Group Abstract Group Abstract

Message Boards Message Boards

Finding the density function for the absolute difference between two random variables

I have two random variables: x1 and x2. Each random variable has the density function: 2*Sqrt[1-x^2]/Pi. Both random variables are semi-circles on the -1 <= x <= 1 interval. The "area" under each random variable equals unity.

When I try to find the density function of Y = Abs[x1 - x2], I get stuck. Does anyone know of any tricks to find the density function for Y? My code is below.

POSTED BY: Patrick McMullen
4 Replies
Posted 14 days ago

Using David Trimas' integral but with slightly simpler integration limits Mathematica 14.3 (Windows 11) results in the following:

pdf2 = Integrate[8 Sqrt[(1 - x^2) (1 - (d + x)^2)]/\[Pi]^2, {x, -1, 1 - d}, 
  Assumptions -> 0 < d < 2]
(* (4 z ((4 + d^2) EllipticE[1 - 4/d^2] - 8 EllipticK[1 - 4/d^2]))/(3 \[Pi]^2) *)

Note that I used d rather than z as in David Trimas' example. I only get the input repeated back using z and even using y. I have quit and restarted the kernel, closed and reopened Mathematica and still the same issue.

Here is a "plot" of the two implementations:

pdf[z_] := If[0 < z < 2, 8/\[Pi]^2 NIntegrate[Sqrt[(1 - x^2) (1 - (x - z)^2)], {x, Max[z - 1, -1], Min[z + 1, 1]}], 0]
Plot[{pdf2, pdf[d]}, {d, 0, 2}, PlotStyle -> {Thickness[0.02], {Thickness[0.005], Yellow}}]

Two equivalent density functions

POSTED BY: Jim Baldwin

The documentation page for PDF used to point to the probability distribution function. With 14.3 it points to Import and Export format.

POSTED BY: Gianluca Gorni

A numerical method to calculate the PDF:

pdf[z_] := 
 If[0 < z < 2, 
  8/π^2 NIntegrate[ 
    Sqrt[(1 - x^2) (1 - (x - z)^2)], {x, Max[z - 1, -1], Min[z + 1, 1]}],
   0]

Maybe there's some kind of trig substitution that would make $$ \frac{8}{\pi^2}\int _{\max (-1,z-1)}^{\min (1,z+1)}\sqrt{\left(1-x^2\right) \left(1-(x-z)^2\right)}dx$$ solvable?

POSTED BY: David Trimas

I suppose this is the correct syntax:

dist = ProbabilityDistribution[2*Sqrt[1 - x^2]/Pi, {x, -1, 1}];
trasf = TransformedDistribution[
  Abs[a - b], {a \[Distributed] dist, b \[Distributed] dist}]
NProbability[x > 1, x \[Distributed] trasf]
Plot[NProbability[x > y, x \[Distributed] trasf], {y, 0, 2}]

Unfortunately, it seems that the calculation of the PDF is very slow.

POSTED BY: Gianluca Gorni
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard