- Score (statistics)
In

statistics , the**score**or**score function**is thepartial derivative , with respect to some parameter $heta$, of thelogarithm (commonly thenatural logarithm ) of thelikelihood function .If the observation is $X$ and its likelihood is $L(\; heta;X)$, then the score $V$ can be found through thechain rule ::$V=frac\{partial\}\{partial\; heta\}\; log\; L(\; heta;X)=frac\{1\}\{L(\; heta;X)\}\; frac\{partial\; L(\; heta;X)\}\{partial\; heta\}.$

Note that $V$ is a function of $heta$ and the observation $X$, so that, in general, it is not a

statistic . Note also that $V$ indicates the "sensitivity" of $L(\; heta;X)$ (its variation normalized by its value).**Mean**The

expected value of $V$ with respect to the observation $x$, given $heta$, written $mathbb\{E\}(V|\; heta)$, is zero.To see this, rewrite the definition of expectation, using the fact that the probability mass function is just $L(\; heta;\; x)$, which is conventionally denoted by $f(x;\; heta)$ (in which the dependence on $x$ is more explicit). The correspondingcumulative distribution function is denoted as $F(x;\; heta)$. With this change of notation and writing $f\text{'}\_\{\; heta\}(x;\; heta)$ for the partial derivative with respect to $heta$,:$mathbb\{E\}(V|\; heta)=int\_\{\; [0,1]\; \}frac\{f\text{'}\_\{\; heta\}(x;\; heta)\}\{f(x;\; heta)\}dF(x;\; heta)=int\_X\; frac\{f\text{'}\_\{\; heta\}(x;\; heta)\}\{f(x;\; heta)\}\; f(x;\; heta)dx=int\_X\; frac\{partial\; f(x;\; heta)\}\{partial\; heta\}\; ,\; dx$

where the integral runs over the whole of the probability space of "X" and a prime denotes partial differentiation with respect to $heta$. If certain differentiability conditions are met, the integral may be rewritten as

:$frac\{partial\}\{partial\; heta\}\; int\_X\; f(x;\; heta)\; ,\; dx=frac\{partial\}\{partial\; heta\}1\; =\; 0.$

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**$heta$, then the mean value of the scores would tend to zero as the number of repeat samples approached infinity.**Variance**The variance of the score is known as the

Fisher information and is written $mathcal\{I\}(\; heta)$. Because the expectation of the score is zero, this may be written as:$mathcal\{I\}(\; heta)=mathbb\{E\}left\{left.\; left\; [\; frac\{partial\}\{partial\; heta\}\; log\; L(\; heta;X)\; ight]\; ^2\; ight|\; heta\; ight\}.$

Note that the Fisher information, as defined above, is not a function of any particular observation, as the random variable $X$ has been averaged out.This concept of information is useful when comparing two methods of observation of some

random process .**Example**Consider a

Bernoulli process , with "A" successes and "B" failures; the probability of success is θ.Then the likelihood "L" is

:$L(\; heta;A,B)=frac\{(A+B)!\}\{A!B!\}\; heta^A(1-\; heta)^B,$

so the score "V" is given by taking the partial derivative of the log likelihood function as follows:

:$V=frac\{partial\}\{partial\; heta\}logleft\; [L(\; heta;A,B)\; ight]\; =frac\{1\}\{L\}frac\{partial\; L\}\{partial\; heta\}.$

This is a standard calculus problem: "A" and "B" are treated as constants. Then

:$V=frac\{A\}\{\; heta\}-frac\{B\}\{1-\; heta\}.$

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

:$E(V)=\; frac\{n\; heta\}\{\; heta\}\; -\; frac\{n(1-\; heta)\}\{1-\; heta\}=\; n\; -\; n\; =\; 0.$

We can also check the variance of $V$. 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

:$operatorname\{var\}(V)=operatorname\{var\}left(frac\{A\}\{\; heta\}-frac\{n-A\}\{1-\; heta\}\; ight)\; =operatorname\{var\}left(Aleft(frac\{1\}\{\; heta\}+frac\{1\}\{1-\; heta\}\; ight)\; ight)=left(frac\{1\}\{\; heta\}+frac\{1\}\{1-\; heta\}\; ight)^2operatorname\{var\}(A)=frac\{n\}\{\; heta(1-\; heta)\}.$

**ee also***

Fisher information

*Information theory

*Score test

*Support curve **References***cite book

last = Schervish

first = Mark J.

title = Theory of Statistics

publisher =Springer

date =1995

location =New York

pages = Section 2.3.1

isbn = 0387945466

*Wikimedia Foundation.
2010.*