Permutation algorithm to simulate $X$, $Y$, $Z$ uniform on $(0,1)$ with $X+Y+Z = 1$

Question

Permutation algorithm to simulate $X$, $Y$, $Z$ uniform on $(0,1)$ with $X+Y+Z = 1$

Edit:

Sorry, I tend to jump back and forth between math notation and computer science notation....often to the chagrin of my more rigorous colleagues (and Math.SE folks ;-)

Also, I accidentally copied a formula, so the problem was overspecified. The last variable is determined by the previous two.

And yes $\pi_i$ is the send element of the random permutation group $\pi$.

I hope this clears things up.

---

Is it possible to satisfy the title condition if I use the following algorithm?

Let $V:=(X,Y,Z)$ be a vector of unspecified variables.

1. Generate a random permutation of $\{1,2,3\}$, call this $\pi$
2. Let $V_{\pi_1}\sim U(0,1),V_{\pi_2}\sim U(0,1-V_{\pi_1}),V_{\pi_3}=1-V_{\pi_1}-V_{\pi_3}$
3. Go back to step 1

I'm hoping that the pdf of $V$ will have standard uniform marginal. It seems like the permutation operation ensures that the components of $V$ have standard uniform marginal over many iterations, since I am uniformly sampling over the permutations of the components.

Is there a subtle flaw to this? I can't seem to find one.

Answer 1

The conditions $X+Y+Z=1$ and $X$, $Y$, $Z$ uniform on $(0,1)$, are incompatible since the first one implies $E(X)+E(Y)+E(Z)=1$ and the second one implies $E(X)=E(Y)=E(Z)=\frac12$.