1
$\begingroup$

I am stuck in showing that Bayesian linear regression can be viewed as a Gaussian Process. Can anybody show me how to get from step (4) to step (5)?

$$ \tilde f_i= f(\mathbf{x}^{(i)}) := \mathbf{w}^T\mathbf{x}^{(i)}+b, \quad \mathbf{w}\sim\mathcal{N}(0,\sigma_w^2\mathbb{I}), \; b\sim\mathcal{N}(0,\sigma_b^2) $$

$\begin{align} \text{cov}(\tilde f_i,\tilde f_j) &= \mathbb{E}[\tilde f_i \tilde f_j] - \mathbb{E}[\tilde f_i]\mathbb{E}[\tilde f_j] \\ &= \mathbb{E}[\tilde f_i \tilde f_j] - 0\\ &=\mathbb{E}\left[(\mathbf{w}^T\mathbf{x}^{(i)}+b)^T(\mathbf{w}^T\mathbf{x}^{(j)}+b)\right] \tag{4}\\ &= \sigma_w^2 {\mathbf{x}^{(i)}}^T\mathbf{x}^{(j)} + \sigma_b^2 \tag{5} \\ &= k(\mathbf{x}^{(i)},\mathbf{x}^{(j)}). \end{align}$

These calculations come from here : https://www.inf.ed.ac.uk/teaching/courses/mlpr/2019/notes/w5b_gaussian_process_kernels.pdf

$\endgroup$

1 Answer 1

4
$\begingroup$

Well, to get from (4) to (5), it's essentially algebra :

$\begin{align} \text{cov}(\tilde f_i,\tilde f_j) &=\mathbb{E}\left[(\mathbf{w}^T\mathbf{x}^{(i)}+b)^T(\mathbf{w}^T\mathbf{x}^{(j)}+b)\right] \\ &= \mathbb{E}\left[({\mathbf{x}^{(i)}}^T\mathbf{w}+b^T)(\mathbf{w}^T\mathbf{x}^{(j)}+b)\right] \\ &= \mathbb{E}\left[{\mathbf{x}^{(i)}}^T\mathbf{w}\mathbf{w}^T\mathbf{x}^{(j)}+b^Tb + {\mathbf{x}^{(i)}}^T\mathbf{w}b + b^T\mathbf{w}^T\mathbf{x}^{(j)}\right] \\ &= \mathbb{E}\left[{\mathbf{x}^{(i)}}^T\mathbf{w}\mathbf{w}^T\mathbf{x}^{(j)}+b^Tb\right] + \mathbb{E}\left[{\mathbf{x}^{(i)}}^T\mathbf{w}b\right] + \mathbb{E}\left[b^T\mathbf{w}^T\mathbf{x}^{(j)}\right] \\ &= \mathbb{E}\left[{\mathbf{x}^{(i)}}^T\mathbf{w}\mathbf{w}^T\mathbf{x}^{(j)}+b^2\right] + {\mathbf{x}^{(i)}}^T\mathbb{E}\left[\mathbf{w}b\right] + \mathbb{E}\left[b^T\mathbf{w}^T\right]\mathbf{x}^{(j)} \tag{1'} \\ &= \mathbb{E}\left[{\mathbf{x}^{(i)}}^T\mathbf{w}\mathbf{w}^T\mathbf{x}^{(j)}\right] + \mathbb{E}\left[b^2\right] + 0 + 0 \tag{2'}\\ &= {\mathbf{x}^{(i)}}^T\mathbb{E}\left[\mathbf{w}\mathbf{w}^T\right]\mathbf{x}^{(j)} + \sigma_b^2 \tag{3'}\\ &= {\mathbf{x}^{(i)}}^T \sigma_w^2 \, \mathbb{I}\mathbf{x}^{(j)} + \sigma_b^2 = \sigma_w^2 {\mathbf{x}^{(i)}}^T\mathbf{x}^{(j)} + \sigma_b^2 \end{align} $

Where we used the independence of the predictor from the weights and bias in (1') and (3'), and the assumption that the weights and bias are mutually independent centered normal random variables in (2').

$\endgroup$
2
  • 1
    $\begingroup$ Thank you very much! This is also what I came up with but why is $\mathbb{E}[b^2] = \sigma_b^2$ and $\mathbb{E}[w w^T ] = \sigma_w^2$? Given that $\sigma_w^2$ is the variance of the weights and $\sigma_b^2$ of the bias and $\mathbb{V}[X] = \mathbb{V}[X] - \mathbb{V}[X^2] - E[X]^2$ Edit: Nevermind, I just realized that $\mathbb{E}[X]^2$ has to be 0 in both cases. Thank you! $\endgroup$
    – samu
    Feb 1, 2021 at 17:05
  • 1
    $\begingroup$ It is because we made the assumption that $\mathbf{w}$ is normally distributed with zero mean and covariance matrix $\sigma_w^2\mathbb{I}$, and similarly $b$ is gaussian centered with $\sigma_b^2$ variance. $\endgroup$ Feb 1, 2021 at 17:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.