How do I find a conditional pdf?

I want to find the pdf of l1? as l1<l0+l2

if any one can help me

The distribution BS* and BS+ in 2D space are given as follows: The BSs* are randomly distributed in a 2D area. The user is located in the center of the area and will connect with the nearest BS*. The pdf of l0 is f(l0) = 2 * (λBS*) * π * l0 * exp(-(λBS*) * π * l0^2), where l0 is the distance between the user and the nearest BS*. The BSs+ are randomly distributed in the same area. The user will connect with the nearest BSs+. The pdf of l2 is f(l2) = 2 * (λBSs+) * π * l2 * exp(-(λBSs+) * π * l2^2), where l2 is the distance between the user and the nearest (BS+). I want to find the pdf of the distance l1 between the nearest BS* and nearest BS+ to the user, where l1=sqrt(l0^2 + l2^2-l0*l2* cos(φ)) as shown in figure below The possible positioning of the nearest BS+ within the captured annular of the inner circle with a radius of l2, centred at the origin.

thank you so much

Answer 1

[Torsten]

Generate random numbers for l0 and l2 that follow the distributions given. Duduce l1 and use histogram to plot its pdf.

Look up "Monte Carlo method" for more details.

I don't understand why you are talking about a conditional pdf: Given l0 and l2, l1=sqrt(l0^2 + l2^2-l0*l2* cos(φ)) will always be smaller than l0+l2 because the points USER, BS+ and BS* form a triangle.

[MOHAMMED MEHDI SALEH]

so with using this l1=sqrt(l0^2 + l2^2-l0*l2* cos(φ)) how i can find pdf as i need it to find expectation of l1^-4

[Torsten]

As I wrote:

Generate N random numbers for l0 and l2 from the given distributions.

Compute l1=sqrt(l0.^2 + l2.^2-l0.*l2* cos(φ))

Compute 1/N * sum(l1.^(-4)) as expectation for l1^(-4).