The measured radius of a circle, R, has pdf
$$ \begin{aligned} f(r) &= 6r(1-r),\; 0 < r < 1 \end{aligned} $$
(a) Find the distribution of the circumference.
(b) Find the distribution of the area of the circle.
Solution
(a) The circumference $C$ of a circle with radius $r$ is $c= 2\pi r$.
So $r= \dfrac{c}{2\pi}$, $\Rightarrow \dfrac{dr}{dc} =\dfrac{1}{2\pi}$.
For $r=0$, $c= 0$ and for $r=1$, $c=2\pi$.
Thus, the distribution of the circumference is
$$ \begin{aligned} f(c) &= f(r) \bigg|\frac{dr}{dc}\bigg| \\ &= 6\frac{c}{2\pi}\big(1- \frac{c}{2\pi}\big)\times \frac{1}{2\pi}\\ &= \frac{c(2\pi-c)}{\pi^3}, 0< c< 2\pi \end{aligned} $$
(b) The area $A$ of a circle with radius $r$ is $A= \pi r^2$.
So $r= \sqrt{\dfrac{a}{\pi}}$, $\Rightarrow \dfrac{dr}{da} =\dfrac{1}{2\sqrt{a\pi}}$.
For $r=0$, $a= 0$ and for $r=1$, $a=\pi$.
Thus, the distribution of the area is
$$ \begin{aligned} f(a) &= f(r) \bigg|\frac{dr}{da}\bigg| \\ &= 6\sqrt{\frac{a}{\pi}}\bigg(1- \sqrt{\frac{a}{\pi}}\bigg)\times \frac{1}{2\sqrt{a\pi}}\\ &= \frac{3}{\pi}\bigg(1- \sqrt{\frac{a}{\pi}}\bigg),\; 0 < a < \pi \end{aligned} $$
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
