Question

If X and Y are independent random variables both uniformly distributed over $(0,1)$, find the joint density function of $R=\sqrt{X^2+Y^2},\theta ={\tan }^{-1}Y/X$.

Short answer

The joint probability density function of $R=\sqrt{X^2+Y^2}$and $\theta ={\tan }^{-1}Y/X$is $r$and uniformly distributed from$0to1.$

Step-by-step solution

Probability density function : 

The probability density function is defined as the integral of the variable density density over a certain range. It is represented by the letter f(x).

Explanation :

If X and Y are independent random variables both uniformly distributed over $(0,1)$, then the joint density of X and Y is,

$f(x,y)=1$

Assume that R and $\theta$have the same probability density function,

$f(R,\theta )=f(x,y){\left|J(x,y)\right|}^{-1}..............\left(1\right)$

Let,

$$\begin{gathered} R=g_1\left(x,y\right)=\sqrt{x^2+y^2}.....\left(2\right) \\ \theta =g_2(x,y)={\tan }^{-1}\left(\frac{y}{x}\right).......\left(3\right) \end{gathered}$$

Now, differentiating $\left(2\right)$with respect to x then,

$$\begin{gathered} \frac{\partial g_1(x,y)}{\partial x}=\frac{\partial }{\partial x}\left(\sqrt{x^2+y^2}\right) \\ =\frac{x}{\sqrt{x^2+y^2}} \end{gathered}$$

Now, differentiating $\left(2\right)$with respect to x then,

$$\begin{gathered} \frac{\partial g_1(x,y)}{\partial y}=\frac{\partial }{\partial y}\left(\sqrt{x^2+y^2}\right) \\ =\frac{y}{\sqrt{x^2+y^2}} \end{gathered}$$

Now, differentiating $\left(3\right)$with respect to x then,

role="math" localid="1647256694571" $$\begin{gathered} \frac{\partial g_2(x,y)}{\partial x}=\frac{\partial }{\partial x}\left[{\tan }^{-1}\left(\frac{y}{x}\right)\right] \\ =\frac{-y}{x^2+y^2} \end{gathered}$$

Differentiating $\left(3\right)$with respect to y then,

$$\begin{gathered} \frac{\partial g_2(x,y)}{\partial y}=\frac{\partial }{\partial y}\left[{\tan }^{-1}\left(\frac{y}{x}\right)\right] \\ =\frac{x}{x^2+y^2} \end{gathered}$$

Explanation :

The Jacobian transformation of X and Y is,

$$\begin{gathered} J(x,y)=\frac{x^2}{{\left(x^2+y^2\right)}^{{\displaystyle \frac{3}{2}}}}+\frac{y^2}{{(x^2+y^2)}^{{\displaystyle \frac{3}{2}}}} \\ =\frac{1}{\sqrt{x^2+y^2}} \\ =\frac{1}{r} \end{gathered}$$

Therefore, the joint probability density function of R and $\theta$is,

$$\begin{gathered} f(R,\theta )=1\times {\left(\frac{1}{r}\right)}^{-1} \\ =1\times r \\ =r \end{gathered}$$

The joint density function is$r$ranging from $0to1.$