a***r
发帖数: 146 | 1 Given two i.i.d uniform random points x and y on the interval [0,1] , what
is the average size of the smallest of the three resulting intervals? | j*****4
发帖数: 292 | 2 1/9
【在 a***r 的大作中提到】
: Given two i.i.d uniform random points x and y on the interval [0,1] , what
: is the average size of the smallest of the three resulting intervals?
| l*****a
发帖数: 559 | 3 verified by simulation.
【在 j*****4 的大作中提到】
: 1/9
| I**A
发帖数: 2345 | 4 怎么算的？
【在 j*****4 的大作中提到】
: 1/9
| j*****4
发帖数: 292 | 5 Let x,y,z be the length of each interval.
So we have x+y+z=1 (0
(see attachment).If we just want to know the average of each interval,that
is the centroid of x(or y,z)(1/3, according to middle school knowledge).
The average(expectation) of the smallest interval will be x-coordinate of
the centroid of shadowed small triangle.(x
That is 1/3*1/3.
【在 I**A 的大作中提到】
: 怎么算的？
| I**A
发帖数: 2345 | 6 thanks, though i still didn't get it..
【在 j*****4 的大作中提到】
: Let x,y,z be the length of each interval.
: So we have x+y+z=1 (0: (see attachment).If we just want to know the average of each interval,that
: is the centroid of x(or y,z)(1/3, according to middle school knowledge).
: The average(expectation) of the smallest interval will be x-coordinate of
: the centroid of shadowed small triangle.(x: That is 1/3*1/3.
| c**********e
发帖数: 2007 | 7 Your answer is correct. It is 1/9. I verified by integration.
A solution with 2-D integration is easier for people to understand.
It is about [0,1]*[0,1]. Look at the upper left triangle xy
part can be done by symmetricty.)
It can be divided by 3 triangular parts, centered at (1/3,2/3).
In each part, the expectation of the smallest of the 3 interval
lengths min{x, y-x, 1-y} is 1/9. So is the average on the upper left
triangle x
【在 j*****4 的大作中提到】
: Let x,y,z be the length of each interval.
: So we have x+y+z=1 (0: (see attachment).If we just want to know the average of each interval,that
: is the centroid of x(or y,z)(1/3, according to middle school knowledge).
: The average(expectation) of the smallest interval will be x-coordinate of
: the centroid of shadowed small triangle.(x: That is 1/3*1/3.
| v********w
发帖数: 136 | 8 awesome
joson's sol is coll too, but the shaded area is not correct
【在 c**********e 的大作中提到】
: Your answer is correct. It is 1/9. I verified by integration.
: A solution with 2-D integration is easier for people to understand.
: It is about [0,1]*[0,1]. Look at the upper left triangle xy
: part can be done by symmetricty.)
: It can be divided by 3 triangular parts, centered at (1/3,2/3).
: In each part, the expectation of the smallest of the 3 interval
: lengths min{x, y-x, 1-y} is 1/9. So is the average on the upper left
: triangle x:
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