# Sigmoidal neuron - linear separability

I am struggling to write a simple proof for the following statement:

The neuron's inputs are proportional to the probability of the respective feature in the input layer

In particular, I am not sure what $$y'$$ means here and how we get to an expression of $$p(\mathbb{y}|x)$$ that depends on $$p(\mathbb{x}|y)/p(\mathbb{x}|y')$$. Also not 100% sure what $$n$$ means.

EDIT 1 I found this question to be a duplicate of this older question. There, the author of the referenced paper points the reader in the right direction.

EDIT 2 In light of the answer provided, I feel that the only missing piece of the puzzle is equation (2) - below.

• I assume you mean p(y|x) in terms of ... – gunes Apr 2 '19 at 14:57
• I agree, it's a bit awkward to use n as proportionality constant. – Tom Apr 2 '19 at 15:21
• In another thread (added link to Q, see Edit 1), the author of the referred paper says that $n$ is the sample size and points to some sort of Signal to Noise ratio and the assumption that the SNR grows like $\sqrt{n}$. I still don't get where eq (2) comes from: although I get the conditional independence part, I don't understand the $n p(x_j)$ exponent. – IcannotFixThis Apr 3 '19 at 8:31

$$y'$$ is the other class, i.e. y-not. If you get rid of the log and exp in the denominator of Eq. (1), you'll have: $$p(y|x)=\frac{1}{1+\exp\left(\log\left(\frac{p(x,y')}{p(x,y)}\right)\right)}=\frac{p(x,y)}{p(x,y)+p(x,y')}$$ The denominator needs to be $$p(x)$$ due to Bayes Rule, so $$y'$$ is really the negative class. $$n$$ is just a proportionality constant, which is also related to your yellow-quoted line. i.e. they actually mean something like $$h_j\propto p(x_j)$$.