I am trying to prove that the random variable:

$Z = 1-X-Y$

where $X\sim U[0,1]$ and $Y\sim U[0, 1-X]$

is $Z \sim U[0,1]$. I'm not sure this is even true, but it feels correct. I'm having trouble with the formalities of how to go about attempting to prove this, especially since $Y$ depends on $X$. Any guidance would be much appreciated.

  • $\begingroup$ $Z$ will not be uniform: take a stick length $1$, spit it uniformly at random along its length, take the right-hand piece and spit it uniformly at random along its length, and take the right-hand piece of that: its length can be anything in $[0,1]$ but its expected length is $\frac14$ $\endgroup$
    – Henry
    Feb 28, 2021 at 12:30

1 Answer 1


$Z$ is not uniform. For $0<t<1$ we have $$P(Z \leq t)$$ $$=P(1-X-Y \leq t)=E(P(Y \geq 1-X-t|X)$$ $$=E\frac {(1-X)-(1-X-t)} {1-X} I_{0\leq 1-X-t \leq 1}+P(X>1-t)$$ $$=E\frac t {1-X} I_{X\leq 1-t}=\int_0^{1-t} \frac t {1-x} dx+t=t-t \ln t$$.


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