# Relation between photometric redshift in a given bin and conditional probability

I take over an old question that is not yet clear for me. Indeed, I would like to formulate the equation (1) below under the form of "conditional probability" but I can't manage to do that. So I decided to ask for the question again since i think I can reformulate this relation with this conditional probalbility, and moreover since I have enough (for the instant I hope) score to launch a bounty if necessary (in case if my question wouldn't have enough attention for most of you).

In the context of photometric probe of surveys (like LSST), I need to understand the relation I have to use for photometric bins.

Considering $$p_{ph}(z_p|z)$$ the probability to measure a photometric redshift equal to $$z_p$$ knowing the real redshift is $$z$$ and given $$n(z)$$ the density distribution of objects, I have the following formula which gives the density $$n_i$$ of galaxies into $$i$$-th bin and that I would like to understand :

$$n_{i}(z)=\frac{\int_{z_{i}^{-}}^{z_{i}^{+}} \mathrm{d} z_{\mathrm{p}}\,n(z)\,p_{\mathrm{ph}}(z_{\mathrm{p}} | z)}{\int_{z_{min}}^{z_{max}} \int_{z_{i}^{-}}^{z_{i}^{+}}\,\mathrm{d} z \,\mathrm{d} z_{\mathrm{p}} \,n(z) \,p_{\mathrm{ph}}(z_{\mathrm{p}} | z)} \quad(1)$$

with $$\left(z_{i}^{-}, z_{i}^{+}\right)$$ which are the values of both sides (+ and -) of redshift $$i$$ bin.

I wonder if I can express this relation with the equivalent of Bayes theorem but with density functions (called conditional density), like this :

$$g\left(x | y_{0}\right)=\frac{f\left(x, y_{0}\right)}{\int f\left(t, y_{0}\right) \mathrm{d} t}\quad(2)$$

Or maybe by making appear the conditional probability (it depends if $$X$$/$$Y$$ are discrete/continuous, i.e (from https://en.wikipedia.org/wiki/Bayes%27_theorem#Random_variables) :

Simple form $$[$$ If $$X$$ is continuous and $$Y$$ is discrete, $$f_{X \mid Y=y}(x)=\frac{P(Y=y \mid X=x) f_{X}(x)}{P(Y=y)}$$ where each $$f$$ is a density function. If $$X$$ is discrete and $$Y$$ is continuous, $$P(X=x \mid Y=y)=\frac{f_{Y \mid X=x}(y) P(X=x)}{f_{Y}(y)}$$ If both $$X$$ and $$Y$$ are continuous, $$f_{X \mid Y=y}(x)=\frac{f_{Y \mid X=x}(y) f_{X}(x)}{f_{Y}(y)}$$

EDIT: I would like to better understand the eq. (1) and in the same time to be able to interpret it as a "Bayesian" formulation knowing z_p seems to be a discret variable (I am not really sure of that since we integrate over dz_pin bottom) and z a continuous variable

Any help is welcome.

• Not quite sure what you are asking. Are you looking to understand what eq. (1) is saying? Or are you looking to derive eq. (1)? – student Sep 14 at 22:53