Logistic regression — the sigmoid reflection identity

Background

The sigmoid is point-symmetric about $(0,\,0.5)$. Reflecting the score flips the probability to its complement:

$$\sigma(-z) = 1 - \sigma(z), \qquad \sigma(0) = 0.5$$

This is exactly why $\mathbb{P}(y=0\mid\mathbf{x}) = 1 - \sigma(z) = \sigma(-z)$: negating the logit gives you the other class's probability. The curve's centre at $z=0$ is the 0.5 decision threshold.

Problem statement

Implement sigmoid_reflection(z) that returns $\sigma(-z)$ for each value in z.

Input

• z — array-like of real-valued scores (logits).

Output

Returns an np.ndarray the same length as z, where element $i$ is $\sigma(-z_i) = \dfrac{1}{1 + e^{z_i}}$.

Examples

Example 1

Input:  z = [0, 2, -3]
Output: [0.5        0.11920292 0.95257413]

Explanation: $\sigma(-0)=0.5$; $\sigma(-2)=1-\sigma(2)\approx0.1192$; $\sigma(3)=1-\sigma(-3)\approx0.9526$.

Example 2

Input:  z = [0]
Output: [0.5]

Explanation: the symmetry centre — $\sigma(0)=0.5$ and $\sigma(-0)=0.5$ agree.

Constraints

• Return $\sigma(-z)$, which must equal $1 - \sigma(z)$ elementwise.
• Vectorise over z; no Python loop.

Notes

• Reading $\sigma(-z)$ as $\mathbb{P}(y=0)$ is the bridge from last lesson's complement rule to this lesson's sigmoid — one identity ties them together.