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Is the likelihood a function of random variables ($X_i$) or a function of realisations of random variables ($x_i$) ?

Is there a consensus on this in the statistics community ?

Mathieu Krisztian
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1 Answers1

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A likelihood is a function of the parameters for a fixed realization of a sample. For instance, a sample $(x_1, x_2, \ldots, x_m)$ drawn from binomial distributions with common parameter $p$ but possibly different (but fixed and known) $n_i$ will have likelihood for $p$ proportional to $$\mathcal L(p \mid x_1, \ldots, x_m) \propto \prod_{i=1}^m p^{x_i} (1-p)^{n_i - x_i} = p^{\sum x_i} (1-p)^{\sum (n_i - x_i)},$$ and this is regarded as a function of the parameter $p$. That is not to say that $n_i$ and $x_i$ must take on specific values, but rather, the emphasis is on how $\mathcal L$ varies with $p$.

heropup
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  • Isn't it incompatible with the following ? (a) the Fisher information is a variance of a function of the likelihood (or an expected value of a function of the derivative of the likelihood). (b) the variance is defined for a random variable (not for a realisation of a random variable). Then, how the fact that (A) the likelihood is a function of realization of random variables* be compatible with the fact that (B) a function of the likelihood (Fisher information) is a variance, so a function of random variables ? – Mathieu Krisztian Jan 07 '22 at 22:47
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    @MathieuKrisztian As I clearly stated, the likelihood is a function of the parameter(s). The Fisher information is the variance of the score. These are not contradictory statements. The score is the gradient (derivative) of the log-likelihood with respect to the parameter(s). – heropup Jan 07 '22 at 23:08
  • Thank you for your kind help. – Mathieu Krisztian Jan 07 '22 at 23:34