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Drill 7 Sigmoid with binary cross entropy

MT2 numerical practice from direct to mixed

Question

Question. A binary classifier has logit z = 1, target 1. Find probability, BCE and the gradient with respect to z.

Answer — step by step

Given

Logit z = 1; binary target y = 1.

Output: sigmoid; loss: binary cross-entropy.

To find

Find probability, BCE and the loss derivative with respect to z.

Formula — which one and why

Sigmoid and its slope. Use sigmoid when the neuron is specified to squash its preactivation into (0,1). Once its output is known, its derivative needs only output × (1−output).

z = preactivation; p = sigmoid output; e is the exponential base, not a prediction error in this formula.

$p=\frac{1}{1+e^{-z}}$ LaTeX: p=\frac{1}{1+e^{-z}}

$\frac{dp}{dz}=p(1-p)$ LaTeX: \frac{dp}{dz}=p(1-p)

Yaad rakhne ka cue: Sigmoid output p; uska slope p × (1−p).

Binary cross-entropy. For a binary target, choose the probability of the correct outcome and take its negative natural log. Sigmoid plus BCE has combined logit derivative p−y.

y is 0 or 1; p is the predicted probability of class 1; z is the sigmoid logit. The p−y derivative already includes the sigmoid derivative.

$L=-[y\ln p+(1-y)\ln(1-p)]$ LaTeX: L=-[y\ln p+(1-y)\ln(1-p)]

$\frac{\partial L}{\partial z}=p-y$ LaTeX: \frac{\partial L}{\partial z}=p-y

Yaad rakhne ka cue: Binary target 1 ho toh −ln(p); target 0 ho toh −ln(1−p).

Solution — step by step

Meaning and formulas

A logit is the raw score before sigmoid. Sigmoid converts it to a probability p. The target y = 1 says the positive class is correct. Binary cross-entropy measures how little probability the model assigned to the correct binary outcome.

$p=\frac1{1+e^{-z}}$ LaTeX: p=\frac1{1+e^{-z}}

$L=-\left[y\ln p+(1-y)\ln(1-p)\right]$ LaTeX: L=-\left[y\ln p+(1-y)\ln(1-p)\right]

For sigmoid combined with this loss, the gradient with respect to the logit has the simplified form:

$\frac{\partial L}{\partial z}=p-y$ LaTeX: \frac{\partial L}{\partial z}=p-y

Step 1: convert logit 1 into probability

$e^{-1}\approx0.367879441$ LaTeX: e^{-1}\approx0.367879441

$1+e^{-1}\approx1.367879441$ LaTeX: 1+e^{-1}\approx1.367879441

Step 2: substitute the target in the full loss formula

$L=-\left[1\ln(0.731058579)+(1-1)\ln(1-0.731058579)\right]$ LaTeX: L=-\left[1\ln(0.731058579)+(1-1)\ln(1-0.731058579)\right]

The second term has multiplier zero, so it contributes zero.

Step 3: calculate the combined gradient

Final answer: probability ≈ 0.731059; BCE ≈ 0.313262; logit gradient ≈ −0.268941. The simplified gradient already includes sigmoid's effect. Multiplying by sigmoid's slope again would apply it twice.