Logistic regression — probability of the true label

Background

Maximum likelihood estimation (MLE) asks: under the model, how probable is the label we actually observed? For a binary example with predicted $p = \mathbb{P}(y=1\mid\mathbf{x})$, the probability assigned to the true label is

$$\mathbb{P}(\text{observed } y) = \begin{cases} p & y = 1 \\ 1 - p & y = 0 \end{cases}$$

A good model makes this number large — close to $1$ — for every training point. That is precisely what MLE pushes for.

Problem statement

Implement label_likelihood(y, p) returning, for each example, the probability the model assigns to its observed label.

Input

• y — array-like of true labels, each 0 or 1.
• p — array-like of predicted probabilities $\mathbb{P}(y=1)$, same length, each in $[0, 1]$.

Output

Returns an np.ndarray the same length as y: p where y == 1, and 1 - p where y == 0.

Examples

Example 1

Input:  y = [1, 0, 1], p = [0.9, 0.2, 0.6]
Output: [0.9 0.8 0.6]

Explanation: for the class-1 points the likelihood is $p$ itself ($0.9$, $0.6$); for the class-0 point it is $1-0.2 = 0.8$.

Example 2 — a confident mistake scores low

Input:  y = [1], p = [0.05]
Output: [0.05]

Explanation: the truth is class 1 but the model gave it only $0.05$, so the observed label was deemed very unlikely.

Constraints

• Use np.where(y == 1, p, 1 - p) or equivalent — one vectorised expression, no loop.
• Assume p is already a valid probability; do not clip.

Notes

• This per-example number is the building block of the dataset likelihood (the product over all examples) and, after a log, of binary cross-entropy.