Probability and Statistical Inference. Robert Bartoszynski

Probability and Statistical Inference

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generally by images

. The choice is such that if images

(

has measure images

, then the probability of the chosen point falling into images

is proportional to images

. Identifying images

with the sample space, we can then write images

To better see this, suppose that in shooting at a circular target images , one is certain to score a hit, and that the point where one hits images is assumed to be chosen at random in the way described above. What is the probability that the point of hit is farther from the center than half of the radius of the target?

From Figure 2.1, it is clear that the point of hit must lie somewhere in the shaded annulus images . Its area is images so that images . Of course, the assumption under which this solution is obtained is not very realistic: typically sets closer to the center are more likely to be hit than sets of the same area located closer to the perimeter.

The concept of “random choice” from an uncountable set is sometimes ambiguous. This is illustrated by the next example.

Example 2.2 Bertrand's Paradox

A chord is chosen at random in a circle. What is the probability that the length of the chord will exceed the length of the side of an equilateral triangle inscribed in the circle?

This problem was originally posed in 1888 by Joseph Bertrand, a French mathematician, who provided three solutions, all valid, but yielding inconsistent results.

Solution 1

The chord is uniquely determined by the angle images (see Figure 2.2). These angles are chosen at random from the interval images . It is clear that the length of the chord exceeds the side of the equilateral triangle if images lies between images and images , so the answer to the question is images .

Geometry depicting the first solution of Bertrand's problem. The length of the chord exceeds the side of the equilateral triangle, uniquely determined by the angle a.

Figure 2.2 First solution of Bertrand's problem.

Solution 2

Let us draw a diameter images (see Figure 2.3) perpendicular to the chord images . Then, the length of the chord exceeds the side of the equilateral triangle if it intersects the line images between points images and images . Elementary calculations give images , so the answer is images .

Geometry depicting the second solution of Bertrand's problem. The length of the chord exceeds the side of the equilateral triangle if it intersects the line QQ′ between points B and B′.

Figure 2.3 Second solution of Bertrand's problem.

Solution 3

The location of the chord is uniquely determined by the location of its center (except when the center coincides with the center of the circle, which is an event with probability zero). For the chord to be longer than the side of the equilateral triangle inscribed in the circle, its center must fall somewhere inside the shaded circle in Figure 2.4. Again, by elementary calculations, we obtain probability images .

Geometry of the third solution of Bertrand's problem. The chord is longer than the side of the equilateral triangle inscribed in the circle, its center falling somewhere inside the shaded circle.

Figure 2.4 Third solution of Bertrand's problem.

The discovery of Bertrand's paradox was one of the impulses that made researchers in probability and statistics acutely aware of the need to clarify the foundations of the theory, and ultimately led to the publication of Kolmogorov's book (1933). In the particular instance of the Bertrand “paradoxes,” they are explained simply by the fact that each of the solutions refers to a different sampling scheme: (1) choosing a point on the circumference and then choosing the angle between the chord and the tangent at the point selected, (2) choosing a diameter perpendicular to the chord and then selecting the point of intersection of the chord with this diameter, and (3) choosing a center of the chord. Random choice according to one of these schemes is not equivalent to a random choice according to the other two schemes.

To see why it is so, we will show that the first and second scheme are not equivalent. The analogous arguments for the other two possible pairs of schemes are left as an exercise.

Geometry explaining Bertrand's paradox, depicting that the angle AOB is a, and that a device chooses angles a at random to produce more intersections of the diameter that are farther from the center.

Figure 2.5 Explanation of Bertrand's paradox.

Example 2.3

Imagine different devices

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