2
$\begingroup$

I'm starting to work through Bishop's "Pattern Recognition and Machine Learning" book, and have run into unfamiliar notation.

Eq. 160 is prefaced by saying "For this purpose, we shall assume that, given the value of $x$, the corresponding value of $t$ has a Gaussian distribution with a mean equal to the value [of the polynomial curve $y(x, \mathbf{w}) = \sum_{j=0}^M w_j x^j$]. Thus we have"

$$p(t|x, \mathbf{w}, \beta) = {\cal N}(t|y(x, \mathbf{w}), \beta^{-1}) \tag{1.60}$$

I understand the LHS to read "the probability density as a function of $t$, given $x$, $\mathbf{w}$ and $\beta$". The RHS is indicating a normal distribution with variance $\beta^{-1}$. The bit I don't understand is what is going on with the mean of this distribution - how should I interpret $\mu = t|y(x, \mathbf{w})$? Why isn't this just written as ${\cal N}(y(x, \mathbf{w}), \beta^{-1})$? What is the bar signifying? I only know its meaning within a probability. (This notation continues to be used, e.g. in Eqs. 1.61, 1.64, 1.69, so I'd like to make sure I understand it!)

(I don't know if there are multiple editions of this book; the section I am reading here is 1.2.5 "Curve fitting re-visited", and Eq. 1.60 is the first equation in this section.)

$\endgroup$

1 Answer 1

1
$\begingroup$

I expect this notation is just meant to emphasize which variables are random and which are fixed.

Earlier in the book Bishop uses the notation $N(x|\mu, \sigma^2)$ for a generic Gaussian distribution (1.46), where $x$ is the random variable, and $\mu$ and $\sigma^2$ are the fixed parameters of the distribution. So it's not that the mean is being written as $t|y(x,w)$, but rather that "$t|$" gives a name to the random variable.

(I'm looking at the 2006 edition)

$\endgroup$

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.