In Rotnizky et al (2000) I read the following
Let $Y$ be a random variable with density $f(Y,\theta)$, write $\ell(\theta) = \sum_i \ell(Y_i,\theta) = \sum_i \log f(Y_i,\theta)$ as the log-likelihood function.
Now define $s$ such that the first $s-1$ derivatives of $\ell$ are zero at $\theta = \theta^*$. In other words: $$\ell^{(1)}(\theta^*) = \ldots = \ell^{(s-1)}(\theta^*)= 0 \qquad \text{and} \qquad \ell^{(s)}(\theta^*) \not = 0$$
Then on page 249 she writes:
For $s\leqslant j \leqslant 2s-1$ then $\ell^{(j)}(Y,\theta^*) = \dfrac{f^{(j)}(Y,\theta^*)}{f(Y,\theta^*)}$. Thus $\ell^{(j)}(Y,\theta^*)$ is a mean-zero random variable.
I wonder why this is the case though. I've tried the following:
Looking at $j=s$ then $$\dfrac{\partial^s}{\partial \theta^s}\ell(\theta) = \dfrac{\partial^{s-1}}{\partial \theta^{s-1}}\left(\dfrac{\partial}{\partial \theta} \ell(\theta)\right) = \dfrac{\partial^{s-1}}{\partial \theta^{s-1}}\left( \dfrac{\frac{\partial}{\partial}f(Y,\theta)}{f(Y,\theta)}\right)$$
Repeating the procedure results in
$$\dfrac{\partial^{s-1}}{\partial \theta^{s-1}}\left( \dfrac{\frac{\partial}{\partial}f(Y,\theta)}{f(Y,\theta)}\right) = \dfrac{\partial^{s-2}}{\partial \theta^{s-2}}\left( \dfrac{\frac{\partial^2}{\partial \theta^2}f}{f(Y,\theta)}-\frac{\left(\frac{\partial}{\partial \theta}f(Y,\theta)\right)^2}{f(Y,\theta)^2}\right)$$
Repeating the procedure $s-2$ times will result in $$= \dfrac{\frac{\partial^s}{\partial \theta^s}f(Y,\theta)}{f(Y,\theta)} + \ldots$$ where $\ldots = 0$ at $\theta^*$ because of the definition of $s$ (it contains derivatives below $s$).
Now consider the mean-value of this
$$E\left[\dfrac{\frac{\partial^s}{\partial \theta^s}f(Y,\theta^*)}{f(Y,\theta^*)}\right] = \int_\Omega \dfrac{\frac{\partial^s}{\partial \theta^s}f(y,\theta^*)}{f(y,\theta^*)} \cdot f(y,\theta^*) \operatorname dy = \int_{\Omega} \dfrac{\partial^s}{\partial \theta^s}f(y,\theta^*) \stackrel{?}{=} \dfrac{\partial^{s-1}}{\partial \theta^{s-1}}f(y,\theta^*)$$
I don't really see why this would be zero at $\theta= \theta^*$
Got it!It originates from the definition of a density function.
Under certain regularity condition the following is valid: $$\int_\Omega \dfrac{\partial^s}{\partial \theta^s} f(y,\theta^*) \operatorname d y = \dfrac{\partial^s}{\partial \theta^s}\int_\Omega f(y,\theta^*) \operatorname d y = \dfrac{\partial^s}{\partial \theta^s} 1 = 0$$