Why does the score test work for values longer in the tail that have a small log-likelihood derivative?

Question

The score test says that we take the derivative of the log-likelihood at $H_0$ and divide it by the fisher information at $H_0$.

$U(\theta )={\frac {\partial \log L(\theta \mid x)}{\partial \theta }}.$

$I(\theta )=-\operatorname {E} \left[\left.{\frac {\partial ^{2}}{\partial \theta ^{2}}}\log f(X;\theta )\,\right|\,\theta \right]\,$

The intuition here mostly makes sense. If $H_0$ is near the most likely point in the data, then the derivative of the log-likelihood will be 0, at the peak of the pdf. If $H_0$ is further away from the peak -- say along the sharp upwards curve of the pdf, as it approaches its maximum -- then the derivative will be large. Therefore, a larger derivative indicates the value is farther from the maximum, and the probability of rejecting the null increases.

But what about farther away from the peak, still.

If we are out three or four sigma from the peak of the curve, then our derivative of the log-likelihood under the null will be very small. However, this contradicts the logic of the prior intuition.

So how does the score test actually work in practice? I'm not clear on how the Fisher information could be counteracting this effect.

Edit:

x <- seq(from=-100, to=100, by=0.10)
y <- sapply(x, function (z) log(dnorm(z, 0, 1)))
plot(x, y)

We can see by simulation that with the $L(x)$, that the general principle holds up.

Answer 1

Remember that we're differentiating the log-likelihood. If $L(\theta;x)$ goes to 0 as $\theta$ moves away from the peak, then $\log L(\theta;x)$ will go to $-\infty$.

Try sketching the log-likelihood and its derivative for the simple case where we draw a single observation $X$ from a $N(\theta,1)$ distribution.