Suppose we have a density function $f \sim N(\mu, \sigma^2)$. Then, the score function is defined to be:

$$ \frac{d}{d\mu}\log f\left(x;\mu,\sigma^2\right) $$

The answer to this for the normal is:

$$ \frac{d}{d\mu}\log f\left(x;\mu,\sigma^2\right) = \frac{x-\mu}{\sigma^2} $$

which looks uncannily like the z-score. Is there a deeper relation here? Thanks!

  • $\begingroup$ Could you elaborate on what you mean by "deeper relation"? The score functions give first-order differential equations for the densities, thereby defining the family of distributions. Thus, different distributions will give rise to different score functions and vice versa. Consequently, the meaning of your question seems to rest on how we are supposed to interpret "uncannily." $\endgroup$ – whuber Mar 30 '17 at 14:16

There is a set of tests based on the score function. For a normal distribution, it gives the standard z-test. I think with parametric tests for normal distributions you always going to end up with the same test, it's fundamentally the only reasonable thing to do (without going non-parametric).

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