仿射变换

若 $X\sim N(\mu,\Sigma)$，$A$ 为可逆矩阵，$Y=AX+b$，则 $Y\sim N(A\mu+b,A\Sigma A^T)$

证明： $F_ {Y}(y)=F_ {X}(A^{-1}(y-b))$，因此 $f_ {Y}(y)dy=f_ {X}(A^{-1}(y-b))dx$， $\begin{aligned} &f_ {Y}(y)\\ =&f_X(A^{-1}(y-b))\frac{1}{\det A}\\ =&\frac{1}{\sqrt{(2\pi)^{n}\det(A\Sigma A^T)}}e^{-\frac{1}{2}(A^{-1}(y-b)-\mu)^T\Sigma^{-1}(A^{-1}(y-b)-\mu)}\\ =&\frac{1}{\sqrt{(2\pi)^{n}\det(A\Sigma A^T)}}e^{-\frac{1}{2}(y-(A\mu+b))^T(A\Sigma A^T)^{-1}(y-(A\mu+b))} \end{aligned}$

进一步地，可得结论：

若 $X\sim N(\mu,\Sigma)$，$A$ 为行满秩的 $m\times n$ 实矩阵，$Y=AX+b$，则 $Y\sim N(A\mu+b,A\Sigma A^T)$

更进一步，可得结论：

若 $X_1,\cdots,X_n$ 相互独立，且 $X_i\sim N(\mu_i,\sigma_i^2)$，则对于任意一组不全为零的 $a_1,\cdots,a_n$，有 $\sum_ {i=1}^{n}a_i X_i+b\sim N(\sum_ {i=1}^n a_i\mu_i+b, \sum_ {i=1}^n a_i^2\sigma_i^2)$

独立性

设 $\left(\begin{array}{c} X_1 \\ X_2 \end{array}\right) \sim N\left( \left(\begin{array}{c} \mu_1 \\ \mu_2 \end{array}\right), \left(\begin{array}{c c} \Sigma_ {11} & 0\
0 & \Sigma_ {22} \end{array}\right) \right)$，则 $X_i\sim N(\mu_i,\Sigma_ {ii})$，且 $X_1$、$X_2$ 独立。

证明： $\begin{aligned} &f_ {X_1,X_2}(x_1,x_2)\\ =&\frac{1}{\sqrt{(2\pi)^{n_1+n_2}\det\Sigma_ {11}\det\Sigma_ {22}}}e^{-\frac{1}{2}\left[(x_1-\mu_1)^T\Sigma_ {11}^{-1}(x_1-\mu_1)+(x_2-\mu_2)^T\Sigma_ {22}^{-1}(x_2-\mu_2)\right]}\\ =&\frac{1}{\sqrt{(2\pi)^{n_1}\det\Sigma_ {11}}}e^{-\frac{1}{2}(x_1-\mu_1)^T\Sigma_ {11}^{-1}(x_1-\mu_1)} \frac{1}{\sqrt{(2\pi)^{n_2}\det\Sigma_ {22}}}e^{-\frac{1}{2}(x_2-\mu_2)^T\Sigma_ {22}^{-1}(x_2-\mu_2)} \end{aligned}$ 结论可见。

设 $\left(\begin{array}{c} X_1 \\ X_2 \end{array}\right) \sim N\left( \left(\begin{array}{c} \mu_1 \\ \mu_2 \end{array}\right), \left(\begin{array}{c c} \Sigma_ {11} & \Sigma_ {12} \
\Sigma_ {21} & \Sigma_ {22} \end{array}\right) \right)$，则 $X_i\sim N(\mu_i,\Sigma_ {ii})$，且 $X_1$、$X_2$ 独立等价于 $\Sigma_ {12}=\Sigma_ {21}=0$。

证明： 取 $A=\left(\begin{array}{c c}I_1 & 0 \\ -\Sigma_ {21}\Sigma_ {11}^{-1} & I_2\end{array}\right)$，则 $AX= \left(\begin{array}{c} X_1 \\ X_2-\Sigma_ {21}\Sigma_ {11}^{-1} X_1 \end{array}\right) \sim N \left( \left(\begin{array}{c} \mu_1 \\ \mu_2-\Sigma_ {21}\Sigma_ {11}^{-1} \mu_1 \end{array}\right), \left(\begin{array}{c c} \Sigma_ {11} & 0 \\ 0 & \Sigma_ {22}-\Sigma_ {21}\Sigma_ {11}^{-1} \Sigma_ {12} \end{array}\right) \right)$。故有 $X_i\sim N(\mu_i,\Sigma_ {ii})$，且 $X_1$ 和 $X_2-\Sigma_ {21}\Sigma_ {11}^{-1} X_1$ 独立。

因为 $X_1$ 和 $X_2-\Sigma_ {21}\Sigma_ {11}^{-1} X_1$ 独立且 $X_2-\Sigma_ {21}\Sigma_ {11}^{-1} X_1\sim N\left(\mu_2-\Sigma_ {21}\Sigma_ {11}^{-1} \mu_1, \Sigma_ {22}-\Sigma_ {21}\Sigma_ {11}^{-1} \Sigma_ {12}\right)$，可得条件分布 $X_2 | X_1\sim N\left(\mu_2+\Sigma_ {21}\Sigma_ {11}^{-1}(X_1-\mu_1), \Sigma_ {22}-\Sigma_ {21}\Sigma_ {11}^{-1} \Sigma_ {12}\right)$