I was hoping someone could help explain a step? I thought I got it, but reading it today, it's not making sense, and I'm wondering if I'm missing something after all. They go from Eq.1 to Eq.2 (given below) using the knowledge that $(x_k - \mu(x,y))^2$ is the variance and equal to the expected value of the Poisson process, which I believe $\mu(x,y)$ is.
$$Eq.1: I_{ij}(\theta) = E[\sum_k(x_k-\mu_k(x,y))^2\frac{1}{\mu_k(x,y)^2}\frac{\delta\mu_k(x,y)}{\delta\theta_i}\frac{\delta\mu_k(x,y)}{\delta\theta_j}]$$
$$Eq.2: I_{ij}(\theta) = \sum_k\frac{1}{\mu_k(x,y)}\frac{\delta\mu_k(x,y)}{\delta\theta_i}\frac{\delta\mu_k(x,y)}{\delta\theta_j}$$
Originally I was thinking that the expected value of an expression multiplied by its variance... is just the expression itself, but that wouldn't account for the $\mu(x,y)$ being removed from the denominator--while substituting $(x_k - \mu(x,y))^2$ for the expected value, I don't think yields equation 2? Or am I wrong on that?
Extra information:
$$\mu_k(x,y)=\theta_{I_0}\int_{A_k}\frac{1}{2\pi\sigma^2}e^{\frac{-(x-\theta_x)^2-(y-\theta_y)^2}{2\sigma^2}}+\theta_{bg}$$
Where $\theta_{I_0}$, $\theta_{bg}$, $\theta_{I_0}$, and $\theta_{bg}$ are all parameters being fit through MLE. $A_k$ is a unit square defining the area over which the 2D gaussian is being integrated over. The likelihood function, which is for a Poisson process, is defined as:
$$L(X|\theta) = \prod_k \frac{\mu_k(x,y)^{x_k}e^{-\mu_k(x,y)}}{x_k!}$$
If $X_k$ has mean $\mu_k:=\mu_k(x,y)$, then for any non-random factors $c_k$, \begin{align} E\left[\sum_k (X_k - \mu_k)^2 c_k\right] = \sum_k c_k E[(X_k - \mu_k)^2] = \sum_k c_k \text{Var}(X_k). \end{align} If furthermore $X_k$ is Poisson, then its variance is the same as its mean, so the above equals $\sum_k c_k \mu_k$. Your expression is this with $c_k := \frac{1}{\mu_k^2} \frac{\partial \mu_k}{\partial \theta_i} \frac{\partial \mu_k}{\partial \theta_j}$.