# How to get solutions for this system of inequalities?

I am new to Mathematica and I am struggling finding the solution for the following problem. I am studying this piecewise function:

Piecewise[{{pdf1[x]/pdf[x] (h + (1 - h) 2 cdf1[x])/(
h + (1 - h) 2 cdf[x]),
x < a}, {pdf1[x]/pdf[x] (h + (1 - h) 2 (1 - cdf1[x] + cdf1[a]))/(
h + (1 - h) 2 (1 - cdf[x] + cdf[a])),
a <= x < b}, {pdf1[x]/pdf[x] (
h + (1 - h) 2 (cdf1[x] - cdf1[b] + cdf1[a]))/(
h + (1 - h) 2 (cdf[x] - cdf[b] + cdf[a])),
b <= x < c}, {pdf1[x]/pdf[x] (
h + (1 - h) 2 (1 - cdf1[x] + cdf1[c] - cdf1[b] + cdf1[a]))/(
h + (1 - h) 2 (1 - cdf[x] + cdf[c] - cdf[b] + cdf[a])), x >= c}}]


where

pdf1[x_] = PDF[NormalDistribution[1, 1], x];
pdf[x_] = PDF[NormalDistribution[0, 1], x];
cdf1[x_] = CDF[NormalDistribution[1, 1], x];
cdf[x_] = CDF[NormalDistribution[0, 1], x];


I would like the first and the third piece of the piecewise function to be smaller than a certain value $$l$$, say $$l=0.48$$; and the second and fourth piece of the function to be greater than the same $$l$$.

I managed to represent graphically the piecewise function with the following code:

Manipulate[Plot[{Piecewise[{{pdf1[x]/pdf[x] (h + (1 - h) 2 cdf1[x])/(
h + (1 - h) 2 cdf[x]),
x < a}, {pdf1[x]/pdf[x] (
h + (1 - h) 2 (1 - cdf1[x] + cdf1[a]))/(
h + (1 - h) 2 (1 - cdf[x] + cdf[a])),
a <= x < b}, {pdf1[x]/pdf[x] (
h + (1 - h) 2 (cdf1[x] - cdf1[b] + cdf1[a]))/(
h + (1 - h) 2 (cdf[x] - cdf[b] + cdf[a])),
b <= x < c}, {pdf1[x]/pdf[x] (
h + (1 - h) 2 (1 - cdf1[x] + cdf1[c] - cdf1[b] + cdf1[a]))/(
h + (1 - h) 2 (1 - cdf[x] + cdf[c] - cdf[b] + cdf[a])),
x >= c}}], l}, {x, -5, 5}, PlotRange -> 3], {a, -10, 10,
Appearance -> "Labeled"}, {b, a, 5, Appearance -> "Labeled"}, {c, b,
5, Appearance -> "Labeled"}, {h, 0, 1,
Appearance -> "Labeled"}, {l, 0, 1, Appearance -> "Labeled"}]


The picture is an example of a solution where, for $$h=0$$, I set $$a=0.05$$, $$b=0.27$$ and $$c=0.41$$ (the yellow line corresponds to $$l=0.48$$) What I am interested in is finding for each $$h$$ the set of combinations of $$a$$, $$b$$ and $$c$$ such that the system of inequalities is satisfied and I need to give a graphical representation of the set of solution somehow. I have tried something also by defining a domain for $$a$$, $$b$$ and $$c$$ such that $$-2, but I feel my skills with Mathematica are too poor to handle the task. Any piece of advice would be highly appreciated. Thank you.

• Given your description of the problem, {a, b, c} are not unique. For instance, for the parameters in the question, a could just as well be 0. Please describe more carefully what you are seeking. By the way,, the last block of code in the question does not execute as written. – bbgodfrey Sep 27 '18 at 1:21
• Thank you for your answer, I edited the last part of the code and now it should work: the image represented is obtained using the values of the parameters I reported in the post and $h=0$. As you correctly noticed ${a, b, c}$ are not unique: I would like to know, given a certain $h$, all the possible ${a, b, c}$ such that the four inequalities are satisfied, this is what I meant when I said 'the set of combination of $a$, $b$ and $c$. Thank you again – Api Sep 27 '18 at 21:37

The following seems to work, but I have not tested it exhaustively. Define the four curves in the Piecewise function as

Clear[a, b, c, f1, f2, f3, f4, f]; h = 0; l = .48; SeedRandom;
f1[x_] := pdf1[x]/pdf[x] (h + (1 - h) 2 cdf1[x])/(h + (1 - h) 2 cdf[x])
f2[x_] := pdf1[x]/pdf[x] (h + (1 - h) 2 (1 - cdf1[x] + cdf1[a]))/
(h + (1 - h) 2 (1 - cdf[x] + cdf[a]))
f3[x_] := pdf1[x]/pdf[x] (h + (1 - h) 2 (cdf1[x] - cdf1[b] + cdf1[a]))/
(h + (1 - h) 2 (cdf[x] - cdf[b] + cdf[a]))
f4[x_] := pdf1[x]/pdf[x] (h + (1 - h) 2 (1 - cdf1[x] + cdf1[c] - cdf1[b] +
cdf1[a]))/(h + (1 - h) 2 (1 - cdf[x] + cdf[c] - cdf[b] + cdf[a]))


Then a can be no smaller that the value of x for which f2 is equal to l, and no larger than the value of x for which f1 is equal to l. Compute these points and place a randomly in that range.

{amin = a /. FindRoot[f2[a] == l, {a, -1/2}], amax = a /. FindRoot[f1[a] == l, {a, 1/2}],
a = RandomReal[{amin, amax}]}
(* {-0.233969, 0.544866, 0.337909} *)


Next, b can be no smaller than a and no larger than the value of x for which f3 is equal to l. Compute these points and place b randomly in that range.

{bmin = a, bmax = b /. FindRoot[f3[b] == l, {b, 1/2}], b = RandomReal[{bmin, bmax}]}
(* {0.337909, 0.678207, 0.432968} *)


Finally, c is calculated in a similar manner.

{cmin = b, cmax = c /. FindRoot[f3[c] == l, {c, 1/2}], c = RandomReal[{cmin, cmax}]}
(* {0.432968, 0.574512, 0.465049} *)

f[x_] := Piecewise[{{f1[x], x < a}, {f2[x], a <= x < b},
{f3[x], b <= x < c}, {f4[x], x >= c}}]
Plot[{f[x], l}, {x, -1, 1}, PlotRange -> All, PlotPoints -> 1000,
ImageSize -> Large, AxesLabel -> {x, f}, LabelStyle -> {Bold, Black, Medium}] The need for reasonable guesses when computing amin, etc. with FindRoot can be eliminated by using NArgMin and NArgMax as follows. (Subsequently edited by including b < c in the computation of cmax to avoid a singularity in f3 for some parameters.)

{amin = NArgMin[{f2[a], f2[a] > l}, a], amax = NArgMax[{f1[a], f1[a] < l}, a],
a = RandomReal[{amin, amax}]}
{bmin = a, bmax = NArgMax[{f3[b], f3[b] < l}, b], b = RandomReal[{bmin, bmax}]}
{cmin = b, cmax = NArgMax[{f3[c], f3[c] < l, b < c}, c], c = RandomReal[{cmin, cmax}]}


Incidentally, I have obtained results similar to those above with h = 0.1 and h = 0.5, the latter shown below. • Thank you for your answer, it has been very helpful in proceeding with my work! I am stuck right now on another issue: I would like to represent on a 3d graph the set of the combination of ${a,b,c)$ that satisfies the four inequalities. Would you be able to provide any piece of advice? everything I have tried does not seem to work. Thank you again – Api Oct 4 '18 at 17:48
• I apologize that I had not accepted your answer until now. I have also posted a new question about the 3d graph. Thank you very much. – Api Oct 4 '18 at 23:03