# Drawing numbers on a plane *uniformly*

I am not being very precise here, because I do not know what would be the more precise terminology. I would definitely appreciate any comments on my method and my terminology.

By choosing uniformly from a part of a plane I mean that the probability of a point lying in a particular "sub area" of the whole area of choice should be the ratio of the "areas/ volume/ higher dimensional volume" of the sub area and the whole area.

Suppose I have to draw $x_1,x_2,x_3\geq 0$ uniformly from the plane $a_1 x+a_2 y +a_3 z=1$ with $a_i>0$

Can I draw $x$ from $U(0,\frac 1 {a_1})$, then $y$ from $U(0,\frac 1 {a_2} (1-a_1 x))$ and then $z$ would just be $\frac {1-a_1 x-a_2 y} {a_3}$. Does that sound correct?

As a followup question, if I am planning to choose any point from the sub plane of $a_1 x+a_2 y +a_3 z=1$ where $x>y>z\geq0$, should I draw $x$ from $U(\frac 1 {\sum_i a_1},\frac 1 {a_1})$, then $y$ from $U(\frac 1 {a_2+a_3} (1-a_1 x),\frac 1 {a_2} (1-a_1 x))$ and then $z$ would just be $\frac {1-a_1 x-a_2 y} {a_3}$.

Would the exercises be similar in higher dimensions?

• Hello @JohnColeman, thanks for the response! I thought that once I restrict x,y,z to be non-negative, I do have a finite volume below the plane, what am I missing? – Juanito Jul 4 '16 at 19:26
• "By choosing uniformly from a part of a plane". I called it a part of a plane, but I am glad to call it a bounded triangle if that helps the conversation! – Juanito Jul 4 '16 at 19:31
• You are absolutely correct. I meant $a_i>0$. Thanks. – Juanito Jul 4 '16 at 19:32
• This seems relevant: cs.stackexchange.com/q/3227 . This does as well: mathoverflow.net/q/76255/89084 – John Coleman Jul 4 '16 at 19:33
• This is great. I am guessing that I can calculate $a1 x, a2 y, a3 z$ from their algorithm first, and then back calculate $x,y,z$ . Would that be sufficient? Also they do not comment on what happens when I restrict the area of choice to $x>y>z$. – Juanito Jul 4 '16 at 20:00

The triangle $$T:=\{(x_1,x_2,x_3)\>|\>x_k\geq0,\ a_1x_1+a_2x_2+a_3x_3=1\}$$ carries a natural measure, namely area. This area is related by a constant factor with the natural measure on its orthogonal projection $T'$ onto the $(x_1,x_2)$-plane, given by $$T'=\{x_1,x_2)\>|\>x_k\geq0,\ a_1x_1+a_2x_2\leq1\}\ .$$ We therefore have to choose a uniformly distributed point $(x_1,x_2)\in T'$, and then $x_3$ is given by $x_3={1\over a_3}(1-a_1x_1-a_2x_2)$.
In order to obtain such a point $(x_1,x_2)\in T'$ choose $x_1$ uniformly in $\bigl[0,{1\over a_1}\bigr]$, and independently $x_2$ uniformly in $\bigl[0,{1\over a_2}\bigr]$. If $a_1x_1+a_2x_2>1$ replace $(x_1,x_2)$ by its mirror point $(x_1',x_2'):=\bigl({1\over a_1}-x_1, \>{1\over a_2}-x_2\bigr)$.
Your proposal "Can I draw $x_1$ from $U(0,\frac 1 {a_1})$, then $x_2$ from $U(0,\frac 1 {a_2} (1-a_1 x))$, and then $x_3$ would just be $\frac {1-a_1 x_1-a_2 x_2} {a_3}$" does not lead to a uniform distribution in $T'$.