TITLE: Projection of Gaussian distribution along a vector.
QUESTION [9 upvotes]: Can anyone help me understand how to compute the projection of a 2D gaussian distribution along a vector. I intuitively realize that the projection will result in a 1D Gaussian, but I want to be sure. Can someone help me understand/show a proof/direct me to a proof where a 2D gaussian projected along a vector gives a line. 
Eg. Consider a Gaussian $\mathbf{X} \sim N (\mu,\Sigma)$ where $\mu = [3,2]^T$ and $\Sigma = \begin{bmatrix} 4 & 0 \\ 0 & 7 \end{bmatrix}$, what is the projection along the vector $v = 2i + 4j$ ? 
Any help would be much appreciated!! Thanks

REPLY [5 votes]: Let $\mathbf{x}\sim\mathcal{N}(\mu_x, \Sigma_x)$ be an $n$-dimensional Gaussian distribution. Then, if $y=\mathbf{v}^\top\mathbf{x}$, where $\mathbf{v}\in\Bbb{R}^n$, it holds that
$$
y\sim\mathcal{N}(\mu_y, \sigma_y^2),
$$
where $\mu_y=\mathbf{v}^\top\mu_x$ and $\sigma_y^2=\mathbf{v}^\top\Sigma_x\mathbf{v}$, since
$$
\mu_y=\Bbb{E}[y]=\Bbb{E}[\mathbf{v}^\top\mathbf{x}]=\mathbf{v}^\top\Bbb{E}[\mathbf{x}]=\mathbf{v}^\top\mu_x
$$
and
$$
\sigma_y^2 = \Bbb{E}[(y-\mu_y)^2]
=
\Bbb{E}[(\mathbf{v}^\top\mathbf{x}-\mathbf{v}^\top\mu_x)^2]
=
\Bbb{E}[(\mathbf{v}^\top(\mathbf{x}-\mu_x))^2]
=
\Bbb{E}[\mathbf{v}^\top(\mathbf{x}-\mu_x)\mathbf{v}^\top(\mathbf{x}-\mu_x)]
=
\Bbb{E}[\mathbf{v}^\top(\mathbf{x}-\mu_x)(\mathbf{x}-\mu_x)^\top\mathbf{v}]
=
\mathbf{v}^\top\Bbb{E}[(\mathbf{x}-\mu_x)(\mathbf{x}-\mu_x)^\top]\mathbf{v}
=
\mathbf{v}^\top\Sigma_x\mathbf{v}.
$$