In Bishop's book, he talks about Bayesian Logistic regression and how to compute it. I understood approximating posterior distribution using Laplace approximation. But I don't quite understand predictive distribution part. Specifically:

The predictive distribution for class $C_1$, given a new feature vector $\phi(x)$, is obtained by marginalizing with respect to the posterior distribution $p(w|t)$... $$p(C_1|\phi,t) = \int{p(C_1|\phi,w)p(w|t)dw} \approx \int{\sigma(w^T\phi)q(w)dw} \ \ \cdots (1)$$ To evaluate the predictive distribution, we first note the function $\sigma(w^T\phi)$ depends on $w$ only through its projection onto $\phi$. Denoting $a=w^T\phi$, we have $$\sigma(w^T\phi) = \int{\delta(a-w^T\phi)\sigma(a)da}$$ where $\delta$ is the dirac delta function. From this we obtain $$\int{\sigma(w^T\phi)q(w)dw} =\int{\sigma(a)p(a)da}$$ where $$p(a) = \int{\delta(a-w^T\phi)q(w)dw}$$ ... Thus our variation approximation to the predictive distribution becomes $$p(C_1|t) = \int{\sigma(a)p(a)da} = \int{\sigma(a)N(a|\mu_a, \sigma^2_a)da}\ \ \cdots (2)$$

What exactly is the problem with equation (1) that we need to substitute it with dirac functions? At the end, he gets equation (2) but both (1) and (2) have a sigmoid function and a normal distribution. Why all the hassle to change from (1) to (2)?

I guess I don't quite understand what it means when he says

function $\sigma(w^T\phi)$ depends on $w$ only through its projection onto $\phi$.


1 Answer 1


(1) involves a multivariate Gaussian with potentially off-diagonal covariance matrix terms. You can see this from the previous section of Bishop, which gives the form of $q(w)$ as $$ q(w) = \mathcal{N}(w \mid w_{\text{MAP}}, S_N), $$ where $S_N$ can generally have off-diagonal terms. This is replaced by an integral over a single variable (as opposed to $D$ variables) with a univariate Gaussian. You see that even then, the integral isn't analytically tractable, which the author fixes by approximating the sigmoid function with a probit, $\Phi(\lambda a)$ and makes use of the identity, $$ \int \Phi(\lambda a) \mathcal{N}(a \mid \mu \sigma^2) da = \Phi\left(\frac{\mu}{(\lambda^{-2} + \sigma^2)^{1/2}}\right), $$ which wouldn't have been possible had he stuck with the original multivariate Gaussian with an unknown covariance. (I'm not sure if an analogous identity exists for a Gaussian with arbitrary covariance, though if you are able to reduce the expression to one that makes use of a simpler identity, then why not?)

As for the projection comment. Consider the following integral, $$ S = \int_{-\infty}^\infty \int_{-\infty}^\infty f(a x + by) g(x,y) dx dy $$ where $a$ and $b$ are constants. The integral is over the entire two-dimensional plane. Imagine, on the plane, the directional vector $(a,b)'$. You can define a line spanning the space along this direction. The inner product of any vector with $(a,b)'$ is a projection of that vector on this line. For example, suppose we have a vector $(c,d)'$ that's orthogonal to $(a,b)'$, so that $(c,d) (a,b)' = 0.$ Then $(x+c, y+d) (a,b)' = (x,y)(a,b)'.$ This means that the value of $f(ax + by)$ is determined entirely by the projection of $(x,y)'$ on the line defined by the directional vector $(a,b)'.$ Any change in a direction orthogonal to that line has no impact on the value of $f$.

We can then perceive this integral as an integral of $f$ over the line parameterized by a single variable $\lambda,$ given by a weighting function over that parameter, where the weight is given by $$p(\lambda) = \int \delta(\lambda - (a x + b y)) g(x,y) dx dy.$$ Consider what this integral is, for a given value of $\lambda.$ This is an integral of $g$ only over the line defined by $a x + b y = \lambda,$ which is perpendicular to the vector $(a,b)'.$ If $g$ is a two-dimensional probability distribution then you can think of this as marginalizing the distribution $g(x,y)$ over the direction orthogonal to $(a,b)'.$ If $g$ is a Gaussian, this marginalization is easy to do, and is done by Bishop in the passage you skipped over.

$S$ is an integral over $x$ and $y$, though you can re-imagine it as an integral over the line given by the direction $(a,b)'$, where at each incremental step, another integral is taken over the line orthogonal to $(a,b)'$ at that given point (so a rotation in the directions in which we integrate). This can be generalized in $D$ dimensions to an integral over a one-dimensional line, where at each increment a $D-1$ dimensional integral is taking place. Any function that depends only on a point's projection on the the line is a constant of the $D-1$ dimensional integral.

  • $\begingroup$ What do you mean by off-diagonal? It's kind of odd that Bishop never mentioned that $S_n$ can be off-diagonal(or it's likely I missed it). After some thinking, I thought the reason behind using Direc delta was that $\sigma(w^T\phi)$ was in fact scalar (as opposed to a sigmoid function in equation (2)) and needed to be converted into sigmoid function in order to do write it as a convolution of sigmoid and gaussian. $\endgroup$
    – MoneyBall
    Commented Apr 20, 2017 at 17:09
  • $\begingroup$ Just look at the form of $S_N$ in 4.143. In general it can have off-diagonal terms. The only way it's completely diagonal is if the transformed feature space is completely uncorrelated (which could be the case if, say, it was the result of a PCA transformation). Also I don't understand what you mean. $\sigma(x)$ is a scalar function no matter what its argument is. $\endgroup$ Commented Apr 20, 2017 at 17:19

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