(3rd Time) Given random positive numbers A, B, if B > A, what is the expected value for A/B?

Question

This question is being posed again because some people have more arguments to make with this one.

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)... = 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.

The answer that i had posted is the definite integral of Tan((π/4)x) between 0 < x < 1, which yields the value 0.441271 approximately.

zanti3 has posted his answer, but his answer presumes that A, B are selected at random from a square array of numbers. Is an infinite array of numbers "square"? That is the real question.

Answer 1

As before, consider the lower "half" of the first quadrant, formed by the lines x=0, y=0, and y=x. Then, as before, the value of A/B is Tan[z], where z is the angle the line from (0,0) to (B,A) makes with the x-axis. This angle is uniformly distributed from 0 to pi/4. Then the integral from 0 to pi/4 of Tan[z]*C, C a constant, should be 1. We find C = 2 / ln(2). The expected value, then, of Tan[z], or A/B, is the integral from 0 to pi/4 of z*Tan[z]*C. I haven't done this integral.

The reason you can't simply integrate z*Tan[z] from 0 to pi/4 is because you must first normalize Tan[z]. This is the same effect as realizing that letting B > A is restricting the original probability to a conditional probability. The integral above, tan(pi/4 * x), also doesn't take into account this need for normalization. It is also not an expected value calculation, and is not the same as letting the angle vary between 0 and pi/4 uniformly.

Good question!

Steve

EDIT - In response to Jeffrey W: Yes, that is an assumption I am making. But I have given a lot of thought to this question, and I now firmly believe that all the answers on this page are correct. Whether it is 1/2, 0, or the integral I gave. Looking at this abstractly, this question does not give explicitly the probability laws governing this space. After much meditation, this problem reminds me most of Bertrand's Paradox: an ambiguous or vague probabilistic universe where different interpretations yield different results. It makes sense, both intuitive and otherwise, for the expected value to be 1/2. But the reasoning for 0 is also sound. What does it mean that A and B are uniformly distributed in an uncountable set? Reducing the problem to one of continuity, as in using the Tan function, is logical, but flawed. So is assuming that A and B are independent in this sample space: does A truly vary uniformly between 0 and B? Is it identical to select A and B simultaneously, casting out erroneous samples, or to first pick B and then limit A? And while the argument for 0 is also logically correct, consider this: what if we were just selecting one variable, say X, uniformly dist. between 0 and infinity. What is the probability X is less than 1000? Zero. Less than 1,000,000? Zero. These intervals are full of uncountable number of elements, and yet have probability zero. It would seem X has probability zero of lying in any interval. This apparent paradox, as well as the one in the original question, is, I believe, because it makes no sense in probability theory to talk of such random variables. I will continue to consider this question, and look forward to other's continued responses.

Answer 2

I'm going to try to answer this question without referring to probability distributions too much, but it will be hard because probability on continuous sets doesn't quite follow the same rules as probability on discrete sets. (But if you're familiar with probability distributions, you'll hopefully be able to make my argument rigorous for yourself.)

I don't think that this question has a meaningful answer, because I don't think the statement "two positive reals are selected at random from 0 to infinity" has any meaningful interpretation. I'll argue why I feel this way at the end of my answer, but for now let's accept that A and B are taken uniformly randomly on [0, ∞), and it turns out that B > A. I'll show that the expected value of A/B is zero.

What is the probability that B lies in the interval [A, 2A]? If we say this probability is p, where p > 0, then by uniformity, the probability that A lies in the twice-as-large interval [A, 3A] will have to be 2p; the probability that A lies in the thrice-as-large interval [A, 4A] will have to be 3p; and so on. No matter how small a positive number I make p (say p = 1/k for some HUGE k), then by uniformity the probability of B lying in [A, (k + 1)A] will be 1. This is absurd; we can't have this probability be 1, because it would mean the probability that B lies in [(k + 1)A, ∞) is 0, and that's hardly uniform.

So the probability that B lies the interval [A, 2A] is zero. So the probability that B > 2A is 1, so the probability A/B < 1/2 is 1.

What is the probability that B lies in the interval [A, 3A]? If we say this probability is p, where p > 0, then by uniformity, the probability that A lies in the twice-as-large interval [A, 5A] will have to be 2p; the probability that A lies in the thrice-as-large interval [A, 7A] will have to be 3p; and so on. No matter how small a positive number I make p (say p = 1/k for some HUGE k), then by uniformity the probability of B lying in [A, (2k + 1)A] will be 1. This is absurd; we can't have this probability be 1, because it would mean the probability that B lies in [(2k + 1)A, ∞) is 0, and that's hardly uniform.

So under the only meaningful interpretation, the probability that B lies the interval [A, 3A] is zero. So the probability that B > 3A is 1, so the probability A/B < 1/3 is 1.

I'm sure you see the pattern--I can prove that the probability B > 2A is 0; I can prove that the probability B > 3A is 0; and, similarly, for any positive N, I can prove that the probability B > NA is 0. So, for any N, the probability A/B < 1/N is 1

Since the probability A/B < 1/N is 1 for any N, then the expected value of A/B is zero. (If it were larger than zero, say the expected value is E, then the probability A/B > E would necessarily have to be greater than 0, but I've shown it's not.)

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Now I'll state why I don't think the statement "Two positive reals are selected at random from 0 to infinity" has any meaningful interpretation.

Let me first consider what would happen on a bounded interval. It's possible to discuss selecting A and B uniformly randomly from [0, 1]--we just say that the probability A lies in [0, 1/2] is the same as the probability it lies in [1/2, 0]; the probabilities that A lies in [0, 1/3], [1/3, 2/3], or [2/3, 1] are the same; and so on. For the selection to be uniformly random, the probabilities that A lies in [0, 1/k], [1/k, 2/k], ..., or [(k-1)/k, 1] should each be 1/k. I can do this; no problems here.

Now let's think about how I would select A from the interval [0, ∞). Fix some positive number n. What is the probability that A lies in the interval [0, n]? If we say this probability is p, where p > 0, then by uniformity, the probability that A lies in the interval [0, 2n] will have to be 2p; the probability that A lies in [0, 3n] will have to be 3p; and so on. No matter how small a positive number I make p (say p = 1/k for some HUGE k), then by uniformity the probability of A lying in [0, kn] will be 1. This is absurd; it's not uniform to say the probability of A lying in [0, kn] is 1 but the probability of A lying in [kn, ∞) is zero.

So under the only meaningful interpretation, the probability that A lies the interval [0, n] is zero, for ANY positive integer n. This means that the probability distribution must be identically zero (except on a set of measure zero). But then the sum (integral) of the entire probability distribution isn't 1, so this isn't really a well-defined probability distribution at all.

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I think I can guess what you were trying to do with the tangent function--map the interval [0, 1] (from which we CAN select numbers uniformly) to [0, ∞). However, this won't work--in order for this to preserve probabilities, your function would have to map, say, all subintervals of length 1/2 in [0,1] to subintervals of the same fixed length in [0, ∞). (It isn't possible to do this--the argument as to why not is similar to things I've said above.)

Answer 3

Wow, watch how fast I can type....


OK, I have a thought on how to look at this question. First, I am going to change the question a little to simplify the analysis, although I don't think the modified question loses any generality or changes the answer. The new question is:

Randomly select two integers between 1 and n. If a ≠ b, let a be the smaller integer and b the larger integer. (If a = b, a/b = 1.) Then, what is the expected value of a/b?

There are n² possible values. They can be arranged in a grid as follows:

1/1 + 1/2 + 1/3 + 1/4 + ... + 1/(n-1) + 1/n

1/2 + 2/2 + 2/3 + 2/4 + ... + 2/(n-1) + 2/n

1/3 + 2/3 + 3/3 + 3/4 + ... + 3/(n-1) + 3/n

1/4 + 2/4 + 3/4 + 4/4 + ... + 4/(n-1) + 4/n

...

1/n + 2/n + 3/n + 4/n + ... + (n-1)/n + n/n


You'll notice along one long diagonal you have 1/1, 2/2, 3/3, etc. up to n/n -- all ones. So, the sum of this long diagonal is 1 * n = n.

Also, you'll notice the numbers on either side of the long diagonal always match -- the grid is symmetrical. Focusing on the upper right piece, we can arrange the numbers as follows:

1/2

1/3 + 2/3

1/4 + 2/4 + 3/4

1/5 + 2/5 + 3/5 + 4/5

...

1/n + 2/n + 3/n + 4/n + ... + (n-1)/n

That accounts for all the number in the upper right part of the grid. The sums of each respective row are 1/2, 1, 3/2, 2, 5/2, 3, etc. There is an obvious pattern, and we can prove each row sums to (n-1)/2 by using the summation formula 1 + 2 + ... + (n-1) = n(n-1)/2.

So, the sum of the upper right part is ½(1 + 2 + 3 + ... + (n-1)) = n(n-1)/4. Since the lower left part is symmetrical with the upper right part, the lower left section also adds up to n(n-1)/4.

Now we can get the sum of the whole grid -- it is n(n-1)/4 + n(n-1)/4 + n = (n² + n)/2. And, finally, the expected value of the two randomly chosen numbers is (n² + n)/2n² = ½ + 1/2n.


Now, using the above result, let's approach this like a calculus question and ask the following: What is the expected value of a/b as n approaches infinity? Well, as n → ∞, (n² + n)/2n² approaches ½. So, I'd say the answer to your question is ½.

Answer 4

SOLUTION - This is a very simple way of looking at this. No matter what B value we choose, the range of A is 0
EXAMINATION OF OTHER SOLUTIONS (which were excellent, by the way)

In response to Stephen Dalton.
"This angle is uniformly distributed from 0 to pi/4. Then the integral from 0 to pi/4 of Tan[z]*C, C a constant, should be 1. We find C = 2 / ln(2). The expected value, then, of Tan[z], or A/B, is the integral from 0 to pi/4 of z*Tan[z]*C. I haven't done this integral."
Doesn't this suggest that the probability of having a certain ratio is proportional to the tangent of the angle? I may be wrong, but I think that's what you said.

What we do want to find is the average of the tangent values, which is NOT tan(pi/8). I really don't know since I'm not familiar enough with polar calculus. I'm thinking the answer would be integral from 0 to pi/4 of tan(x) divided by pi/4, but I have issues with this reasoning, since I'm not very comfortable with doing things this way (but mostly since you don't get 1/2 as an answer). I'm not sure if the densities of polar rays or something skews the results... I dunno. The method actually seems totally legit though...

I also like zanti's approach, especially since it avoids calculus, where when integrating you have to worry about so many constants and crap, AND it got the same answer as mine. :)

Answer 5

properly the respond could surely be between a million > answer > 0 and in case you're searching for for an certainly envisioned fee i could assume it may be .5 or top around the midsection. it truly is because of the fact once you extra up and divided all the possibilites .5 is the mean of a million and 0 and all the opportunities could desire to be in between those 2 numbers. additionally there does not be from now on solutions closer to 0 or closer to a minimum of one. So .5 could be envisioned fee yet you recognize for a actuality the limitations of the respond could be decrease than one and extra desirable than 0.

Answer 6

0>x>1 no matter the values of A and B, if B > A

Answer 7

Intuitively, I believe that Zanti is right. It just seems that if A and B are random, then A/B has to be 1/2. That's what the average of all possibilities would boil down to.