11
$\begingroup$

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

$\endgroup$

3 Answers 3

7
$\begingroup$

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 $$ \sigma_y^2 = \Bbb{E}[(y-\mu_y)^2] = \Bbb{E}[(\mathbf{v}^\top\mathbf{x}-\mathbf{v}^\top\mu_x)^2] = \Bbb{E}[(\mathbf{v}^\top(\mathbf{x}-\mu_x))^2] = \Bbb{E}[\mathbf{v}^\top(\mathbf{x}-\mu_x)\mathbf{v}^\top(\mathbf{x}-\mu_x)] = \Bbb{E}[\mathbf{v}^\top(\mathbf{x}-\mu_x)(\mathbf{x}-\mu_x)^\top\mathbf{v}] = \mathbf{v}^\top\Bbb{E}[(\mathbf{x}-\mu_x)(\mathbf{x}-\mu_x)^\top]\mathbf{v} = \mathbf{v}^\top\Sigma_x\mathbf{v}. $$

$\endgroup$
2
  • $\begingroup$ I think you are calculating the inner product but not the projection. The projection is $v^T x / \| x \|$ $\endgroup$ Sep 26, 2018 at 3:37
  • 2
    $\begingroup$ @nullgeppetto probably assumes that $\mathbf{v}$ is a unit vector $\endgroup$
    – fuji
    Nov 6, 2018 at 12:13
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks for the reply. Yes, I understand its a linear combination of the variables, but I do not quite understand how to get to the equivalent 1D gaussian for the example case I showed. What I am trying to understand is what that linear combination is for the example I just stated. Could you help me perhaps with the steps that I need to follow to get to the linear combination for the given direction vector that I gave. I am fairly new to statistics, so I would appreciate a little pointers as to how to proceed with the actual analysis. $\endgroup$
    – rrr
    Aug 5, 2015 at 19:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .