UPDATE (1 April): So far, there have been no complete
solutions for the March riddle. The riddle will therefore run one extra month,
but with several modifications and extra hints, as described below.
A new April riddle will be posted, as usual, in parallel to this one.
- Part 1, Part 2 and Part 3 for
*k*=*n* - Parts 1+2
- Part 3
- Part 1 + Part 2 for the case
*k*=1 - Part 1 + Part 2 for the case
*k*=2 - Part 1 + Part 3 for the case
*k*=1 - Part 3, but only
*Z*_{1}and*Z*_{2}are required to be uniformly distributed in [0,1] and independent of each other. - Part 3, but only
*Z*_{1}through*Z*_{k}are required to be uniformly distributed in [0,1] and independent of each other, for the maximum possible*k*(with proof of impossibility for higher*k*)
Hint for Part 2: Consider what the sum of all these variables is.
Credit for the invention of this month's riddle goes to Terry Soo, Ander Holroyd, Oded Schramm, Jim Propp and Omer Angel. They refer to it as "fun with uniform distributions". Answer all parts to appear on the solvers' list. ## Part 1:X,Y and Z are independent random variables, all uniformly
distributed between a and b. However, a and b are
not known. Suggest how to calculate from these a new random variable, uniformly
distributed between 0 and 1.
## Part 2:X_{1}, ..., X_{n} are n random
variables uniformly distributed between 0 and 1, satisfying that their sum is
a constant.
For which values of
As a function of ## Part 3:Suppose thatX_{1}, ..., X_{n} are n
independent random variables uniformly distributed between 0 and 1.
Further, suppose that Y_{1}, ..., Y_{n} are
random variables with the same values as the X_{i}
variables, only sorted from smallest to largest. Not only has
the original order been lost here, the variables Y_{i}
are now neither uniformly distributed nor independent. (For example, despite
the fact that both Y_{1} and Y_{n}
share the same range,
Y_{1} ≤ Y_{n} always holds, so the
two variables have a dependency and their expectations are different, not
both equal 0.5.)
Let
For which values of
As a function of
Note that ( |
## List of solvers:## Partial solvers:## Part 1:Mark Tilford (31 March 06:04)Hongcheng Zhu (6 April 00:17) ## Part 2:## Part 2 for
Mark Tilford (31 March 06:04) |

Elegant solutions can be submitted to the puzzlemaster at __riddlesbrand.scso.com__.
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The solution will be published at the end of the month.

Enjoy!