By using a general Likelihood computed from a theoretical model and $\lambda_{i}, \lambda_{j}$ as cosmological parameters, We have the following definition of an element $(i, j)$ of Fisher matrix $F$ : $$ F_{i j}=\left\langle-\frac{\partial^{2} \ln (\mathcal{L})}{\partial \lambda_{i} \lambda_{i}}\right\rangle=\left\langle\frac{\partial \ln (\mathcal{L})}{\partial \lambda_{i}} \frac{\partial \ln (\mathcal{L})}{\partial \lambda_{j}}\right\rangle $$
We make here a strong assumption by considering that all Likelihoods are Gaussian, relating this latter and the $\chi^{2}$ by:
$$\chi^{2}=\sum_{i=1}^{n}\left(\frac{x_{i}-\mu}{\sigma}\right)^{2}\quad(1)$$
$$\Rightarrow \ln (\mathcal{L})=-\frac{1}{2} \chi^{2}+K$$
with $K$ a constant, where one has taken general notations with data vector :
$$\mathbf{X} \equiv \left\{x_{1}, . ., x_{n}\right\}$$.
We Considering a model with
$$\mu=\overline{\mathbf{X}}\quad(2)$$
from the maximum likelihood estimator where the mean of data is represented by vector $\overline{\mathbf{X}}$.
Thus, we can write : $$ -\frac{\partial \ln \mathcal{L}}{\partial \lambda_{i}}=-\sum_{k=1}^{n} \frac{\left(x_{k}-\mu\right)}{\sigma} \frac{\partial \mu}{\partial \lambda_{i}} $$
so :
$$ \begin{aligned} F_{i j} &=\sum_{k=1}^{n} \sum_{k^{\prime}=1}^{n}\left\langle\frac{\left(x_{k}-\mu\right)\left(x_{k^{\prime}}-\mu\right)}{\sigma^{4}} \frac{\partial \mu}{\partial \lambda_{i}} \frac{\partial \mu}{\partial \lambda_{j}}\right\rangle \\ &=\sum_{k=1}^{n} \sum_{k^{\prime}=1}^{n} \delta_{k k^{\prime}} \frac{1}{\sigma^{4}}\left\langle\left(x_{k}-\mu\right)\left(x_{k^{\prime}}-\mu\right)\right\rangle \frac{\partial \mu}{\partial \lambda_{i}} \frac{\partial \mu}{\partial \lambda_{j}} \end{aligned} $$ since $: \delta_{k k^{\prime}}\left\langle\left(x_{k}-\mu\right)\left(x_{k^{\prime}}-\mu\right)\right\rangle=\sigma^{2}$ Following :
$$ F_{i j}=\sum_{k=1}^{n} \frac{1}{\sigma^{2}}\left\langle\frac{\partial \mu}{\partial \lambda_{i}} \frac{\partial \mu}{\partial \lambda_{j}}\right\rangle=\sum_{k=1}^{n} \frac{1}{\sigma^{2}} \frac{\partial \mu}{\partial \lambda_{i}} \frac{\partial \mu}{\partial \lambda_{j}}\quad(3) $$
Question : I don't know if I have to consider a unique value for $\mu$ and $\sigma$ like I did in eq$(1)$ or maybe should I rather write eq$(1)$ like this :
$$\chi^{2}=\sum_{i=1}^{n}\left(\frac{x_{i}-\mu_{i}}{\sigma_{i}}\right)^{2}\quad(4)$$ ?
and then consider eq$(2)$ with the $\mu$ as a vector of different means :
$$\mu=\overline{\mathbf{X}}\quad(5)$$
with
$$\overline{\mathbf{X}} \equiv \left\{\bar{x}_{1}, \bar{x}_{2}, . ., \bar{x}_{n}\right\}=\left\{\mu_{1}, \mu_{2}, . ., \mu_{n}\right\}\quad(6)$$
and not : $$\overline{\mathbf{X}} \equiv \left\{\mu, \mu, . ., \mu\right\}$$
In the case of expression $(6)$, the final expression of Fisher element $F_{ij}$ would be :
$$F_{i j}=\sum_{k=1}^{n} \frac{1}{\sigma_{k}^{2}}\left\langle\frac{\partial \mu_{k}}{\partial \lambda_{i}} \frac{\partial \mu_{k}}{\partial \lambda_{j}}\right\rangle=\sum_{k=1}^{n} \frac{1}{\sigma_{k}^{2}} \frac{\partial \mu_{k}}{\partial \lambda_{i}} \frac{\partial \mu_{k}}{\partial \lambda_{j}}\quad(7)$$
As you can see, I make confusions in $\chi^2$ definition between the expected value of a model and its generalization when we consider a vector of data, which seems to assume that the means are different and not equal to a same single value $\mu$.
This is the same issue about a unique $\sigma$ instead of $\sigma_k$ : indeed, $\sigma_k$ would mean that I have multiple measures for the same point $k$.
If have had only one measure for each point point $k$, I think that correct expression is eq$(3)$. The same thing for single $\mu$. Associated expression of $\chi^2$ would be eq(1) in this case.
If have had multiple measures for each point point $k$, I think that correct expression is eq$(7)$ since I can define a $\sigma_k$ from these multiple data. $\sigma_k$ is the error on each point $k$, i.e on each multiple measure for point $k$. The same thing for multiple distincts $\mu_k$ which means that we would have an expected value for each point $k$. Associated expression of $\chi^2$ would be eq(4) in this case.
So finally, is my reasoning on point 1. and 2. correct ?
The appropriate expression equation$(3)$ or equation$(7)$ depends on these 2 cases 1. and 2. , doesn't it ?