# Why does the parameter estimator not depend on the parameter?

In the proof for the Cramer Rao Inequality, my book writes:

$$E[\hat{\theta}] = \int{\hat{\theta}(\textbf{x})L(\theta;\textbf{x})d\textbf{x}=\theta}$$

Then differentiating both sides of this equation with respect to $$\theta$$, and interchanging the order of integration and differentiation, gives

$$\int\hat{\theta}\frac{\delta L}{\delta \theta}d \textbf{x}=1$$

This is because $$\hat{\theta}$$ does not depend on $$\theta$$.

But why does $$\hat{\theta}$$ not depend on $$\theta$$? Because i think $$\hat{\theta}$$ depends on the random sample which depends on $$\theta$$

The estimator is a function of the random variables $$x_i$$, which means the estimator is itself a random variable. In a frequentist framework the parameter, however, is not a random variable; it is just a fixed but unknown constant.

It is in this sense that $$\hat{\theta}$$ is not a function of $$\theta$$ (even though you are right that each $$x_i$$ “depends on” $$\theta$$ in the sense that the $$x_i$$ are drawn from a distribution with parameter $$\theta$$).

For the sake of concreteness, say $$X$$ has a normal distribution with parameters $$\mu$$ and $$\sigma$$. Say we draw an iid sample of size $$n$$ to estimate $$\mu$$. Then our estimator, the sample mean $$\bar{x}=1/n\Sigma x_i$$, is a function of the $$x_i$$, which are each identically distributed $$N(\mu, \sigma^2)$$ distributions, but it is not a function of $$\mu$$ or $$\sigma$$.

This can all be made absolutely precise in the formalism of measure theory, where the estimator is a measurable function, but the parameters describe the underlying probability measure itself.

It's because $$\partial _x$$ denotes "explicit" differentiation. For example, Say you have a function $$f=n^2$$. Then $$\partial _n f = 2n$$, even if $$n=z^2$$. In that case, $$\partial _z f$$ would still yield $$0$$. $$D_x$$ and $$\partial_x$$ do not mean the same thing.

One thing to notice. Although the derivative on the left hand side might be a total derivative ($$D_x$$, in this case), by application of the Leibniz Integral Rule, the differentiation becomes partial inside the integral sign.

• This is not really the reason. It is true the $x$ depends on the parameters, but you are conflating a random variable with a probability measure. – symplectomorphic Feb 25 at 7:49

The estimator $$\hat\theta$$ is first of all a statistic, which by definition, is a function of the sample observations $$X_1,X_2,\ldots,X_n$$ independent of the parameter of interest ($$\theta$$ in this case). How can an estimator of an unknown parameter be dependent on the very parameter you are estimating?