Dan Gifford edited Introduction.tex  about 10 years ago

Commit id: 6b10d751282d33dcdbba21fe2539b57771181e57

deletions | additions      

       

There are several key probabilities we need to know in order to accurately predict the expectation value of some observable with mass. In more statistical language, we would like to know $P(\hat{\theta}|M)$. That is, given a cluster of mass $M$, what is the probability the observable is detected as $\hat{\theta}$. For example, the observable $\hat{\theta}$ can be velocity dispersion, some richness, or total luminosity.   Let's Let's  take velocity dispersion as an example here. There are two relevant velocity dispersions in our studies. The first is the true 3D/2D 3D/2D  velocity dispersion $\sigma$ which we cannot measure in the real universe. \citet{Evrard08} measured $\sigma$ for halos in N-body simulations and showed that they relate to the critical mass $M_{200}$ of the host halo on a very tight relation \begin{equation}  \langle \sigma | M \rangle = 1093 \left(\frac{h(z) M}{1e15 M_{\odot}}\right)^{0.34}  \end{equation} 

\begin{equation}  P(\hat{\sigma} | M) = \sum_{\sigma} P(\hat{\sigma} | \sigma) P(\sigma | M)  \end{equation}  Really there are completeness and purity terms in there as well, but lets ignore those for a second. So that is our expected distribution of $\hat{\sigma}$ for a given $M$. The other term is equally important $P(\hat{\sigma} | \sigma)$. This represents the probability of observing a velocity dispersion $\hat{\sigma}$ given $\sigma$. Why is this important? When we observe clusters in the real universe, we don't don't  measure the ``Evrard" ``Evrard"  velocity dispersion $\sigma$. We are randomly drawing from a distribution where the $\sigma$ is the expectation value. This is what \citet{Gifford13a} means by l.o.s effects. So what is that distribution? It's It's  approximately lognormal with $S_{\log(\hat{\sigma}) | \log(\sigma)}\sim 25\%$. So: \begin{equation}  P(\hat{\sigma} | \sigma) = \frac{1}{\sqrt{2\pi}S_{\ln\hat{\sigma} | \ln\sigma}} e^{\frac{(\ln\hat{\sigma} - \ln\sigma)^{2}}{2 S_{\ln\hat{\sigma} | \ln\sigma}^2}}  \end{equation}