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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

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.

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 $vin mathbb{R}^2$, 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.

Let $mathbf{x}simmathcal{N}(mu_x, Sigma_x)$ be an $n$-dimensional Gaussian distribution. Then, if $y=mathbf{v}^topmathbf{x}$, where $mathbf{v}inBbb{R}^n$, it holds that
$$
ysimmathcal{N}(mu_y, sigma_y^2),
$$
where $mu_y=mathbf{v}^topmu_x$ and $sigma_y^2=mathbf{v}^topSigma_xmathbf{v}$, since
$$
mu_y=Bbb{E}[y]=Bbb{E}[mathbf{v}^topmathbf{x}]=mathbf{v}^topBbb{E}[mathbf{x}]=mathbf{v}^topmu_x
$$
and
$$
sigma_y^2 = Bbb{E}[(y-mu_y)^2]
=
Bbb{E}[(mathbf{v}^topmathbf{x}-mathbf{v}^topmu_x)^2]
=
Bbb{E}[(mathbf{v}^top(mathbf{x}-mu_x))^2]
=
Bbb{E}[mathbf{v}^top(mathbf{x}-mu_x)mathbf{v}^top(mathbf{x}-mu_x)]
=
Bbb{E}[mathbf{v}^top(mathbf{x}-mu_x)(mathbf{x}-mu_x)^topmathbf{v}]
=
mathbf{v}^topBbb{E}[(mathbf{x}-mu_x)(mathbf{x}-mu_x)^top]mathbf{v}
=
mathbf{v}^topSigma_xmathbf{v}.
$$