# Is the score-function (gradient of likelihood function) a random-variable?

Wikipedia:The score is the gradient (the vector of partial derivatives) of $$\log \mathcal{L}(\theta)$$, the natural logarithm of the likelihood function, with respect to an $$m$$-dimensional parameter vector $$\theta$$.

The score has a mean (zero) and variance (fisher info).

So, is the score a random variable?

• It can be considered a random variable if viewed as a function of the random variable "X", which is for example the case when we're talking about population Fisher information. Other times, I've head people refer to the score function after having observed data, and conditional on it, where it is not a random variable. Jul 31 at 18:46
• Does it make sense to derive the moment generating function,pdf,pmf of score-function? A sum of exponentials is erlang. Does the score of a sum of exponentials converge in distribution to a score of the corresponding erlang? Jul 31 at 18:55
• It certainly has a distribution, and hence may have an MFG or density. If by "make sense" you mean "have a practical application" I haven't personally run into it: I have used score only to derive fisher info or treated is with fixed data. Aug 1 at 14:04
• I cannot figure out what probability density of mass function it has. Do you know? Aug 1 at 20:06

Both the log-likelihood function and the score function can be considered as functions of the parameter vector, based on a fixed observed data vector. However, we can broaden our interpretation of these functions to consider them to also be functions of the data vector (seeis th related answer). Taking this broader view of these functions means that the score function can be considered as a random variable by substituting the random data vector into the data $$\mathbf{x}$$ for the function. To do this, we take the log-likelihood to be a function of both the data and the parameter:
$$\ell_\mathbf{x}(\theta) = g(\mathbf{x},\theta).$$
$$s_\mathbf{x}(\theta) = h(\mathbf{x},\theta) = \nabla_\theta g(\mathbf{x},\theta).$$
Using this mapping, we then obtain the random version of the score function by taking $$\mathbf{X}$$ to be a random vector and using:
$$s_\mathbf{X}(\theta) = h(\mathbf{X},\theta).$$