逻辑损失梯度永不为零是否无法取得最小值？逻辑回归场景求证

Great question—let’s unpack this clearly for your specific logistic regression setup.

1. First: Can the gradient ever be exactly zero? #

Your intuition about individual terms in the gradient being non-zero is correct: for any single sample where $x^{(n)} \neq 0$ (which is almost always the case, since zero-vector samples carry no predictive information), the term $\frac{- y^{(n)}x{(n)} }{1 + e^{y{(n)}w^{\top} x^{(n)}}}$ will never be zero. The denominator $1 + e^{y{(n)}w^\top x^{(n)}}$ is always greater than 1, and the numerator $-y^{(n)}x{(n)}$ is non-zero.

But here’s the key: the gradient is the average of all these individual terms. Multiple non-zero terms can cancel each other out to produce a zero average. Let’s take a simple concrete example:

• Suppose we have two samples: $x^{(1)} = [1], y^{(1)} = +1$; $x^{(2)} = [1], y^{(2)} = -1$.
• The gradient becomes:
  $$
  \nabla_w J_{train}(w) = \frac{1}{2} \left( \frac{-11}{1 + e^{w1}} + \frac{-(-1)1}{1 + e^{-w1}} \right)
  $$
• Simplify the second term: $\frac{1}{1 + e^{-w}} = \frac{e^w}{1 + e^w}$. Adding the two terms together:
  $$
  \frac{1}{2} \left( \frac{-1}{1+e^w} + \frac{e^w}{1+ew} \right) = \frac{1}{2} \cdot \frac{e^w - 1}{1 + e^w}
  $$
• When $w=0$, this equals $\frac{1}{2} \cdot \frac{1-1}{1+1} = 0$. So the gradient is exactly zero here!

At this $w=0$, the loss function reaches its minimum value: $J_{train}(0) = \frac{1}{2}(\log(1+e^0) + \log(1+e^0)) = \log(2)$. If you try $w \to +\infty$ or $w \to -\infty$, the loss will blow up to infinity, confirming this is indeed the minimum.

2. When does the gradient never reach zero? #

This happens when your training data is linearly separable. Let’s say there exists some $w^$ such that for all samples, $y^{(n)}(w)^\top x^{(n)} > 0$. If you scale $w^$ by a positive constant $t$ (so $w = t w^$), then $y^{(n)}w\top x^{(n)} = t \cdot y^{(n)}(w*)^\top x^{(n)}$, which grows to $+\infty$ as $t \to +\infty$.

In this scenario:

• Each term in the gradient becomes $\frac{-y^{(n)}x{(n)}}{1 + e^{t \cdot C_n}}$, where $C_n = y^{(n)}(w*)^\top x^{(n)} > 0$. As $t \to +\infty$, $e^{t C_n} \to +\infty$, so each term approaches 0.
• The gradient approaches zero, but never actually hits it (since $e^{t C_n}$ is always finite for finite $t$).
• The loss function approaches 0 (since $\log(1 + e^{-\infty}) = 0$), but never actually reaches 0—because you can’t have an infinitely large $w$.

In this case, the loss function has an infimum (greatest lower bound) of 0, but no actual minimum value. Gradient descent will keep updating $w$ to make it larger and larger, getting closer and closer to zero loss but never reaching it.

3. Key Takeaways #

• Linear inseparable data: There exists a finite $w$ where the gradient is exactly zero, and this $w$ achieves the minimum loss.
• Linear separable data: The gradient never reaches zero (only approaches it as $w$ grows infinitely), and the loss function has no minimum—only an infimum that we can approximate arbitrarily closely with iterative methods like gradient descent.

内容的提问来源于stack exchange，提问作者Charlie Parker