Projection of Gaussian distribution along a vector. 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
 A: 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
$$
\sigma_y^2 = \Bbb{E}[(y-\mu_y)^2]
=
\Bbb{E}[(\mathbf{v}^\top\mathbf{x}-\mathbf{v}^\top\mu_x)^2]
=
\Bbb{E}[(\mathbf{v}^\top(\mathbf{x}-\mu_x))^2]
=
\Bbb{E}[\mathbf{v}^\top(\mathbf{x}-\mu_x)\mathbf{v}^\top(\mathbf{x}-\mu_x)]
=
\Bbb{E}[\mathbf{v}^\top(\mathbf{x}-\mu_x)(\mathbf{x}-\mu_x)^\top\mathbf{v}]
=
\mathbf{v}^\top\Bbb{E}[(\mathbf{x}-\mu_x)(\mathbf{x}-\mu_x)^\top]\mathbf{v}
=
\mathbf{v}^\top\Sigma_x\mathbf{v}.
$$
A: In general for $\mathbb{R}^n$ space, given a column matrix $V$ where each column $V^j$ is a vector in $\mathbb{R}^n$, the projection to the subspace generated by $V^j$ is $V(V^tV)^{-1}V^t$ (let's assume $V^j$ are all independent so we don't have any issue with matrix rank). That is, for any vector $b$, it's orthogonal projection into the subspace generated by $V^j$ is 
$$p^{V}(b) = [V(V^tV)^{-1}V^t]b$$
That is the same for a Gaussian vector, when project to a subspace of $\mathbb{R}^n$, one just need to write the projection matrix, and then 
$$p^{V}(X) = [V(V^tV)^{-1}V^t]X$$
Back to your example, you have the subspace generated by a single vector $v\in \mathbb{R}^2$, its projection matrix will be
$$p^{v} = v (v^{t}v)^{-1} v = \begin{bmatrix}2 \\4 \end{bmatrix}([2,4],\begin{bmatrix}2 \\4 \end{bmatrix})^{-1}[2,4]=\frac{1}{5}\begin{bmatrix}1 & 2 \\2&4 \end{bmatrix}$$
Finally 
$$p^{v}(X) = \frac{1}{5}\begin{bmatrix}1 & 2 \\2&4 \end{bmatrix}\begin{bmatrix}X_1 \\X_2 \end{bmatrix}$$
It's a linear transformation of $X$, so you can easily calculate the expectation and variance.
A: See https://en.wikipedia.org/wiki/Multivariate_normal_distribution where it is stated that a multi-variate distribution is multi-variate normal if and only if every linear combination of the variables is normally distributed. If I understand correctly, your "projection" defines a linear combination that you are interested in of the variables, so that is indeed normal. Let me know if you meant something else by "projection" though.
