Given random positive numbers A, B, if B > A, what is the expected value for A/B?

Question

Two positive reals are selected at random from 0 to infinity. Let A be the smaller of the two, B the larger. What's the expected value for A/B?

Given a set of values x, with each value having a probability p of occurring, such that ∑p = 1. Then the expected value is ∑xp. For example, the expected value for a typical thrown 6-sided dice is

(1)(1/6)+(2)(1/6)+(3)(1/6)+(4)(1/6)+(5)(1/6)+(6)(1/6) = 7/2. = 3.5

It doesn't work to first work this out for 0 < A,B < N for some finite N, and then let N -> ∞, because that introduces a sampling bias.

Yes, it can be, floodtl. It can be used to help solve my OTHER posted problem about obtuse triangles. And there are problems in statistical physics where this would be a help.

Gypsy doctor, this is the same as first assuming 0 < A,B < N, for some finite N.

tedfisher17, a more practical variation of this problem is to first suppose a random scattering of points in space (ideal gas). Pick any two distances between them. What's the expected ratio?

mathsmanretired, you have the right idea, but A, B can have any value. You THEN choose the smaller of the two as the numerator. In other words, there is no restriction on the choice of (A,B). Your bisector idea suggests that there is.

quadrillerator, if you read my other posted question about obtuse triangles, the question of the probability of 3 random points forming an obtuse triangle critically depends on avoiding introducing bias in your probability calculations. Problems like this can actually be approximated by computerized Monte Carlo methods, so, for example, if you are going to use an ideal gas as a model, how do you determine this expected value for the ratio if the ideal gas is in interstellar space? I.E., virtually boundless?

Several have suggested that the ratio should have a "flat" distribution from 0 to 1. If you ASSUME that, then the expectation value is 1/2. However, it doesn't necessarily follow from (A,B) being selected at random, where 0 < A,B < infinity.

Answer 1

0.5

If you select B from 0 to infinity, there is a limited set of equally probably values of A, from 0 to B. The average of these values is B/2. So the expected value would be (B/2) / B = (B/(2*B)) = 1/2 or .5

Answer 2

Here are my thoughts but no guarantee of correctness.

Let A be put along the x axis and B along the y axis of a coordinate first quadrant. The allowed values will be between y = x and the y axis.
If there were an upper limit say N then this area would be bounded by the line y = N at the top.
Now here's the doubtful bit. Draw a line through the origin dividing this area into equal halves. Each area has half the total probability and will be independent of N.
Could the expected value be B/A anywhere on this line?
Since this line to give equal areas must be the angle bisector it will be 67.5 degrees from the x axis.
Therefore B/A = tan 67.5

Edit. Surely if A > B then you have the area between y = x and the x axis and the same reasoning applies. An upper limit will correspond to x = N and the triangular area contains points all with equal probability so divide the area into two.
I still think that expected value (larger/smaller) = 1 + sqrt2

Further edit. I've just used a random number generator to find 50 pairs of numbers up to 9999 and I got the average ratio to be 4.91 so now I'm not so confident of my result.
But! The median value was 2.15. I now think that is what my method gave; the median not the mean value.

Answer 3

I maintain my objection to an infinite probability distribution:

It is impossible to have a uniform probability distribution over an infinite domain.

But....

Choose A first. Now pick some finite value M. What is the probability that B will fall within M*A of the origin? Given the infinite domain, that probability ought to be zero. Thus the probability that B will fall *outside* that domain is 1, and the expected value for the ratio must be at least M.

Since this is true for any value of M, the expected value of the ratio is infinite. I don't stand by this answer, however, since I still believe the initial premise of an infinite distribution is either poorly specified or flatly impossible (depending on your interpretation).

Answer 4

A/B = N

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