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!