# How to generate two groups of $n$ random numbers in $U(0,1)$ such that sum of these two groups equal?

I want to have two groups of $n$ random numbers $u_i$ and $v_i$ in $U(0,1)$, such that $\sum u_i = \sum v_i$

What I tried is:

I can firstly get $u_i$ by $U\sim U(0,1)$, make $s=\sum u_i$.

Then I found it is very difficult to generate another $n$ uniformly distributed random numbers $v_i$ from $U(0,1)$ that sum to $s$, where $s$ can be any real value in $[0,n]$

Try to make the question clearer, my original problem is:

The random variables, of course, is not independent! But my question requires the sampled values for each parameter roughly distributed in range [0,1] such that the Monte Carlo sampling will effectively go over the whole parameter space for the system.

I have $8$ parameters $\kappa_i, i=1,\ldots,8$ from a system, each parameter $\kappa_i$ can be any value (better be uniformly distributed) in $[0,1]$. But I have a constraint on my parameters which is $\kappa_1+\kappa_2+\kappa_3+\kappa_4=\kappa_5+\kappa_6+\kappa_7+\kappa_8$. Now I want to sample the whole parameter space (is this counted as Monte Carlo?) with such constraint. What should I do?

• You want samples of numbers $\kappa_i$ such that $\sum_{i=1}^4 \kappa_i = \sum_{i=5}^8 \kappa_i$. It sounds like you want the marginal distribution of each $\kappa_i$ to be uniform on $[0, 1]$. But what conditions do you want to impose on their joint distribution? – Adrian Feb 6 '16 at 15:50
• There appears to be an interesting and important question here, but it is asked in a very confusing way. Please edit it to address the comments. – whuber Feb 6 '16 at 16:14
• I think it is crucial what kind of dependence you allow for. Your constraint implies a dependence structure, so the four samples $\kappa_i: 1 \leq i \leq 8$ cannot be independent from each other. (To see this, note that if all $\kappa_i$ are random and independent, the probability of your constraint holding is equal to $0$.) 'How independent' do they have to be? Which dependence structure do you allow for? – Jeremias K Feb 6 '16 at 18:10
• It's on hold so I can't post this as a solution but why not first sample $\kappa_1,\kappa_2,\kappa_3,\kappa_5,\kappa_6,\kappa_7\sim U(0,1)$. Then simulate one more $\kappa_i\sim U(0,1)$ and add it to whichever set $A=\{\kappa_1,\kappa_2,\kappa_3\}$ or $B=\{\kappa_5,\kappa_6,\kappa_7\}$ makes it such that $A>B$ or $B>A$. So either $i=4$ and $\kappa_4$ gets added to $A$ or $i=8$ and $\kappa_8$ gets added to $B$. Then, for whichever $\kappa$ is left, set the last $\kappa_8=\kappa_1+\kappa_2+\kappa_3+\kappa_4-(\kappa_5+\kappa_6+\kappa_7)$ or (ctd)... – RustyStatistician Feb 6 '16 at 19:54
• The problem of partitioning a set into two equal-in-sum (/as equal as possible) subsets is the partition problem, but it's not quite clear to me how your conditions on the distribution are supposed to work. – Glen_b -Reinstate Monica Feb 7 '16 at 10:30

Here is a procedure to generate two $n$tuples of random variables $(u_k)$ and $(v_k)$, both i.i.d. uniform on $(0,1)$, with $u_1+\cdots u_n=v_1+\cdots+v_n$.

For every $k$, let $f_k$ denote the PDF of the sum of $k$ i.i.d. random variables uniformly distributed on $(0,1)$. Generate $n$ i.i.d. random variables $u_i$ uniform on $(0,1)$, and consider $s_n=u_1+\cdots+u_n$. Then generate $v_n$ distributed as $u_n$ conditionally on $u_1+\cdots+u_n=s_n$, that is, with density $$f_n(s_n)^{-1}\cdot f_{n-1}(s_n-x)\cdot\mathbf 1_{(0,1)}(x).$$ Next, define $s_{n-1}=s_n-v_n$ and generate $v_{n-1}$ distributed as $u_{n-1}$ conditionally on $u_1+\cdots+u_{n-1}=s_{n-1}$, that is, with density $$f_{n-1}(s_{n-1})^{-1}\cdot f_{n-2}(s_{n-1}-x)\cdot\mathbf 1_{(0,1)}(x),$$ and so on, until $v_2$ distributed as $u_2$ conditionally on $u_1+u_2=s_2$, that is, with density $$f_2(s_2)^{-1}\cdot f_{1}(s_2-x)\cdot\mathbf 1_{(0,1)}(x)=f_2(s_2)^{-1}\cdot \mathbf 1_{(s_2-1,s_2)}(x)\cdot\mathbf 1_{(0,1)}(x).$$ Finally define $v_1=s_2-v_2$. Then $(v_k)$ is i.i.d. uniform on $(0,1)$ and $u_1+\cdots u_n=v_1+\cdots+v_n$ almost surely.

To sum up, the procedure is exact but it requires to compute $n-1$ PDFs, each PDF $f_k$ being a polynomial of degree $k-1$ on each interval $(i-1,i)$ with $1\leqslant i\leqslant k$.

• Can you elaborate on the notation for the $f_n(s_n)$ and $f_{n-1}(s_{n-1})^{-1}$, and / or in general computing the pdf $f_k$? – Antoni Parellada Feb 14 '16 at 21:19
• @AntoniParellada The factor is $$\frac1{f_{n-1}(s_{n-1})}.$$ – Did Feb 16 '16 at 5:39

I have 8 parameters $κ_i,i=1,…,8$ from a system, each parameter $κ_i$ can be any value (better be uniformly distributed) in $[0,1]$. But I have a constraint on my parameters which is $κ_1+κ_2+κ_3+κ_4=κ_5+κ_6+κ_7+κ_8$. Now I want to sample the whole parameter space (is this counted as Monte Carlo?) with such constraint. What should I do?

• There is actually a neat little exam question for introductory courses on theoretical probability here. The text of the question could read as follows. Consider seven independent random variables $T$, $U$, $V$, $W$, $X$, $Y$, $Z$, and compute the distribution of $S=(U+V+W+T)-(X+Y+Z)$ conditionally on the event $[0<S<1]$. Check that this distribution is not the uniform distribution on $(0,1)$. Hint: Show that $S+3$ is (unconditionally) distributed like $U+V+W+T+X+Y+Z$. // @OP This page might interest you. – Did Mar 7 '16 at 18:10