Given sum of uniform random variables on $[0,1]$, $Z_1 + Z_2 + \dots + Z_n = 1$, what is the probability that exactly k random variables are at least $\frac{1}{n}$?

In other words, what's

$Pr[\text{exactly k random variables are at least }\frac{1}{n} \mid Z_1 + Z_2 + \ldots + Z_n = 1]$

where $Z_i$s are uniform random variables on $[0,1]$?


  • $\begingroup$ How are $Z_i$ sampled in a way that they are uniform random variables and they satisfy that sum criteria? The sum criteria makes them non uniform. $\endgroup$ – Hugh May 30 '16 at 10:45
  • 1
    $\begingroup$ @Hugh I think it is conditional probability. $\endgroup$ – Element118 May 30 '16 at 10:49
  • $\begingroup$ The posts means to say something like $P(\text{at least $k$ $Z_i$ are at least $1/n$}\mid Z_1+\dotsb+Z_n = 1)$ $\endgroup$ – Em. May 30 '16 at 10:50

The probability for $j$ particular variables to be at least $\frac1n$ is

$$ \left(1-\frac jn\right)^{n-1}\;, $$

so by inclusion-exclusion the probability for exactly $k$ variables to be at least $\frac1n$ is

$$ \sum_{j=k}^n(-1)^{j-k}\binom nj\binom jk\left(1-\frac jn\right)^{n-1}\;. $$

I'm not aware of any way to simplify this.


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