I am reading Kevin Murphy's Machine Learning: A Probabilistic Perspective and I am having difficulties understanding Equation 1.5.

Murphy introduces the linear function $$ y(\mathbf{x}) = \mathbf{w}^T\mathbf{x} + \epsilon = \sum_{j=1}^D w_j x_j + \epsilon, $$ where $\mathbf{w}^T\mathbf{x}$ represents the inner product between the input vector $\mathbf{x}$ and the model's weight vector $\mathbf{w}$, and $\epsilon$ is the residual error. Then Murphy mentions that we often assume that $\epsilon$ has a Gaussian distribution with mean $\mu$ and variance $\sigma^2$.

Murphy makes the connection between linear regression and Gaussians with Equation 1.5: $$ p(y|\mathbf{x, \theta}) = \mathcal{N}(y|\mu(\mathbf{x}), \sigma^2(\mathbf{x})). $$

I read this as the probability of getting output $y$, given a data point $\mathbf{x}$ and some model parameters $\mathbf{w}$ and $\sigma^2$, is normally distributed with mean $y$ given the mean of $\mathbf{x}$ (I don't understand this) and variance of $\mathbf{x}$.

To me, it seems that $y$ should be normally distributed with mean $\mathbf{w}^T\mathbf{x}$ and variance $\sigma^2$ only if $\epsilon$ is normally distributed with mean $0$ variance $\sigma^2$, i.e., if $$ \epsilon \sim \mathcal{N}(0, \sigma^2), $$ then $$ p(y|\mathbf{x, \theta}) = \mathcal{N}(\mathbf{w}^T\mathbf{x}, \sigma^2). $$

  • $\begingroup$ I haven't read the book, but I think you are both right, the mu() bit is a bit confusing, it probably means the expectation of y given the modeled x (less the error) $\endgroup$
    – Dirk N
    May 12, 2021 at 10:11
  • 1
    $\begingroup$ $y | \mu(x)$ is the more general idea, as the mean function $\mu(x)$ can be pretty complicated. But in this context the mean is relatively simple, a linear function of the inputs. So, without reading the book I'd guess the author is just going back and forth a little between general ideas and particular examples. $\endgroup$
    – Nick Cox
    May 12, 2021 at 10:18
  • $\begingroup$ Read $\mu(x)$ like $f(x)$ not like $\bar{x}$ because $\mu$ names whatever function provides the conditional expectation of y as a function of x. $\endgroup$ May 12, 2021 at 10:49
  • $\begingroup$ There are cases where you do not have constant variance $\sigma^{2}$ and you variance is directly affected by your observed $x_{i}$. An example of such case can be found here online.stat.psu.edu/stat462/node/186. So, in order to write the regression in the most general expression it denotes the variance as $\sigma^{2}(x)$. $\endgroup$
    – Fiodor1234
    May 12, 2021 at 11:06
  • $\begingroup$ Looks like a mistake (typo) in the book to me. Conditioning on a constant (like $\mu(x)$) has no effect. The author just put the $\mu$ in the wrong place. The mean of the distribution is really $\mu(y|x)$. $\endgroup$ May 25, 2021 at 23:06

2 Answers 2


As commented by Nick Cox, I'd suspect that you understand the special case of standard linear regression properly, but the equation 1.5 involving $\mu(x)$ and $\sigma^2(x)$ regards a more general case, of which the standard linear model is a special case with $\mu(x)=w^Tx$ and $ \sigma^2(x)=\sigma^2$ constant.


This is because linear regression can be seen in two different ways, an Ordinary Least Squares approach or a Gaussian distribution approach. The second one is related to Generalized Linear Models and you can find a good explanation in the next video.


(had the same question a while ago and this video helped me)

BTW, Kevin has a new book called Probabilistic Machine Learning and you can download it from his github page: https://probml.github.io/pml-book/book1.html


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