I read the likelihood is defined in logistic regression as the probability $$ L(w) = P(y|x, w) = \prod P(y^i| x^i,w) = \prod (\sigma(z^i))^{y^i}(1-\sigma(z^i))^{(1-y^i)} $$ and the log of the last equation is:

$$ log(L(w)) = \sum y^i log((\sigma(z^i)) + (1-y^i)log(1-\sigma(z^i)) $$

I understand (independent probabilities) $$ L(w) = P(y|x, w) = \prod P(y^i| x^i,w) $$ and I understand that from $$ L(w) = \prod (\sigma(z^i))^{y^i}(1-\sigma(z^i))^{(1-y^i)} $$ the log is (basic log properties) $$ log(L(w)) = \sum y^i log((\sigma(z^i)) + (1-y^i)log(1-\sigma(z^i)) $$ However, how do I get $$ \prod P(y^i| x^i,w) = \prod (\sigma(z^i))^{y^i}(1-\sigma(z^i))^{(1-y^i)} $$ This basically means that $$ P(y| x,w) = (\sigma(z))^{y}(1-\sigma(z))^{(1-y)} $$ and I just don't see that.


1 Answer 1


By the definition of the logistic regression model $\mathrm P(y = 1 | x,w) = \sigma\left(\left<w, x\right>\right)$ thus $\mathrm P(y = 0 | x,w) = 1- \sigma\left(\left<w, x\right>\right)$

You should be able to verify (by setting y=0/y=1) that this is equivalent to

$\mathrm P(y | x,w) = \sigma\left(\left<w, x\right>\right)^y \left[ 1- \sigma \left( \left<w, x \right> \right) \right]^{1-y}$

I think you understand the rest?

  • $\begingroup$ Yes, I see, it is just a way to compact everything into a single expression. $\endgroup$
    – user
    May 3, 2016 at 19:37

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