# Finding probability density function: dot in the surface of a disc

I'm beginning to study continuous random variable statistics and I'm having trouble to solve exercises, so I need help. I don't know how to apply definitions in this case. (Also, I'm not a native English speaker, so please correct or ignore grammatical mistakes if you see any of them).

A disc of radius R has a dot somewhere in its surface, being the probability of finding the dot the same anywhere in the disc's surface. If we call X the distance from the center of the disc to that dot, find the expression for the probability density function of the random variable X.

Thanks in advance. I'd prefer if someone gave me the guidelines instead of the actual answers to see if I can work it out myself.

Edit: My initial reasoning was the following one:

The probability of finding the dot depends on how far away you are from the center (because the farther away you are, the bigger the area you've covered), so the distribution function would be $P(X)=\frac{\pi X^2}{\pi R^2}=\frac{X^2}{R^2}$ (so that when you are on the edge of the disc the probabily of the dot being contained in that area is 1). But I have no idea if this makes any sense.

• You should add the self-study tag to your question. Also you could try to tell us what you've tried so far and/or where you get stuck. See stats.stackexchange.com/tags/self-study/info Commented Jun 4, 2017 at 15:50
• Thanks, I just did. I just started studying this so I probably didn't make any sense and I'm pretty lost. Commented Jun 4, 2017 at 16:01
• I think what you're describing in your edit would be the cumulative distribution function. Luckily, you can derive the probability distribution function from the cumulative distribution function... Commented Jun 4, 2017 at 16:36
• If the cumulative distribution function is equal to the integral of the PDF (in this case from X=0 to X=R), should I differenciate? Or would that approach be conceptually wrong in this case? Commented Jun 4, 2017 at 16:44
• @Manuel Go ahead and differentiate unless you think the PDF doesn't exist or something I guess
– BCLC
Commented Jun 5, 2017 at 5:18

One way to start would be to write down the probability that the distance of the dot from the center is less than or equal to $x$

$$P(X\leq x) = ...$$

This will be the cdf of $X$. It can be written down by basic geometric reasoning.

You can then obtain the density from the cdf.

Start with the cdf $F_X(x) = P(X \le x)$

Obviously

1. $P(X \le x) = 0$ if $x \le 0$
2. $P(X \le x) = 1$ if $x \ge R$

Now for

• $P(X \le x)$ if $0 < x < R$

we want to find the probability that the dot will end up somewhere in the annulus with outer circle having radius of $x$ with inner circle having a radius of zero.

Let the polar coordinates of the dot be $(R,\Theta)$ (capital since the coordinates are random). Then we have

$$P(X \le x) = \frac{\int_0^{2 \pi} \int_0^x 1_{(r,\theta) \in D} \cdot r dr d\theta}{\int_0^{2 \pi} \int_0^R 1_{(r,\theta) \in D} \cdot r dr d\theta} = \frac{\pi x^2}{\pi R^2} \tag{*}$$

• where $D = \{(a,b) | d((a,b),C) < R\}$

• where $C$ is the center of the disc.

• (Why do we have $1_{(r,\theta) \in D}$?)

Thus, $f_X(x) = F_X'(x) = 2 \pi x$

Remarks:

1. cdf of X is area of the circle whose radius is X, which is the distance between the point and the centre of the disc, and is concentric with the disc
2. pdf of X is the circumference of the aforementioned circle
3. The point's coordinates are randomly distributed according the the pdf $f_{(R,\Theta)}(r,\theta) = 1_{(r,\theta) \in D}$
4. Area of a circle is something like the sum of the circumferences of all the circles smaller than it but are concentric with it. Recall that this is the idea of integration (and of course for differentiation and the like): sum of infinite rectangles with infinitesimal length or something like that.
5. The formula $(*)$ actually applies to any value of $x$. For example, if $x = R+1$, then we have

$$P(X \le R+1) = \frac{\int_0^{2 \pi} \int_0^{R+1} 1_{(r,\theta) \in D} \cdot r dr d\theta}{\int_0^{2 \pi} \int_0^R 1_{(r,\theta) \in D} \cdot r dr d\theta}$$

$$=\frac{\int_0^{2 \pi} \int_0^{R} 1_{(r,\theta) \in D} \cdot r dr d\theta + \int_0^{2 \pi} \int_R^{R+1} 1_{(r,\theta) \in D} \cdot r dr d\theta}{\int_0^{2 \pi} \int_0^R 1_{(r,\theta) \in D} \cdot r dr d\theta}$$

$$=\frac{\int_0^{2 \pi} \int_0^{R} 1_{(r,\theta) \in D} \cdot r dr d\theta + 0}{\int_0^{2 \pi} \int_0^R 1_{(r,\theta) \in D} \cdot r dr d\theta} = \frac{\pi R^2}{\pi R^2} = 1$$