Why Euler's constant in sigmoid function?

Course Q&A Machine Learning Specialization Supervised ML: Regression and Classification
1

Hello all,

In logistic regression, we use sigmoid activation function to make the predicted output fall between zero and one, because in logistic regression, the output should be the probability. The sigmoid function will return 0.5 if the input (z) to the sigmoid function is zero and the result of the sigmoid function will come close to 0, if input (z) approaches negative infinity, and will come close to 1, if input (z) approaches positive infinity. But why do we have to use the Euler’s constant (~2.718) particularly in this sigmoid function? In my opinion, any positive constant other than the Euler’s constant, for example even pi (~3.14) will also work the same way, yet we still go with the Euler’s constant. Why is it so? I know that for the curve (e to the power x), the slope (the rate of rise over run) will always be the same at any point and is it because of this property, the Euler’s constant is chosen for the sigmoid function or is there any property which is so relatable for it to be chosen?

2

Euler and the natural logarithm are tied at the hip. Since we use the log in the cost, and the exponential in the prediction, these go hand-in-hand when we want to compute the gradients. Euler shows up in may unexpected places, it’s rather like \pi in that way.

And all log functions can be related by a multiplier.