# Bounding parameter satisfying a collection of inequalities

I have a set of equations with some inequality constraints that I expect generally does not have a unique solution.

The equations take the form below:

$$\alpha/N+(1-\alpha)x_1=a_1$$ $$\alpha/N+(1-\alpha)x_2=a_2$$ $$\vdots$$ $$\alpha/N+(1-\alpha)x_N=a_N$$ $$x_1+x_2+\dots+x_N=1$$ $$0 $$1>\alpha>0$$

where $$N$$ is fixed and $$a_i$$ are known and satisfy $$a_i>0$$ and $$a_1+\dots a_N=1$$.

If I toy around in Mathematica I get that if I take $$N=3$$ and $$(a_1,a_2,a_3)=(1/2,3/10,1/5)$$ the feasible solutions are given as a family of solutions for $$x_1$$, $$x_2$$, and $$x_3$$ all depending on $$\alpha<3/5$$. I expect in general that one can get some estimated range for $$\alpha$$ that depends on the values of $$a_i$$, computationally for instance with enough time.

I am wondering if there is any structure I can utilize for this set of equations to simplify things, or if there is a simple reduction of this problem to a simpler one.

It seems from the examples that $$\alpha<1-2\min\{a_i\}$$ is the best bound, but I don't see why this is true immediately.

P.S. The motivation for looking at these equations is related to probability, and estimation of a distribution from its marginals.

• For another data point if one takes $N=4$ and $(a_1,a_2,a_3,a_4)=(1/2,1/6,1/6,1/6)$ you get $\alpha<2/3$.
– asd
Aug 26 at 2:35

Let $$n:=N$$ and $$t:=\alpha$$. We have $$0 for all $$i\in[n]:=\{1,\dots.n\}$$ -- or, equivalently, $$t where $$a_{\max}:=\max_{i\in[n]}a_i$$ and $$a_{\min}:=\min_{i\in[n]}a_i$$.
So, $$t_{n,a}$$ is the best bound on $$t$$.
If, as in your example, $$n=3$$ and $$(a_1,a_2,a_3)=(1/2,3/10,1/5)$$, then $$t_{n,a}=3/5$$, as you found. If, as in your comment, $$n=4$$ and $$(a_1,a_2,a_3,a_4)=(1/2,1/6,1/6,1/6)$$, then $$t_{n,a}=2/3$$, as you also found.