Suppose $\theta$ and $R$ are independent random variables, with $\theta$ being Uniform$(−\pi/2,\pi/2)$ and $R$ having the probability density function given by:

$$f_R(r)=\frac{2r}{(1+r^2)^2} \:\text{for } r>0, \quad f_R(r)=0 \:\text{otherwise}.$$

Determine the joint PDF for $X=R\cos\theta$ and $Y=R\sin\theta$.

It is simple enough to calculate the cumulative distribution functions of $\theta$ and $R$, having

$$F_\theta(x)=\frac{1}{\pi}x+\frac{1}{2} \:\text{ on }\: (-\pi/2,\pi/2),\quad \text{and } \:F_r(x)=\frac{x^2}{x^2+1}\:\text{ on }\:(0,\infty),$$

with my aim being eventually to calculate $F_{XY}(x,y)$ from which, by taking partial derivatives, I would be able to obtain $f_{XY}(x,y).$

However, $F_{XY}(x,y)=\mathbb{P}[X=R\cos\theta\le x,Y=R\sin\theta\le y],$ and no matter how much I try to interpret this expression, I can't seem to find a way to bring it into terms of $F_r$ and $F_\theta$ as would lead (I imagine) to the solution.

Is this the correct way to tackle this sort of problem? If so, any hints?


2 Answers 2


You only want the probability density function.   Thus you do not need to know what the cummulative density function is, just how to differentiate it, w.r.t. $x,y$.

Using $~\arctan:\Bbb R\mapsto (-\pi/2.. \pi/2)$

$$\begin{align}f_{\small X,Y}(x,y)&=\dfrac{\mathrm d^2 F_{\small X,Y}(x,y)}{\mathrm d x~\mathrm dy}\\[1ex]&=\dfrac{\mathrm d^2~F_{\small R,\Theta}(\surd (x^2+y^2),\arctan(x/y))}{\mathrm d x~\mathrm d y}\\[1ex]&=\left\lVert\dfrac{\partial\langle\surd(x^2+y^2),\arctan(x/y) \rangle}{\partial\langle x,y\rangle}\right\rVert \left.\dfrac{\mathrm d^2 F_{_{R,\Theta}(r,\theta)}}{\mathrm d r\,\mathrm d \theta}\right\vert_{r=\surd(x^2+y^2)\\\theta=\arctan(x/y)}\\[1ex]&=\left\lVert\dfrac{\partial\langle\surd(x^2+y^2),\arctan(x/y) \rangle}{\partial\langle x,y\rangle}\right\rVert f_{\small R,\Theta}(\surd(x^2+y^2),\arctan(x/y))\tag{1}\\[1ex]&=\begin{Vmatrix}\dfrac{\partial \surd(x^2+y^2)}{\partial x}& \dfrac{\partial\surd(x^2+y^2)}{\partial y}\\ \dfrac{\partial \arctan(x/y)}{\partial x}&\dfrac{\partial\arctan(x/y)}{\partial y}\end{Vmatrix}\,f_{\small R}(\surd(x^2+y^2))\,f_{\small\Theta}(\arctan(x/y))\\[1ex]&~~\vdots\end{align}$$

(1): This is known as the Jacobian Transformation. When $U=g(X,Y), V=h(X,Y)$ then: $$f_{\small X,Y}(x,y)=\left\lVert\dfrac{\partial \langle g(x,y), h(x,y)\rangle}{\partial\langle x,y\rangle}\right\rVert\,f_{\small U,V}(g(x,y),h(x,y))$$

  • $\begingroup$ Thanks for such a detailed response. I wasn't aware of the Jacobian Transformation beforehand so will have to do some reading, but the result I obtain with your method checks out :). $\endgroup$ Nov 6, 2020 at 2:59

Let me give a different exposition of essentially the same answer as Graham Kemp's.

You want to find $$F_{XY}(x,y) = P[X≤x,Y≤y] = \int_{-\infty}^x\int_{-\infty}^y f_{XY}(\xi, \eta) d\xi d\eta$$ and, because you know the joint density function $f_{R \theta}$ of $R$ and $\theta$, you can compute the probability of an event of the form $\{(R, \theta) \in A\}$ as $$P[(R, \theta) \in A] = \iint_A f_{R\theta}(r, t) dr dt.$$

The key now is to use the change of variables formula, stated for an integrable function $h$ and a domain $U$ as $$\iint_{\phi(U)} h(\mathbf v) d\mathbf v = \iint_{U} h(\phi(\mathbf u)) \det(D\phi)(\mathbf u) d\mathbf u,$$ to calculate the probability of the event $\{X≤x,Y≤y\}$ using the known density $f_{R\theta}$. Thus,

  • Let $\mathbf v = (r, t)$ and $\mathbf u = (\xi, \eta)$
  • Let $U = (-\infty, x] \times (-\infty, y]$, so that $P[X≤x,Y≤y] = P[(X, Y) \in U]$
  • Let $\phi : \mathbb R^2 \to \mathbb R^2$ be such that $\phi(X,Y) = (R, \theta)$. That is, $\phi(\xi, \eta) = (\sqrt{\xi^2 + \eta^2}, \arctan(\eta / \xi))$
  • Let $h=f_{R\theta}$.

Then, $P[X≤x,Y≤y] = P[(X, Y) \in U] = P[\phi(X,Y) \in \phi(U)] = P[(R, \theta) \in \phi(U)]$ and, by definition of density and the change of variables formula, this last expression is equal to $$\iint_{\phi(U)} f_{R\theta}(r, t) dr dt = \iint_{U} f_{R\theta}(\phi(\xi, \eta)) \det(D\phi)(\xi, \eta) d\xi d\eta.$$

Finally, since $U$ is arbitrary, you can conclude that $$f_{XY}(\xi, \eta) = f_{R\theta}(\phi(\xi, \eta)) \det(D\phi)(\xi, \eta) \\ = f_{R}(\sqrt{\xi^2 + \eta^2}) f_\theta(\arctan(\eta / \xi)) \det(D\phi)(\xi, \eta) ,$$ and now all that is left is to compute the Jacobian $\det(D\phi)$.

Note that the description of the set $\phi(U)$ in terms of the variables $r$ and $t$ is not easy to write, but it is not necessary to do so, as it is possible to express probabilities of arbitrary events $P[(R, \theta) \in A]$ as integrals of the density over abstract sets $A$.

  • $\begingroup$ Thank you both, I really appreciate the explanation; I would never have come up with either of these solutions, so they has been really enlightening to see. $\endgroup$ Nov 6, 2020 at 11:44

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