Affine transformation applied to a multivariate Gaussian random variable - what is the mean vector and covariance matrix of the new variable? Given a random vector $\mathbf x \sim N(\mathbf{\bar x}, \mathbf{C_x})$ with normal distribution. $\mathbf{\bar x}$ is the mean value vector and $\mathbf{C_x}$ is the covariance matrix of $\mathbf{x}$.
An affine transformation is applied to the $\mathbf{x}$ vector to create a new random $\mathbf{y}$ vector:
$$
\mathbf{y} = \mathbf{Ax} + \mathbf{b}
$$
Can we find mean value $\mathbf{\bar y}$ and covariance matrix $\mathbf{C_y}$ of this new vector $\mathbf{y}$ in terms of already given parameters ($\mathbf{\bar x}$, $\mathbf{C_x}$, $\mathbf{A}$ and $\mathbf{b}$)?
Can you please show the steps. Once I learn the method, I will use it on several other distributions myself.
 A: We find the mean of $\mathbf{y}$ by using the fact that $\mathbb{E}\{\}$ is a linear operator.
$$
\mathbf{\bar{y}} = \mathbb{E}\{\mathbf{y}\} = \mathbb{E}\{\mathbf{A}\mathbf{x}+\mathbf{b}\} = \mathbf{A}\mathbb{E}\{\mathbf{x}\}+\mathbf{b} = \mathbf{A}\mathbf{\bar{x}}+\mathbf{b}
$$
Then we find covariance of
$$
\begin{array}{rcl}
\mathbf{C_y} & \triangleq & \mathbb{E}\{(\mathbf{y}-\mathbf{\bar{y}})(\mathbf{y}-\mathbf{\bar{y}})^\top\} \\
& = & \mathbb{E} \Big\{ \Big[ (\mathbf{A}\mathbf{x}+\mathbf{b})-(\mathbf{A}\mathbf{\bar{x}}+\mathbf{b}) \Big] \Big[ (\mathbf{A}\mathbf{x}+\mathbf{b})-(\mathbf{A}\mathbf{\bar{x}}+\mathbf{b}) \Big] ^\top \Big\} \\
& = & \mathbb{E} \Big\{ \Big[ \mathbf{A}(\mathbf{x}-\mathbf{\bar{x}}) \Big] \Big[ \mathbf{A}(\mathbf{x}-\mathbf{\bar{x}}) \Big] ^\top \Big\} \\
& = & \mathbb{E} \Big\{ \mathbf{A}(\mathbf{x}-\mathbf{\bar{x}}) (\mathbf{x}-\mathbf{\bar{x}})^\top \mathbf{A}^\top \Big\} \\
& = & \mathbf{A} \mathbb{E} \Big\{ (\mathbf{x}-\mathbf{\bar{x}}) (\mathbf{x}-\mathbf{\bar{x}})^\top  \Big\} \mathbf{A}^\top \\
& = & \mathbf{A}\mathbf{C_x}\mathbf{A}^\top
\end{array}
$$
Then, $\mathbf{y}$ is defined as,
$$
\mathbf{y} \sim \mathcal{N}(\mathbf{A}\mathbf{\bar{x}+\mathbf{b}, \mathbf{A}\mathbf{C_x}\mathbf{A}^\top})
$$
That is,
$$
f_\mathbf{Y}(\mathbf{y)}
= {1 \over \sqrt{\lvert2\pi\mathbf{A}\mathbf{C_x}\mathbf{A}^\top\rvert}}
\exp\left(- {1 \over 2} \big[\mathbf{y}-(\mathbf{A}\mathbf{\bar{x}}+\mathbf{b}) \big]^\top (\mathbf{A}\mathbf{C_x}\mathbf{A}^\top)^{-1} \big[\mathbf{y}-(\mathbf{A}\mathbf{\bar{x}}+\mathbf{b}) \big] \right)
$$
