0
$\begingroup$

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 $k$.

  1. 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.

  2. 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 ?

$\endgroup$
7
  • $\begingroup$ I think you may be mixing some things up in your notation. Let's say we have one probe -- the height of $N$ people. Then our data will be a vector, $x_i$ (where $i=1, \cdots, N$). If we want to measure the mean and standard deviation of the height, then our model with have two parameters, $\mu$ and $\sigma$. The correct likelihood in this case is Eq 1. Now let's say we have 2 probes -- we measure the height and weight of $N$ people. Then our data is a $N\times 2$ array, which you can label by $x^A_i$, where $A$ tells us whether we are talking about weight or height, (...) $\endgroup$
    – Andrew
    Apr 5 at 5:57
  • $\begingroup$ (...) and $i$ runs from $1$ to $N$. If we want to measure the mean and standard deviation of height and weight, then we have four parameters: $\mu^A$ and $\sigma^A$, where $A=\{{\rm height}, {\rm weight}\}$. In this case, neither Eq 1 nor Eq 4 are correct. You would want to use a modified version of Eq 1, where you put an $A$ index on $x_i$, $\mu$, and $\sigma$, and sum over $A$. (This is what I did in my answer). Eq 4 would correspond to a case where you want to fit $2N$ parameters $\mu_i$ and $\sigma_i$ to $N$ data points $x_i$. I can't imagine a situation where you would want to do this. $\endgroup$
    – Andrew
    Apr 5 at 6:00
  • $\begingroup$ @Andrew . Thanks, actually you are right when you make the distinction between 2 probes (If I have well understood by the distinctions of 2 types of measures, height and weight). This is also the same thing in my cosmological context, I have 2 probes (without considering their cross-correlation but this is another story) : first what we call the galaxy clustering angular power spectrum $C_\ell^{GG}$ and on another side the Weak lensing $C_\ell^{WL}$. If I am in a Fisher information context, I have to use these 2 probes inside the Big Fisher matrix and so the $\chi^2$ should run over GG and WL $\endgroup$
    – youpilat13
    Apr 5 at 6:37
  • $\begingroup$ But I have to grasp better the subtiliies in the appropriate interpretation for the $\chi^2$ in my case. The survey on which I work provide only a single data at point "$k$", so at first sight, I would say that $\chi^2$ eq(7) is more suitable since I have the mean $\mu_k$ for the $k$-th observable of the model. Any suggestion is welcome, you did already a lot for me.. Regards $\endgroup$
    – youpilat13
    Apr 5 at 6:44
  • $\begingroup$ I find it very difficult to believe you would want to use Eq 4 (or everything after). It is not meaningful to fit $N$ parameters to $N$ data points. Your model will simply reproduce your data, you will not learn anything. And in this case you'd be fitting $2N$ parameters to $N$ data points!! $\endgroup$
    – Andrew
    Apr 5 at 6:49

1 Answer 1

0
$\begingroup$

The notation in this question is very hard to follow. In defining the Fisher matrix it seems the indices $i$ and $j$ label parameters, but in the $\chi^2$ likelihood the index $i$ is labeling data samples. And it is not at all clear what the underlying data or model are supposed to be.

Nevertheless, given a choice between Eq 3 and 7, I think the answer is pretty clearly that Eq 3 is correct. Eq 3 represents a case where you use $n$ data points to measure 1 parameter $\mu$ and its uncertainty $\sigma$. In Eq 7, you are fitting $n$ parameters, plus uncertainty, to $n$ data samples (I'm assuming each data sample $x_i$ is a single number). I can't imagine a scenario where this is a meaningful thing to do. You would surely be overfitting your data set. I also hope you agree it would be hard to trust an uncertainty estimate of a parameter based on fitting that parameter to a single data point.


As I understand the edit to the question, the data are estimators $\hat{a}_{\ell m}^A$, where $A\in \{G, W\}$, which should be unbiased estimators of the cosmological parameters $a_{\ell m}$. The model parameters are $a_{\ell m}$ and $C_{\ell}^{A B}$. Then I would expect the log likelihood to have the form \begin{equation} \ln \mathcal{L} = -\frac{1}{2} \sum_{A=\{G, W\}} \sum_{B=\{G, W\}}\sum_{\ell=\ell_{\rm min}}^{\ell_{\rm max}} \sum_{m=-\ell}^{\ell} \left(a_{\ell m} - \hat{a}_{\ell m}^A\right) C^{-1}_{AB, \ell} \left(a_{\ell m} - \hat{a}_{\ell m}^B \right) \end{equation}

I'm inspired to make this guess based on the way the question was phrased. However, physically, I would have thought the $a_{\ell m}$ are not interesting physical parameters, since they are essentially random and not predicted by the theory. So I would actually have thought a likelihood of the following form would make more sense \begin{equation} \ln \mathcal{L} = -\frac{1}{2} \sum_{A=\{G, W\}} \sum_{B=\{G, W\}}\sum_{\ell=\ell_{\rm min}}^{\ell_{\rm max}} \sum_{m=-\ell}^{\ell} \left(\hat{a}_{\ell m}^A\right) C^{-1}_{AB, \ell} \left(\hat{a}_{\ell m}^B \right) \end{equation}

$\endgroup$
7
  • $\begingroup$ Thanks, could you take a look at my EDIT please ? I tried to be clearer regarding the possibility of having a mean vector with different values inside and not a unique scalar value $\mu$ characterizing the model (see eq$(8)$). $\endgroup$
    – youpilat13
    Mar 27 at 3:44
  • $\begingroup$ @youpilat13 OK I updated the answer. $\endgroup$
    – Andrew
    Mar 27 at 4:04
  • $\begingroup$ so finally, which equation ( eq$(3)$ or eq$(7)$ ) is correct in my case ? .i.e a mean vector with different element values ? or a mean vector filled with a unique scalar $\mu$ ? Best reagrds $\endgroup$
    – youpilat13
    Mar 27 at 4:24
  • $\begingroup$ I showed another possibility of combination in EDIT2 : the problem remains regarding the expression of $\bar{X}(\theta)$, i.e knowing if I have to consider a unique scalar value or if the expected values of the model are equal to zero to respect the definition of $\chi^2$ with data vector $(X-\bar{X}(\theta))$ ? $\endgroup$
    – youpilat13
    Mar 27 at 6:18
  • 1
    $\begingroup$ @youpilat13 Look, you've got to ask one question and be clear on what it is you want to know. I'm quite lost with what it is you're asking with the edits. If you want to add another probe, then in the equations in my edit you would have $A$ and $B$ loop over all the probes instead of just $G$ and $W$. I have no idea what Eq 3 and 7 are even supposed to mean, if $x_i$ is not a scalar variable. Based on what you've written I think one of the likelihoods in my edit are probably right, but I think you've also got to clarify for yourself what you are trying to do. $\endgroup$
    – Andrew
    Mar 27 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.