Author: Aaron Robotham
Date: 2018-05-28

# NFW Distribution

#### Aaron Robotham

#### 2018-05-28

## QDF Proof

This is a compact version of the paper arXiv 1805.09550.

First we define the NFW PDF for any input scale radius \(q=R/R_{vir}\)

\[ d(q,c) = \frac{c^2}{(c q + 1)^2 \left(\frac{1}{c + 1} + \ln(c + 1) - 1\right)} \]

We define the un-normalised integral up to a normalised radius \(q\) as

\[ p'(q,c) = \ln(1 + c q)-\frac{c q}{1 + c q}. \]

We can define \(p\), the correctly normalised probability (where \(p(q=1,c)=1\) with a domain [0,1]) as

\[ p(q,c) = \frac{p'(q,c)}{p'(1,c)} \Rightarrow p'(q,c)=p(q,c) p'(1,c) \]

Using the un-normalised variant \(p'\) we can state that

\[ p'+1 = \ln(1 + c q) - \frac{c q}{1 + c q} + \frac{1 + c q}{1 + c q} = \ln(1 + c q) + \frac{1}{1 + c q}. \]

Taking exponents and setting equal to \(y\) we get

\[ y = e^{p'+1} = (1 + c q) e^{1/(1 + c q)}. \]

We can the define \(x\) such that

\[ x = 1 + c q, \\ y = x e^{1/x}. \]

Here we can use the Lambert W function to solve for \(x\), since it generically solves for relations like y = x e^{x} (where \(x = W_0(y)\))). The exponent has a \(1/x\) term, which modifies that solution to the slightly less elegant

\[ x = -\frac{1}{W_0(-1/y)} = -\frac{1}{W_0(-1/e^{(p'+1)})}, \\ \text{sub for } q = \frac{x - 1}{c}, \\ q = -\frac{1}{c}\left(\frac{1}{W_0(-1/e^{(p'+1)})}+1\right). \]

Given the above, this opens up an analytic route for generating exact random samples of the NFW for any \(c\) (where we must be careful to scale such that \(p'(q,c)=p(q,c) p'(1,c)\)) via,

\[ r([0,1]; c) = q(p=U[0,1]; c). \]

## Some Example Usage

Both the PDF (dnfw) integrated up to x, and CDF at q (pnfw) should be the same (0.373, 0.562, 0.644, 0.712):

    for(con in c(1,5,10,20)){
      print(integrate(dnfw, lower=0, upper=0.5, con=con)$value)
      print(pnfw(0.5, con=con))
    }

    ## [1] 0.373455
    ## [1] 0.373455
    ## [1] 0.5618349
    ## [1] 0.5618349
    ## [1] 0.6437556
    ## [1] 0.6437556
    ## [1] 0.7116174
    ## [1] 0.7116174

The qnfw should invert the pnfw, returning the input vector (1:9)/10:

    for(con in c(1,5,10,20)){
      print(qnfw(p=pnfw(q=(1:9)/10,con=con), con=con))
    }

    ## [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
    ## [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
    ## [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
    ## [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

The sampling from rnfw should recreate the expected PDF from dnfw:

    for(con in c(1,5,10,20)){
      par(mar=c(4.1,4.1,1.1,1.1))
      plot(density(rnfw(1e6,con=con), bw=0.01))
      lines(seq(0,1,len=1e3), dnfw(seq(0,1,len=1e3),con=con),col='red')
      legend('topright',legend=paste('con =',con))
    }

Happy sampling!