为E ( x 1 ⋅ x 2 ) = E ( x 1 ) ⋅ E ( x 2 ) \mathbb{E}(x_1\cdot x_2)=\mathbb{E}(x_1)\cdot \mathbb{E}(x_2)E (x 1 ⋅x 2 )=E (x 1 )⋅E (x 2 )
证明：随机变量x 1 ⋅ x 2 x_1\cdot x_2 x 1 ⋅x 2 的期望为E ( x 1 ⋅ x 2 ) = E ( x 1 ) ⋅ E ( x 2 ) + C o v ( x 1 , x 2 ) \mathbb{E}(x_1\cdot x_2)=\mathbb{E}(x_1)\cdot\mathbb{E}(x_2)+\mathrm{Cov}(x_1,x_2)E (x 1 ⋅x 2 )=E (x 1 )⋅E (x 2 )+C o v (x 1 ,x 2 )因为随机变量x 1 x_1 x 1 和x 2 x_2 x 2 相互独立，则C o v ( x 1 , x 2 ) = 0 \mathrm{Cov}(x_1,x_2)=0 C o v (x 1 ,x 2 )=0进而可知E ( x 1 ⋅ x 2 ) = E ( x 1 ) ⋅ E ( x 2 ) + 0 = E ( x 1 ) ⋅ E ( x 2 ) \mathbb{E}(x_1\cdot x_2)=\mathbb{E}(x_1)\cdot\mathbb{E}(x_2)+0=\mathbb{E}(x_1)\cdot\mathbb{E}(x_2)E (x 1 ⋅x 2 )=E (x 1 )⋅E (x 2 )+0 =E (x 1 )⋅E (x 2 )证毕。
随机变量乘积的方差： 已知两个随机变量x 1 x_1 x 1 和x 2 x_2 x 2 为相互独立, 则x 1 ⋅ x 2 x_1\cdot x_2 x 1 ⋅x 2 的方差为V a r ( x 1 ⋅ x 2 ) = V a r ( x 1 ) ⋅ V a r ( x 2 ) + V a r ( x 1 ) ⋅ E ( x 2 ) 2 + V a r ( x 2 ) ⋅ E ( x 1 ) 2 \mathrm{Var}(x_1\cdot x_2)=\mathrm{Var}(x_1)\cdot\mathrm{Var}(x_2)+\mathrm{Var}(x_1)\cdot \mathbb{E}(x_2)^2+\mathrm{Var}(x_2)\cdot \mathbb{E}(x_1)^2 V a r (x 1 ⋅x 2 )=V a r (x 1 )⋅V a r (x 2 )+V a r (x 1 )⋅E (x 2 )2 +V a r (x 2 )⋅E (x 1 )2
证明：已知随机变量x 1 x_1 x 1 和x 2 x_2 x 2 相互独立，则随机变量x 1 ⋅ x 2 x_1\cdot x_2 x 1 ⋅x 2 的方差为V a r ( x 1 ⋅ x 2 ) = E ( ( x 1 ⋅ x 2 − E ( x 1 ⋅ x 2 ) ) 2 ) = E ( x 1 2 ⋅ x 2 2 ) − E ( x 1 ⋅ x 2 ) 2 = E ( x 1 2 ) ⋅ E ( x 2 2 ) − E ( x 1 ) 2 ⋅ E ( x 2 ) 2 = ( V a r ( x 1 ) + E ( x 1 ) 2 ) ⋅ ( V a r ( x 2 ) + E ( x 2 ) 2 ) − E ( x 1 ) 2 ⋅ E ( x 2 ) 2 = V a r ( x 1 ) ⋅ V a r ( x 2 ) + V a r ( x 1 ) ⋅ E ( x 2 ) 2 + V a r ( x 2 ) ⋅ E ( x 1 ) 2 \begin{aligned}\mathrm{Var}(x_1\cdot x_2)&=\mathbb{E}\left((x_1\cdot x_2-\mathbb{E}(x_1\cdot x_2))^2\right)\&=\mathbb{E}(x^2_1\cdot x_2^2)-\mathbb{E}(x_1\cdot x_2)^2\&=\mathbb{E}(x^2_1) \cdot \mathbb{E}(x^2_2)-\mathbb{E}(x_1)^2\cdot \mathbb{E}(x_2)^2\&=(\mathrm{Var}(x_1)+\mathbb{E}(x_1)^2)\cdot(\mathrm{Var}(x_2)+\mathbb{E}(x_2)^2)-\mathbb{E}(x_1)^2\cdot \mathbb{E}(x_2)^2\&=\mathrm{Var}(x_1)\cdot \mathrm{Var}(x_2)+\mathrm{Var}(x_1)\cdot \mathbb{E}(x_2)^2+\mathrm{Var}(x_2)\cdot \mathbb{E}(x_1)^2\end{aligned}V a r (x 1 ⋅x 2 )=E ((x 1 ⋅x 2 −E (x 1 ⋅x 2 ))2 )=E (x 1 2 ⋅x 2 2 )−E (x 1 ⋅x 2 )2 =E (x 1 2 )⋅E (x 2 2 )−E (x 1 )2 ⋅E (x 2 )2 =(V a r (x 1 )+E (x 1 )2 )⋅(V a r (x 2 )+E (x 2 )2 )−E (x 1 )2 ⋅E (x 2 )2 =V a r (x 1 )⋅V a r (x 2 )+V a r (x 1 )⋅E (x 2 )2 +V a r (x 2 )⋅E (x 1 )2 证毕。
具体实例
给定两个独立同分布的随机变量x 1 x_1 x 1 和x 2 x_2 x 2 ，且x 1 , x 2 ∼ N ( 0 , 1 ) x_1,x_2\sim \mathcal{N}(0,1)x 1 ,x 2 ∼N (0 ,1 )，根据以上两随机变量乘积的期望公式可知，x 1 ⋅ x 2 x_1\cdot x_2 x 1 ⋅x 2 的期望为E ( x 1 ⋅ x 2 ) = E ( x 1 ) ⋅ E ( x 2 ) = 0 × 0 = 0 \mathbb{E}(x_1\cdot x_2)=\mathbb{E}(x_1)\cdot \mathbb{E}(x_2)=0\times 0 = 0 E (x 1 ⋅x 2 )=E (x 1 )⋅E (x 2 )=0 ×0 =0根据以上两随机变量乘积的方差公式可知x 1 ⋅ x 2 x_1\cdot x_2 x 1 ⋅x 2 的方差为V a r ( x 1 ⋅ x 2 ) = V a r ( x 1 ) ⋅ V a r ( x 2 ) + V a r ( x 1 ) ⋅ E ( x 2 ) 2 + V a r ( x 2 ) ⋅ E ( x 1 ) 2 = 1 × 1 + 1 × 0 + 1 × 0 = 1 \begin{aligned}\mathrm{Var}(x_1\cdot x_2)&=\mathrm{Var}(x_1)\cdot\mathrm{Var}(x_2)+\mathrm{Var}(x_1)\cdot \mathbb{E}(x_2)^2+\mathrm{Var}(x_2)\cdot \mathbb{E}(x_1)^2\&=1\times 1 +1\times 0+ 1\times 0\&=1\end{aligned}V a r (x 1 ⋅x 2 )=V a r (x 1 )⋅V a r (x 2 )+V a r (x 1 )⋅E (x 2 )2 +V a r (x 2 )⋅E (x 1 )2 =1 ×1 +1 ×0 +1 ×0 =1
Original: https://blog.csdn.net/qq_38406029/article/details/122269646
Author: 鬼道2022
Title: 随机变量乘积的期望和方差
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