# How to find an array of probabilities that give equal products

I'm trying to figure out if there is an algorithm which can solve the following:

Given an array of randomly generated values, for example: $$[54.6, 1.96, 5.0, 31.31]$$

Find an array of equal length, in this example $$[x_1, x_2, x_3, x_4]$$ such that:

$$54.6 \cdot x_1=1.96\cdot x_2=5.0\cdot x_3= 31.31\cdot x_4$$

and

$$x_1 + x_2 + x_3 + x_4 = 1.0$$

Any suggestions for an algorithm to solve this?

• The problem, as you have stated it, is a linear system of equations. Since you mentioned that the $x_i$ are probabilities, then you also want the constraints $x_i\geq0$. So, you have a linear programming problem. Some algorithms that solve this problems are the Simplex, and many interior point methods.
– plop
Feb 23, 2021 at 15:16
• Note that the solutions below assume that the input array consists of non-zero numbers. Remember to also look at the case when some of the values in the input array are zero. In that case all $x_i$ for which the corresponding $a_i$ is non-zero, must be zero themselves. So, you get all the solutions in the convex polyhedron $\sum_{i,a_i=0}x_i=1$, with $x_i\geq0$ and $x_i=0$, for $i$ such that $a_i\neq0$.
– plop
Feb 23, 2021 at 15:56

Let the array be $$a_1,\ldots,a_n$$, which we assume are positive. The condition $$a_i x_i = a_j x_j$$ implies that $$x_i = \frac{a_n}{a_i} x_n.$$ Since the $$x_i$$ must sum to $$1$$, $$1 = \sum_{i=1}^n \frac{a_n}{a_i} x_n \Longrightarrow x_n = \frac{\frac{1}{a_n}}{\sum_{i=1}^n \frac{1}{a_i}}.$$ The formula for $$x_i$$ shows that $$x_j = \frac{\frac{1}{a_j}}{\sum_{i=i}^n \frac{1}{a_i}}.$$
If you consider an array $$[a_1, …, a_n]$$, then an easy way to choose the array $$[x_1, …, x_n]$$ is to define $$x_i = \frac{1}{a_i}$$. That way, $$\forall i, a_i\times x_i = 1$$. The problem is that $$\sum x_i$$ is not necessarily equal to $$1$$, but is instead equal to $$S = \sum \frac{1}{a_i}$$.
If you now consider $$x_i' = \frac{1}{S\times a_i}$$, then we still have $$\forall (i, j), a_i x_i' = a_j x_j'$$, but we also have $$\sum x_i' = \frac{1}{S}\sum \frac{1}{a_i} = 1$$.
Since $$S$$ is easy to compute, you have what you wanted.
I supposed here that all $$a_i > 0$$.