# Regularity of Cramer-Rao inequality for unbiased estimators

As I understand it, the use of Cramer-Rao inequality requires a few regularity conditions, including the parametric family being dominated and swapping of differentiation and integration in computation of expectation of the estimator. How do I prove the latter condition for unbiased estimators?

Mathematically, suppose $\hat{\theta}\left(X\right)$ is an unbiased estimator of $\theta$. I want the proof of the following statement $$\frac{\partial}{\partial\theta}\int \hat{\theta}\left(X\right)p(x,\theta)dx = \int \frac{\partial}{\partial\theta} \left(\hat{\theta}\left(X\right)p(x,\theta)\right)dx$$

My progress : Obviously, $\hat{\theta}\left(X\right)$ being unbiased for $\theta$, the left hand side is $1$. I'm having trouble showing that the right hand side is also equal to $1$. I have tried to right $p(x,\theta)$ in terms of the log-likelihood $L(x,\theta)$ as $e^{L(x,\theta)}$ without success. My calculations reduce to showing that $E_{\theta}\left(\hat{\theta}\left(X\right)\frac{\partial}{\partial\theta}L(X,\theta)\right)=1$.

Any help would be much appreciated. Thank you.

$$\frac{\partial}{\partial\theta}\int \hat{\theta}\left(X\right)p(x,\theta)dx = \int \frac{\partial}{\partial\theta} \left(\hat{\theta}\left(X\right)p(x,\theta)\right)dx$$
Is true in general. For example, letting $P[x|\theta] \sim \textrm{Exp}(\lambda)$, then $\hat{\lambda}(X) = \lambda$ is an unbiased oracle estimator.
$$\begin{split} & \frac{\partial}{\partial\lambda}\int \hat{\lambda}\left(X\right)p(x,\lambda)dx = \int \frac{\partial}{\partial\lambda}\left(\hat{\lambda}\left(X\right)p(x,\lambda)\right)dx\\ & \Leftrightarrow 1 = \int\frac{d}{d\lambda}\lambda^2\textrm{Exp}(\lambda x) = \int 2\lambda\textrm{Exp}(\lambda x) + \lambda^2x\textrm{Exp}(\lambda x) dx) = 2 + \lambda^2 \end{split}$$