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.

source: https://arxiv.org/pdf/1503.02406.pdf

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.

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  • $\begingroup$ I assume you mean p(y|x) in terms of ... $\endgroup$
    – gunes
    Apr 2, 2019 at 14:57
  • 1
    $\begingroup$ I agree, it's a bit awkward to use n as proportionality constant. $\endgroup$
    – Tom
    Apr 2, 2019 at 15:21
  • $\begingroup$ 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. $\endgroup$ Apr 3, 2019 at 8:31

1 Answer 1


$y'$ 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 $h_j\propto p(x_j)$.


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