# Distribution of uniform RVs under sum constraint

Suppose I generate $$x_1,x_2,x_3,x_4$$ through the following procedure:

1. Sample $$x_1,x_2,x_3 \sim \text{unif}(0, 1)$$, iid
2. While $$x_1+x_2+x_3 > 1$$, resample them all
3. Let $$x_4 = 1 - x_1 - x_2 - x_3$$

What is the distribution of the $$x_1,x_2,x_3,$$ and $$x_4$$ we end up with afterwards? I empirically found that they all seem to follow the same distribution, but can't figure out how to derive this distribution analytically.

• @Xi'an I don't think that's quite true: $x_4$ will be decidedly non-uniform.
– whuber
Nov 26, 2019 at 23:54
• @Xi'an The question, though, concerns the distribution of $X_4$ and whether it's the same as the distribution of the other $X_i.$
– whuber
Nov 27, 2019 at 17:07
• could you explain "I empirically found that they all seem to follow the same distribution". I think it is not valid. Feb 21, 2020 at 14:19

The answer is:that's not true. it is look like

1. generate $$(X_1,X_2,X_3)$$ such that $$S=X_1+X_2+X_3 \leq 1$$
2. $$X_4=(1-S)$$

or on the other hands $$X_4= (1-S|S<1)$$. now we want to find a conditional distribution $$(1-S|S<1)$$.

$$S=X_1+X_2+X_3$$ and according to https://en.wikipedia.org/wiki/Irwin%E2%80%93Hall_distribution

$$\begin{eqnarray} f_S(a)=\left\{ \begin{array}{cc} \frac{1}{2} a^2 & 0\leq a <1 \\ \frac{1}{2}(-2a^2+6a-3) & 1\leq a<2 \\ \frac{1}{2} (a-3)^2 & 2\leq a<3 \end{array} \right. \end{eqnarray}$$

so $$P(S\leq 1)=\int_{0}^{1} \frac{1}{2} a^2 da=\frac{1}{6}$$

and

$$P( 1-x \leq S\leq 1)=\int_{1-x}^{1} \frac{1}{2} a^2 da$$ $$=\frac{1}{6} a^3|_{1-x}^{1}=\frac{1}{6}(1-(1-x)^3)$$

so $$F_{X_4}(x)=P(X_4\leq x)=P(1-S\leq x|S<1)=P(S\geq 1-x|S\leq 1)$$

$$=\frac{P( 1-x \leq S\leq 1)}{P(S\leq 1)}=1-(1-x)^3$$ $$,x<1$$

its not look like uniform distribution and $$f_{X_4}(x)=3(1-x)^2$$ .($$beta(1,3)$$ distribution)

a simple simulation confirms that the $$X_4$$ does not follow uniform distribution:

x4<-c()
count<-1
simu<-6*1000
for(i in 1:simu){
x<-runif(3)
s<-sum(x)
if(s<1) {x4[count]<-1-s;count<-count+1}
}
plot(density(x4),type="l",xlim=c(0,1))
> ks.test(x4,runif(length(x4)))

Two-sample Kolmogorov-Smirnov test

data:  x4 and runif(length(x4))
D = 0.37026, p-value < 2.2e-16
alternative hypothesis: two-sided


it's simple to check $$1-S|S>1$$ does not have uniform distribution.

• That's a nice proof that x4 doesn't follow the uniform distribution. Can you derive the distributions of x1, x2, and x3 though? Dec 2, 2019 at 17:54
• X_1,...,X_3 follow uniform distribution according to they are generated from U(0,1).(look at step 1) Dec 3, 2019 at 20:34
• It's simple that when $X_1$ generate from uniform distribution,so the distribution of $X_1$ obviously is uniform, because you just created it. but you created $X_4$ from $X_1,...,X_3$. so you just does not know the distribution of it, so based on $X_1,...,X_3$ you can extract the distribution of $X4$. Dec 28, 2019 at 19:04