- Score (statistics)
statistics, the score or score function is the partial derivative, with respect to some parameter , of the logarithm(commonly the natural logarithm) of the likelihood function.If the observation is and its likelihood is , then the score can be found through the chain rule:
Note that is a function of and the observation , so that, in general, it is not a
statistic. Note also that indicates the "sensitivity" of (its variation normalized by its value).
expected valueof with respect to the observation , given , written , is zero.To see this, rewrite the definition of expectation, using the fact that the probability mass function is just , which is conventionally denoted by (in which the dependence on is more explicit). The corresponding cumulative distribution functionis denoted as . With this change of notation and writing for the partial derivative with respect to ,
where the integral runs over the whole of the probability space of "X" and a prime denotes partial differentiation with respect to . If certain differentiability conditions are met, the integral may be rewritten as
It is worth restating the above result in words: the expected value of the score is zero.Thus, if one were to repeatedly sample from some distribution, and repeatedly calculate the score with the true , then the mean value of the scores would tend to zero as the number of repeat samples approached infinity.
The variance of the score is known as the
Fisher informationand is written . Because the expectation of the score is zero, this may be written as
Note that the Fisher information, as defined above, is not a function of any particular observation, as the random variable has been averaged out.This concept of information is useful when comparing two methods of observation of some
Bernoulli process, with "A" successes and "B" failures; the probability of success is θ.
Then the likelihood "L" is
so the score "V" is given by taking the partial derivative of the log likelihood function as follows:
This is a standard calculus problem: "A" and "B" are treated as constants. Then
So if the score is zero, θ = "A"/("A" + "B"). We can now verify that the expectation of the score is zero. Noting that the expectation of "A" is "n"θ and the expectation of "B" is "n"(1 − θ), we can see that the expectation of "V" is
We can also check the variance of . We know that "A" + "B" = "n" (so "B" = "n" - "A") and the variance of "A" is "n"θ(1 − θ) so the variance of "V" is
last = Schervish
first = Mark J.
title = Theory of Statistics
location =New York
pages = Section 2.3.1
isbn = 0387945466
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