# Give the cdf and pdf given that we select a point uniformly on an annulus

Select uniformly a point from the annulus $$\{(x,y):1\leq x^2+y^2\leq 4\}$$. Let R be the distance to $$(0,0)$$. Give the cdf and pdf of R.

I know that the sample space of choosing a point would be the area of the annulus ($$3\pi$$) and I know the method of getting the pdf from the cdf or vice versa. I'm just not sure how to start; How would I represent the distance R from the point (x,y) and would this be the cdf or pdf?

• Why delete part of the question when the answers address that part specifically? – StubbornAtom Mar 26 at 7:53

How would I represent the distance R from the point (X,Y) and would this be the cdf or pdf?

By definition of Euclidean Distance, $$R=\sqrt{X^2+Y^2}$$.

So immediately we know the support is $$\{r: 1\leqslant r^2\leqslant 4\}$$

Now, since the points are uniformly distributed: the pdf for $$R$$, at any point $$r$$ within that support, will be equal to the ratio of the circumference of a circle with radius $$r$$ to the area of the annulus. Call this $$f_R(r)$$.

And the CDF will be the integral $$\displaystyle F_R(r)=\int_1^r f_R(s)\mathrm d s$$ .

• Thanks. Why would it be the circumference of the circle, would it not be the area? – PLC Mar 26 at 4:51
• @Graham Kemp doesn't this give for $r=2$ which lies within that support, $\Bbb{P}(R=2)=\frac{4}{3}>1$? I interpretted the pdf for $R$ at any point within the support as $\frac{2\pi r}{3 \pi} = \frac{2r}{3}$ – Tikak Mar 26 at 5:04
• A Probability Density Function may have values exceeding one. It is not a probability mass. Also, yes , $f_R(r)=\dfrac{2r}{3}\mathbf 1_{1\leqslant r\leqslant 2}$. – Graham Kemp Mar 26 at 12:07
• It is the circumference because that is the derivative of the area of a circle with respect to the radius. (Which is a hint to find the CDF without integration.) – Graham Kemp Mar 26 at 12:10

It is easier to find the CDF first (and then it is easy to find the PDF from there). Try to compute $$P(R \le r)$$ for any real number $$r$$. (This will be a ratio of areas.)