Let's derive the pdf for M(n). (I'll speak loosely here - when I say
"at x", I mean "between x and x + dx", where dx is an infinitesimal
and so forth and we handle things properly. I'll also speak of the RVs
having a given value; I mean a particular observation of them. All
this is to save space and typing, because I'm lazy.)
For M(n) to be at x, we need n individual X_i to be below x, n above
it and the middle one "equal" to it. So, using the notation (a, b) = a
choose b, the pdf is going to be
f(x) = (2n+1, n).x^n . (n+1).(1-x)^n . 1
since there are (2n+1, n) ways of selecting the X_i to be below x,
each has a probability x of being below x, (n+1, n) = n+1 ways of
choosing which of the remaining X_i are to be above x, and each of
those has probability (1-x) of being above x; 1 is the pdf of the
final X_i at the point x. Clearly this is symmetrical about x = 1/2.
It's easy to show that it has a maximum at x = 1/2 and minima at x =
0, 1.
From here you should be able to use approximations for large n to put
some bounds on the probability that M(n) will be more than a given
distance from 1/2, and show that this goes to 0 as n goes to infinity. |