Examples

Event probability - Meeting in the park

A man and a woman arrange to meet in the park between 12:00 and 1:00pm. They both arrive at a random time with uniform distribution over that hour, and do not coordinate with each other.

Find the probability that the first person to arrive has to wait longer than 15 minutes for the second person to arrive.

Solution

Let denote the time the man arrives. Use minutes starting from 12:00, so . Let denote the time the woman arrives, using the same interval.

The probability we seek is:

Because and are symmetrical in probability, these terms have the same value, so we just double the first one for our answer.

Since the arrivals are independent of each other, we have .

Since each arrival time is uniform over the interval, we have:

Therefore the joint density is . Calculate:

Uniform disk: Cartesian vs. polar

Suppose that a point is chosen uniformly at random on the unit disk.

(a) Let and be the Cartesian coordinates of the chosen point. Are and independent?

(b) Let and give the polar coordinates of the chosen point. Are and independent?

Solution

(a)

Write for the joint distribution of and . We have:

Then computing , we obtain:

By similar reasoning, for .

The product is not equal to , so and are not independent. Information about the value of does provide constraints on the possible values of , so this result makes sense.

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(b)

To find the marginals and , the standard method is to integrate the density in the opposite variables.

varies!

The probability density is not constant! The area of a differential sector depends on .

We can take two approaches to finding the density :

(i) Area of a differential sector divided by total area:

So the density is .

(ii) Via the CDF:

Joint PDF from joint CDF: uniform distribution in polar

The region ‘below’ a given point , in polar coordinates, is a sector with area . The factor is a percentage of the circle with area .

The density is a constant across the disk, so the CDF at is this same area times . Thus:

Then in polar coordinates the density is given by taking partial derivatives:

Once we have , integrate to get the marginals:

Check independence:

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In this problem it is feasible to find the marginals directly, without computing the new density, only using some geometric reasoning.

Extra - Infinitesimal method for the marginals

The probability is the area (over ) of a thickened circle with radius and thickness . The circumference of a circle at radius is . So the area of the thickened circle is . So the probability is . This tells us that the marginal probability density is .

The probability is the area (over ) of a thin sector with radius 1 and angle . This area is . So the probability is . This tells us that the marginal probability density is .

These results agree with those of the ‘calculus’ approach above!