It gets a bit messy algebraically, but I recommend that you transform your three random variables (d_1, d_2, theta) into three new variables (r, r_c, phi_1). the variable "r" is computed using the Law of Cosines, r_c is just d_2 and phi_1 is the angle formed between r and d_2. Using the Law of Sines, we have

d_1 / sin(phi_1) = d_2 / sin(phi_2).

phi_2 is the angle between r and d_1.

Realizing that phi_2 = pi - theta - phi_1, we can eliminate phi_2 from the Law of Sines and after some messy manipulation, find phi_1 as a function of d_1, d_2 and theta.

You then construct the Jacobian matrix d(r, rc, phi_1)/d(d_1, d_2, theta). The transformed PDF is given by

p(r, rc, phi_1) = p(d_1, d_2, theta) * det(J)^{-1}

where det(J) is the determinant of the Jacobian J.

The PDF for r requires you to integrate out the rc, phi_1 variables:

p(r) = int(p(r, rc, phi_1)*det(J)^{-1} d.rc d.phi_1)

Hope this helps,

Cheers!

EDIT:

PS - I forgot to mention that you may need to renormalize the new PDF so that it integrates to "1" over your random variables.

EDITED CORRECTION: In some places the wrongly used R_a has been replaced by correct R_b; and 0 has been inserted into line (3.A1).

Currently the formulation of the problem is incomplete, since it does not present the rule of choosing the head and/or the user. Moreover it is not clear if the radii R_b and R_c are given consants.

My trial to omit these incoveniences is to assume that the radii ARE given positive constants and that the choices of the head and its user are conditionally uniform (classical distribution on the finite sets), given the numbers N_b and N_c , respectively.

Under these additional assumptions about the randomness, the distribution of the position of the head is uniform within the disc of radius R_b, independently of the number N_b, which implies that the density function of the rv d_b = d_1 is given by the derivative of the distribution function F_b(x) = x^2/(R_b)^2, for 0< x< R_b, i.e.

(1) . . . . f_b(x) = 2 x / (R_b)^2

Now the much more complex case of evaluation of the pdf of d_c = d_2, since

the conditional pd of d_c i uniform in the common part of the two discs

- the one of radius R_b with center at the origin

- the one of radius R_c with center at the position of the head at the distance d_b from the origin.

Fortunately the implied conditional pd of the distance r of the user wrt the origin depends then on the value of d_b, at most. Further advices in preparation:-)

Hopefully the suggested assumptions are acceptable.

Joachim Domsta

PS. Under a simplifying assumption that R_c < R_b, for d_b < R_b - R_c the conditional pd of r given d_b=x is implied by

A) uniformly distributed 2D vector over the disc of radius R_c with center at any point at distance x from the origin, e.g. at point (x, 0).

B) uniformly distributed 2D vector over the common part of the greater disc and the disc of radius R_c, again with center at any point at distance x from the origin, e.g. at point (x, 0).

A technical problem appears how to perform the integration of these conditional wrt the unconditional pd of r_b; in symbols this would be:

(2) . . . . f(r) = \int_[0 , R_b] f( r | x ) \cdot 2 x dx / (R_b)^2

PS2. In case A, in the typical position x> R_c for the conditional density we have

(3.A1) . . . f(r|x) = L(r|x) / [\pi (R_c)^2], for x - R_c < r < x + R_c (and 0 elsewhere),

where L(r|x) is the length of the common part of the arc of the circle with radius r with center at the origin and the disc of radius R_c with center at (x,0).

Good luck, Joachim Domsta

CORRECTION PERFORMED:

In some places the wrongly used R_a has been replaced by correct R_b; and 0 has been inserted into line (3.A1).

PS3. I have a conjecture, that under the assumption of the above comment, the effective pd of r is the same as the user was chosen from the uniform pf of the disc of radius R_b which means that the density is equal 2 r/ (R_b)^2, for 0