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Based on this question, from one dimensional Gaussian to two dimensional Gaussian, I have the following question. Any help is appreciated.

Suppose $x$ is one-dimensional Gaussian distributed, with mean $\mu_x$ and variance $\sigma_x^2$, such that $$x \sim \mathcal N(\mu_x, \sigma_x^2).$$ Suppose $\mathbf y$ is two-dimensional Gaussian distributed, with mean $\boldsymbol{\mu}_y$, and covariance $\boldsymbol{\Sigma}_y$, such that $$ \mathbf y \sim \mathcal N(\boldsymbol{\mu}_y, \boldsymbol{\Sigma}_y).$$ What would the density function for $$\mathbf z=\mathbf y+\begin{bmatrix} a \\ b \end{bmatrix}x$$ be, given $a$ and $b$ are two constants?

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Assume that $x$ is independent from $\mathbf y$. The vector $\mathbf z$ remains a Gaussian vector since any linear combination of $z_1$ and $z_2$ can be written as $$ \alpha_1y_1+\alpha_1ax+\alpha_2y_2+\alpha_2bx=\underbrace{\alpha_1y_1+\alpha_2y_2}_{\text{Gaussian since $\mathbf y$ is a Gaussian vector}}+\underbrace{(\alpha_1a+\alpha_2b)x}_{\text{Gaussian independent from $\mathbf y$}}. $$ Hence, $\mathbf z$ is characterized by its mean vector and covariance matrix.

The mean vector is given by $$ \boldsymbol\mu_y+\begin{bmatrix} a\mu_x \\ b\mu_x \end{bmatrix}. $$ Additionally, $$ \mathrm{Var}\,z_1=\boldsymbol\Sigma_y({1,1})+a^2\sigma_x^2,\quad\mathrm{Var}\,z_2=\boldsymbol\Sigma_y({2,2})+b^2\sigma_x^2, $$ since $\mathbf y$ and $x$ are independent, and $$ \mathrm{Cov}\,(z_1,z_2)=\mathrm{Cov}\,(y_1,y_2)=\boldsymbol\Sigma_y({1,2}), $$ by bilinearity of $\mathrm{Cov}$. The covariance matrix is thus given by $$ \begin{bmatrix} \boldsymbol\Sigma_y({1,1})+a^2\sigma_x^2 &\boldsymbol\Sigma_y({1,2}) \\ \boldsymbol\Sigma_y({2,1})&\boldsymbol\Sigma_y({2,2})+b^2\sigma_x^2 \end{bmatrix}. $$

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Hint: If $x$ and $y$ are independent, then they are jointly Gaussian. So, you can characterize $z$ as being Gaussian with its mean and covariance matrix (which you can calculate from that of $x$ and $y$).

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