91Ó°ÊÓ

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).

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 $\mathrm{θ}$have the same probability density function,

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

Let,

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

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

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

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

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

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

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

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

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

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 $\mathrm{θ}$is,

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

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