P(y) = exp(y*z)/SUM[y'=0..1](y'*z) = exp(y*z)/(exp(0*z)+exp(1*z)) = exp(y*z)/(1+exp(z)) 6.22

P(y) = Sigmoid((2*y-1)*z) 6.23

where

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PB

Pius Braun30w

P(y) = exp(y*z)/SUM[y'=0..1](y'*z) = exp(y*z)/(exp(0*z)+exp(1*z)) = exp(y*z)/(1+exp(z)) 6.22

P(y) = Sigmoid((2*y-1)*z) 6.23

where

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Jayam Thaker11w

Ya sure. In probablity theory, the sum of all possible outcomes sum over to 1. For example a coin(fair or not) can only land on two sides, heads and tails. So in neural networ if there are only two possible states 0 and 1 the probablity of them occuring add up to 1.

So if prob(1) is say X, the prob(0) would be 1-x.

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Hello EveryOne !!!

I am Abdulrehman working as Graduate Researcher in a University of Engineering and Technology, Lahore Pakistan. I have bee...

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