Probability and Statistical Inference. Robert Bartoszynski

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СКАЧАТЬ href="#ulink_ef2225cf-7d2b-5da5-965a-283ee19fb1ca">Figure 2.1 Hitting a target.

      Example 2.1 Geometric Probability

      One of the first examples of an uncountable sample space is associated with “the random choice of a point from a set.” This phrase is usually taken to mean the following: a point is selected at random from a certain set in a finite‐dimensional space (line, plane, etc.), where has finite measure (length, area, etc.), denoted generally by . The choice is such that if ( has measure , then the probability of the chosen point falling into is proportional to . Identifying with the sample space, we can then write .

      To better see this, suppose that in shooting at a circular target , one is certain to score a hit, and that the point where one hits 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 . Its area is so that . 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 (see Figure 2.2). These angles are chosen at random from the interval . It is clear that the length of the chord exceeds the side of the equilateral triangle if lies between and , so the answer to the question is .

Figure 2.2 First solution of Bertrand's problem.

      Solution 2

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

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 .

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 СКАЧАТЬ