(2nd Time) Given random positive numbers A, B, if B > A, what is the expected value for A/B?

Question

This question is being reposted because it went into voting before I had a chance to conclude it.

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.

It's very tempting for many people to say, "well, the ratio A/B can be anywhere from 0 to 1, so the expected value must be 1/2". But this PRESUMES a flat distribution. If you're going to make this claim, show why.

chancebeaube, you don't integrate the free parameter θ, you integrate the ratio as a function of θ. That is, integrate Tan(θ). Otherwise, every single distribution problem would end up having an expectation value at halfway of the range of the free parameter.

chancebeaube, the expected value for a distribution which is a function of some parameter from 0 to 1 is the definite integral of that distribution from 0 to 1, when all the values of that parameter is known to occur equally likely. You've pointed out correctly that since (A,B) fills a 2D space with uniform probability, the angle ArcTan(A/B) it makes occurs with equal likelihood. But the ratio A/B is Tan(θ), which is the distribution of the values for A/B for 0 < θ < π/4. The definite integral of Tan(θ) is -Log(Cos(θ)), and for limits 0 < θ < π/4 this comes to a numerical value of 0.346574, which is the answer.

Answer 1

The expected value is the squareroot of 2 minus 1 [or, tangent(pi/8) ].

The trick to finding this answer comes from considering all ordered pairs (x,y) in a Cartesian coordinate plane. Draw the picture as you read along...

Since we require positive values, consider just the first quadrant of the plane where x>0, y>0.

Next, since we require one value to be greater than the other (say, x>y) we will select all points in the first quadrant above the line y=0 (the x-axis) AND below the line y=x.

If you shade in this half-quadrant you will see that every point lies on some ray from the origin with an angle of 'theta' where 0 < theta < pi/2 radians.

In degrees this is the same as saying 0 < theta < 45 degrees.

Now here is the formulation requirement: select any point in our shaded probability space (x,y). Then the ratio x/y is simply the slope of the ray from (0,0) through (x,y); that slope is equal to the tangent of 'theta' as described above.

Now 'theta' can be any measure where 0 < theta < pi/4. The expected value of theta is going to be pi/8 (use an integral) and consequently the expected slope of the ray = expected value of x/y is tangent (pi/8).

Answer 2

Well the answer would definitely be between 1 > Answer > 0 and if you are looking for an actual expected value I would assume it would be .5 or right around the middle. This is because after you added up and divided all the possibilites .5 is the mean of 1 and 0 and all the possibilities must be in between those two numbers. Also there would not be any more answers closer to zero or closer to one. So .5 would be expected value but you know for a fact the boundaries of the answer would be less than one and greater than zero.

Answer 3

Wow, watch how rapidly i will type.... ok, I actually have a theory on the thank you to look at this question. First, i visit exchange the question slightly to simplify the diagnosis, in spite of the actuality that i do no longer think of the changed question loses any generality or variations the respond. the recent question is: Randomly pick 2 integers between one million and n. If a ? b, enable a be the smaller integer and b the bigger integer. (If a = b, a/b = one million.) Then, what's the envisioned fee of a/b? There are n² obtainable values. they may be arranged in a grid as follows: one million/one million + one million/2 + one million/3 + one million/4 + ... + one million/(n-one million) + one million/n one million/2 + 2/2 + 2/3 + 2/4 + ... + 2/(n-one million) + 2/n one million/3 + 2/3 + 3/3 + 3/4 + ... + 3/(n-one million) + 3/n one million/4 + 2/4 + 3/4 + 4/4 + ... + 4/(n-one million) + 4/n ... one million/n + 2/n + 3/n + 4/n + ... + (n-one million)/n + n/n you would be conscious alongside one long diagonal you have one million/one million, 2/2, 3/3, etc. as much as n/n -- all ones. So, the sum of this long diagonal is one million * n = n. additionally, you would be conscious the numbers on the two ingredient of the long diagonal continually experience -- the grid is symmetrical. focusing on the better suited piece, we are able to rearrange the numbers as follows: one million/2 one million/3 + 2/3 one million/4 + 2/4 + 3/4 one million/5 + 2/5 + 3/5 + 4/5 ... one million/n + 2/n + 3/n + 4/n + ... + (n-one million)/n That debts for all the quantity interior the better suited area of the grid. The sums of each and every respective row are one million/2, one million, 3/2, 2, 5/2, 3, etc. there is an obvious development, and we are able to instruct each and every row sums to (n-one million)/2 by ability of utilising the summation formulation one million + 2 + ... + (n-one million) = n(n-one million)/2. So, the sum of the better suited area is ½(one million + 2 + 3 + ... + (n-one million)) = n(n-one million)/4. because of the fact the decrease left area is symmetrical with the better suited area, the decrease left area additionally provides as much as n(n-one million)/4. Now we are able to get the sum of the full grid -- it is n(n-one million)/4 + n(n-one million)/4 + n = (n² + n)/2. And, finally, the envisioned fee of the two randomly chosen numbers is (n² + n)/2n² = ½ + one million/2n. Now, utilising the above consequence, enable's approach this like a calculus question and ask here: what's the envisioned fee of a/b as n procedures infinity? properly, as n ? ?, (n² + n)/2n² procedures ½. So, i could say the respond on your question is ½.

Answer 4

well i hav absolutely no idea for the last few problems but for the first one...
if B is greater than A then you can't divide A by B and get a number bigger than one
EX B=2 A=1
1/2
EX B=8 A=4
4/9
so the answer to the expected value is that A/B < 1

Answer 5

It depends on how you distribute the probability of each number. It isn't possible for every number to have an equal probability. If they did, then the sum of their probabilities would be infinite not 1.

There are ways to randomly choose positive integers. For instance: let the probability of choosing x be 1 / 2^x. Then the sum of the probabillites would be 1.

I don't think it would be possible to do something like this for all positive real numbers though because real numbers are uncountable.

Answer 6

Since +/+=+, all the combinations give a positive quotient.
If B = A, then A/B =1. Since B>A, A/B<1, so, all positive numbers less than 1.
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