How to generate repetitive graphs?












2












$begingroup$


I need to create several graph for different values of a parameter $h$.



The code is the following



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]
h = 0; l = .48; SeedRandom[1900];
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]))
amin = NArgMin[{f2[a], f2[a] > l}, a], amax = NArgMax[{f1[a], f1[a] < l}, a]

btab = Interpolation@Table[{a, NArgMax[{f3[b], f3[b] < l}, b]}, {a, amin, amax, .02}];

ctab = Interpolation[Flatten[Table[{{a, b},
Quiet@NArgMax[{f3[c], f3[c] < l, b < c}, c]}, {a, amin, amax, .02}, {b, a, btab[a], .02}], 1], InterpolationOrder -> 1];

Quiet@RegionPlot3D[amin < a < amax && a < b < btab[a] && b < c < ctab[a, b], {a, amin, amax}, {b, amin, btab[amin]}, {c, amin, ctab[amin, btab[amin]]}, AxesLabel -> {a, b, c},
LabelStyle -> {Black, Bold, Medium}, BoxRatios -> Automatic, ImageSize -> Large, PlotPoints -> 500, Mesh -> None]


I would like to generate several graphs for different values of $h$, say for $h$ between 0 and 1, with $dh$=0.01.
How can I do this?










share|improve this question











$endgroup$












  • $begingroup$
    h = Range[0, 1, 0.01] will give you a list of h. Then you can either redefince h as a variable in the functions it is used or you can use a loop (Do, Table) to evaluate along h and produce a list of plots. Map and MapAt can also work but the syntax is slightly trickier.
    $endgroup$
    – Titus
    Dec 8 '18 at 15:23
















2












$begingroup$


I need to create several graph for different values of a parameter $h$.



The code is the following



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]
h = 0; l = .48; SeedRandom[1900];
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]))
amin = NArgMin[{f2[a], f2[a] > l}, a], amax = NArgMax[{f1[a], f1[a] < l}, a]

btab = Interpolation@Table[{a, NArgMax[{f3[b], f3[b] < l}, b]}, {a, amin, amax, .02}];

ctab = Interpolation[Flatten[Table[{{a, b},
Quiet@NArgMax[{f3[c], f3[c] < l, b < c}, c]}, {a, amin, amax, .02}, {b, a, btab[a], .02}], 1], InterpolationOrder -> 1];

Quiet@RegionPlot3D[amin < a < amax && a < b < btab[a] && b < c < ctab[a, b], {a, amin, amax}, {b, amin, btab[amin]}, {c, amin, ctab[amin, btab[amin]]}, AxesLabel -> {a, b, c},
LabelStyle -> {Black, Bold, Medium}, BoxRatios -> Automatic, ImageSize -> Large, PlotPoints -> 500, Mesh -> None]


I would like to generate several graphs for different values of $h$, say for $h$ between 0 and 1, with $dh$=0.01.
How can I do this?










share|improve this question











$endgroup$












  • $begingroup$
    h = Range[0, 1, 0.01] will give you a list of h. Then you can either redefince h as a variable in the functions it is used or you can use a loop (Do, Table) to evaluate along h and produce a list of plots. Map and MapAt can also work but the syntax is slightly trickier.
    $endgroup$
    – Titus
    Dec 8 '18 at 15:23














2












2








2


1



$begingroup$


I need to create several graph for different values of a parameter $h$.



The code is the following



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]
h = 0; l = .48; SeedRandom[1900];
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]))
amin = NArgMin[{f2[a], f2[a] > l}, a], amax = NArgMax[{f1[a], f1[a] < l}, a]

btab = Interpolation@Table[{a, NArgMax[{f3[b], f3[b] < l}, b]}, {a, amin, amax, .02}];

ctab = Interpolation[Flatten[Table[{{a, b},
Quiet@NArgMax[{f3[c], f3[c] < l, b < c}, c]}, {a, amin, amax, .02}, {b, a, btab[a], .02}], 1], InterpolationOrder -> 1];

Quiet@RegionPlot3D[amin < a < amax && a < b < btab[a] && b < c < ctab[a, b], {a, amin, amax}, {b, amin, btab[amin]}, {c, amin, ctab[amin, btab[amin]]}, AxesLabel -> {a, b, c},
LabelStyle -> {Black, Bold, Medium}, BoxRatios -> Automatic, ImageSize -> Large, PlotPoints -> 500, Mesh -> None]


I would like to generate several graphs for different values of $h$, say for $h$ between 0 and 1, with $dh$=0.01.
How can I do this?










share|improve this question











$endgroup$




I need to create several graph for different values of a parameter $h$.



The code is the following



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]
h = 0; l = .48; SeedRandom[1900];
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]))
amin = NArgMin[{f2[a], f2[a] > l}, a], amax = NArgMax[{f1[a], f1[a] < l}, a]

btab = Interpolation@Table[{a, NArgMax[{f3[b], f3[b] < l}, b]}, {a, amin, amax, .02}];

ctab = Interpolation[Flatten[Table[{{a, b},
Quiet@NArgMax[{f3[c], f3[c] < l, b < c}, c]}, {a, amin, amax, .02}, {b, a, btab[a], .02}], 1], InterpolationOrder -> 1];

Quiet@RegionPlot3D[amin < a < amax && a < b < btab[a] && b < c < ctab[a, b], {a, amin, amax}, {b, amin, btab[amin]}, {c, amin, ctab[amin, btab[amin]]}, AxesLabel -> {a, b, c},
LabelStyle -> {Black, Bold, Medium}, BoxRatios -> Automatic, ImageSize -> Large, PlotPoints -> 500, Mesh -> None]


I would like to generate several graphs for different values of $h$, say for $h$ between 0 and 1, with $dh$=0.01.
How can I do this?







plotting equation-solving numerics inequalities procedural-programming






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 9 '18 at 22:17









bbgodfrey

44.6k958109




44.6k958109










asked Dec 8 '18 at 15:14









ApiApi

537




537












  • $begingroup$
    h = Range[0, 1, 0.01] will give you a list of h. Then you can either redefince h as a variable in the functions it is used or you can use a loop (Do, Table) to evaluate along h and produce a list of plots. Map and MapAt can also work but the syntax is slightly trickier.
    $endgroup$
    – Titus
    Dec 8 '18 at 15:23


















  • $begingroup$
    h = Range[0, 1, 0.01] will give you a list of h. Then you can either redefince h as a variable in the functions it is used or you can use a loop (Do, Table) to evaluate along h and produce a list of plots. Map and MapAt can also work but the syntax is slightly trickier.
    $endgroup$
    – Titus
    Dec 8 '18 at 15:23
















$begingroup$
h = Range[0, 1, 0.01] will give you a list of h. Then you can either redefince h as a variable in the functions it is used or you can use a loop (Do, Table) to evaluate along h and produce a list of plots. Map and MapAt can also work but the syntax is slightly trickier.
$endgroup$
– Titus
Dec 8 '18 at 15:23




$begingroup$
h = Range[0, 1, 0.01] will give you a list of h. Then you can either redefince h as a variable in the functions it is used or you can use a loop (Do, Table) to evaluate along h and produce a list of plots. Map and MapAt can also work but the syntax is slightly trickier.
$endgroup$
– Titus
Dec 8 '18 at 15:23










4 Answers
4






active

oldest

votes


















2












$begingroup$

The explicit code needed to answer your question is



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];

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]))

l = .48;

Column@Table[
amin = NArgMin[{f2[a], f2[a] > l}, a]; amax = NArgMax[{f1[a], f1[a] < l}, a];
btab = Interpolation@Table[{a, NArgMax[{f3[b], f3[b] < l}, b]},
{a, amin, amax, (amax - amin)/20}];
fc[a0_, b0_] := Quiet@NArgMax[{f3[c], f3[c] < l, b < c} /. {a -> a0, b -> b0}, c];
Plot3D[{b, fc[a, b]}, {a, amin, amax}, {b, a, btab[a]}, AxesLabel -> {a, b, c},
PlotLabel -> StringForm["h = ``", h], LabelStyle -> {Black, Bold, 15},
BoxRatios -> Automatic, ImageSize -> Large, Mesh -> None,
PlotStyle -> Opacity[.5], PlotPoints -> 10, MaxRecursion -> 0],
{h, 0, .98, .49}]


enter image description here



Note that I used the more accurate second addendum to my earlier answer for a single plot. I also used Plot3D instead of RegionPlot3D, because the latter is very slow here. Even the calculation above is slow, of course.






share|improve this answer











$endgroup$





















    5












    $begingroup$

    Something like



    GraphicsColumn[Table[Plot[some expression with h in it],{h,0,1,0.01}]]


    perhaps.






    share|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for your answer, I have already tried this way, but I was not able to solve my issue: as you can see from the code I posted, to obtain my graph I first need to compute two tables, whose values change with $h$. I would need a procedure that, for each $h$ re-run automatically all the code.
      $endgroup$
      – Api
      Dec 8 '18 at 16:32










    • $begingroup$
      Make a function of h that does that. The way to get the most out of Mathematica is to compose functions of functions of functions, ...
      $endgroup$
      – John Doty
      Dec 8 '18 at 17:51










    • $begingroup$
      I would have done a function to get the graph I need, if I had been able. Unfortunately I could not manage to do so.
      $endgroup$
      – Api
      Dec 9 '18 at 17:41










    • $begingroup$
      What did you try? Why could you not manage? Can you come up with a simple example of the barrier you found?
      $endgroup$
      – John Doty
      Dec 9 '18 at 21:42



















    3












    $begingroup$

    Here is an example in which I use Manipulate to plot two graphs for different parameter values. This example may help you to adapt your code in a similar manner.



    Clear[alfa, newalfa, a1, a2, x, s, chn];
    Manipulate[
    SeedRandom[s];
    alfa = RandomReal[1, 20];
    newalfa = alfa*(1 + chn);
    Manipulate[
    Plot[{Sin[a1 x], Sin[a2 x]}, {x, 0, 10}],
    Row[{Control[{a1, alfa, Animator, AnimationRunning -> False}]}],
    Row[{Control[{a2, newalfa, Animator, AnimationRunning -> False}]}]
    ],
    {{s, 1, "s"}, 1, 100, 1},
    {{chn, 0, "change"}, -0.2, 0.2, 0.02}
    ]


    This code yields the following graphs.



    enter image description here






    share|improve this answer









    $endgroup$





















      3












      $begingroup$

      In all honesty this does nothing more than @John Doty suggested, but it illustrates my point in an earlier comment.



      Define



      h = Range[1, 10, 0.5]
      a = 1
      b = 1
      f[a_, b_, h_, x_] := PDF[NormalDistribution[h, b + a], x]
      g[a_, b_, h_, x_] := PDF[InverseGaussianDistribution[a + h, b], x]


      Then



      Do[{hh = h[[j]], Print[Plot[{f[a, b, hh, x], g[a, b, hh, x]}, {x, -5, 5}, 
      PlotLegends -> "Expressions"]]}, {j, 1, Length[h]}]


      It will print 20 PDF plots given a and b evaluated over range x, for each value of h. From what I understand in the OP, the part of the code below the definition of f4 can go into the brackets above, with the plot put inside a Print. Also, it looks like the brackets around amin, amax are reduntant, but I may be wrong. The operations inside the Do loop can be as complex as needed, so the calculation of the tables can be put in directly. Unfortunately I was not able to run the code in the OP and see what I was supposed to get.






      share|improve this answer









      $endgroup$













      • $begingroup$
        Thank you for answering, I was not able to implement the procedure though
        $endgroup$
        – Api
        Dec 9 '18 at 17:37











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      4 Answers
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      4 Answers
      4






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      votes









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      active

      oldest

      votes









      2












      $begingroup$

      The explicit code needed to answer your question is



      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];

      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]))

      l = .48;

      Column@Table[
      amin = NArgMin[{f2[a], f2[a] > l}, a]; amax = NArgMax[{f1[a], f1[a] < l}, a];
      btab = Interpolation@Table[{a, NArgMax[{f3[b], f3[b] < l}, b]},
      {a, amin, amax, (amax - amin)/20}];
      fc[a0_, b0_] := Quiet@NArgMax[{f3[c], f3[c] < l, b < c} /. {a -> a0, b -> b0}, c];
      Plot3D[{b, fc[a, b]}, {a, amin, amax}, {b, a, btab[a]}, AxesLabel -> {a, b, c},
      PlotLabel -> StringForm["h = ``", h], LabelStyle -> {Black, Bold, 15},
      BoxRatios -> Automatic, ImageSize -> Large, Mesh -> None,
      PlotStyle -> Opacity[.5], PlotPoints -> 10, MaxRecursion -> 0],
      {h, 0, .98, .49}]


      enter image description here



      Note that I used the more accurate second addendum to my earlier answer for a single plot. I also used Plot3D instead of RegionPlot3D, because the latter is very slow here. Even the calculation above is slow, of course.






      share|improve this answer











      $endgroup$


















        2












        $begingroup$

        The explicit code needed to answer your question is



        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];

        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]))

        l = .48;

        Column@Table[
        amin = NArgMin[{f2[a], f2[a] > l}, a]; amax = NArgMax[{f1[a], f1[a] < l}, a];
        btab = Interpolation@Table[{a, NArgMax[{f3[b], f3[b] < l}, b]},
        {a, amin, amax, (amax - amin)/20}];
        fc[a0_, b0_] := Quiet@NArgMax[{f3[c], f3[c] < l, b < c} /. {a -> a0, b -> b0}, c];
        Plot3D[{b, fc[a, b]}, {a, amin, amax}, {b, a, btab[a]}, AxesLabel -> {a, b, c},
        PlotLabel -> StringForm["h = ``", h], LabelStyle -> {Black, Bold, 15},
        BoxRatios -> Automatic, ImageSize -> Large, Mesh -> None,
        PlotStyle -> Opacity[.5], PlotPoints -> 10, MaxRecursion -> 0],
        {h, 0, .98, .49}]


        enter image description here



        Note that I used the more accurate second addendum to my earlier answer for a single plot. I also used Plot3D instead of RegionPlot3D, because the latter is very slow here. Even the calculation above is slow, of course.






        share|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          The explicit code needed to answer your question is



          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];

          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]))

          l = .48;

          Column@Table[
          amin = NArgMin[{f2[a], f2[a] > l}, a]; amax = NArgMax[{f1[a], f1[a] < l}, a];
          btab = Interpolation@Table[{a, NArgMax[{f3[b], f3[b] < l}, b]},
          {a, amin, amax, (amax - amin)/20}];
          fc[a0_, b0_] := Quiet@NArgMax[{f3[c], f3[c] < l, b < c} /. {a -> a0, b -> b0}, c];
          Plot3D[{b, fc[a, b]}, {a, amin, amax}, {b, a, btab[a]}, AxesLabel -> {a, b, c},
          PlotLabel -> StringForm["h = ``", h], LabelStyle -> {Black, Bold, 15},
          BoxRatios -> Automatic, ImageSize -> Large, Mesh -> None,
          PlotStyle -> Opacity[.5], PlotPoints -> 10, MaxRecursion -> 0],
          {h, 0, .98, .49}]


          enter image description here



          Note that I used the more accurate second addendum to my earlier answer for a single plot. I also used Plot3D instead of RegionPlot3D, because the latter is very slow here. Even the calculation above is slow, of course.






          share|improve this answer











          $endgroup$



          The explicit code needed to answer your question is



          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];

          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]))

          l = .48;

          Column@Table[
          amin = NArgMin[{f2[a], f2[a] > l}, a]; amax = NArgMax[{f1[a], f1[a] < l}, a];
          btab = Interpolation@Table[{a, NArgMax[{f3[b], f3[b] < l}, b]},
          {a, amin, amax, (amax - amin)/20}];
          fc[a0_, b0_] := Quiet@NArgMax[{f3[c], f3[c] < l, b < c} /. {a -> a0, b -> b0}, c];
          Plot3D[{b, fc[a, b]}, {a, amin, amax}, {b, a, btab[a]}, AxesLabel -> {a, b, c},
          PlotLabel -> StringForm["h = ``", h], LabelStyle -> {Black, Bold, 15},
          BoxRatios -> Automatic, ImageSize -> Large, Mesh -> None,
          PlotStyle -> Opacity[.5], PlotPoints -> 10, MaxRecursion -> 0],
          {h, 0, .98, .49}]


          enter image description here



          Note that I used the more accurate second addendum to my earlier answer for a single plot. I also used Plot3D instead of RegionPlot3D, because the latter is very slow here. Even the calculation above is slow, of course.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Dec 9 '18 at 22:38

























          answered Dec 9 '18 at 22:02









          bbgodfreybbgodfrey

          44.6k958109




          44.6k958109























              5












              $begingroup$

              Something like



              GraphicsColumn[Table[Plot[some expression with h in it],{h,0,1,0.01}]]


              perhaps.






              share|improve this answer









              $endgroup$













              • $begingroup$
                Thank you for your answer, I have already tried this way, but I was not able to solve my issue: as you can see from the code I posted, to obtain my graph I first need to compute two tables, whose values change with $h$. I would need a procedure that, for each $h$ re-run automatically all the code.
                $endgroup$
                – Api
                Dec 8 '18 at 16:32










              • $begingroup$
                Make a function of h that does that. The way to get the most out of Mathematica is to compose functions of functions of functions, ...
                $endgroup$
                – John Doty
                Dec 8 '18 at 17:51










              • $begingroup$
                I would have done a function to get the graph I need, if I had been able. Unfortunately I could not manage to do so.
                $endgroup$
                – Api
                Dec 9 '18 at 17:41










              • $begingroup$
                What did you try? Why could you not manage? Can you come up with a simple example of the barrier you found?
                $endgroup$
                – John Doty
                Dec 9 '18 at 21:42
















              5












              $begingroup$

              Something like



              GraphicsColumn[Table[Plot[some expression with h in it],{h,0,1,0.01}]]


              perhaps.






              share|improve this answer









              $endgroup$













              • $begingroup$
                Thank you for your answer, I have already tried this way, but I was not able to solve my issue: as you can see from the code I posted, to obtain my graph I first need to compute two tables, whose values change with $h$. I would need a procedure that, for each $h$ re-run automatically all the code.
                $endgroup$
                – Api
                Dec 8 '18 at 16:32










              • $begingroup$
                Make a function of h that does that. The way to get the most out of Mathematica is to compose functions of functions of functions, ...
                $endgroup$
                – John Doty
                Dec 8 '18 at 17:51










              • $begingroup$
                I would have done a function to get the graph I need, if I had been able. Unfortunately I could not manage to do so.
                $endgroup$
                – Api
                Dec 9 '18 at 17:41










              • $begingroup$
                What did you try? Why could you not manage? Can you come up with a simple example of the barrier you found?
                $endgroup$
                – John Doty
                Dec 9 '18 at 21:42














              5












              5








              5





              $begingroup$

              Something like



              GraphicsColumn[Table[Plot[some expression with h in it],{h,0,1,0.01}]]


              perhaps.






              share|improve this answer









              $endgroup$



              Something like



              GraphicsColumn[Table[Plot[some expression with h in it],{h,0,1,0.01}]]


              perhaps.







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Dec 8 '18 at 15:23









              John DotyJohn Doty

              6,95811024




              6,95811024












              • $begingroup$
                Thank you for your answer, I have already tried this way, but I was not able to solve my issue: as you can see from the code I posted, to obtain my graph I first need to compute two tables, whose values change with $h$. I would need a procedure that, for each $h$ re-run automatically all the code.
                $endgroup$
                – Api
                Dec 8 '18 at 16:32










              • $begingroup$
                Make a function of h that does that. The way to get the most out of Mathematica is to compose functions of functions of functions, ...
                $endgroup$
                – John Doty
                Dec 8 '18 at 17:51










              • $begingroup$
                I would have done a function to get the graph I need, if I had been able. Unfortunately I could not manage to do so.
                $endgroup$
                – Api
                Dec 9 '18 at 17:41










              • $begingroup$
                What did you try? Why could you not manage? Can you come up with a simple example of the barrier you found?
                $endgroup$
                – John Doty
                Dec 9 '18 at 21:42


















              • $begingroup$
                Thank you for your answer, I have already tried this way, but I was not able to solve my issue: as you can see from the code I posted, to obtain my graph I first need to compute two tables, whose values change with $h$. I would need a procedure that, for each $h$ re-run automatically all the code.
                $endgroup$
                – Api
                Dec 8 '18 at 16:32










              • $begingroup$
                Make a function of h that does that. The way to get the most out of Mathematica is to compose functions of functions of functions, ...
                $endgroup$
                – John Doty
                Dec 8 '18 at 17:51










              • $begingroup$
                I would have done a function to get the graph I need, if I had been able. Unfortunately I could not manage to do so.
                $endgroup$
                – Api
                Dec 9 '18 at 17:41










              • $begingroup$
                What did you try? Why could you not manage? Can you come up with a simple example of the barrier you found?
                $endgroup$
                – John Doty
                Dec 9 '18 at 21:42
















              $begingroup$
              Thank you for your answer, I have already tried this way, but I was not able to solve my issue: as you can see from the code I posted, to obtain my graph I first need to compute two tables, whose values change with $h$. I would need a procedure that, for each $h$ re-run automatically all the code.
              $endgroup$
              – Api
              Dec 8 '18 at 16:32




              $begingroup$
              Thank you for your answer, I have already tried this way, but I was not able to solve my issue: as you can see from the code I posted, to obtain my graph I first need to compute two tables, whose values change with $h$. I would need a procedure that, for each $h$ re-run automatically all the code.
              $endgroup$
              – Api
              Dec 8 '18 at 16:32












              $begingroup$
              Make a function of h that does that. The way to get the most out of Mathematica is to compose functions of functions of functions, ...
              $endgroup$
              – John Doty
              Dec 8 '18 at 17:51




              $begingroup$
              Make a function of h that does that. The way to get the most out of Mathematica is to compose functions of functions of functions, ...
              $endgroup$
              – John Doty
              Dec 8 '18 at 17:51












              $begingroup$
              I would have done a function to get the graph I need, if I had been able. Unfortunately I could not manage to do so.
              $endgroup$
              – Api
              Dec 9 '18 at 17:41




              $begingroup$
              I would have done a function to get the graph I need, if I had been able. Unfortunately I could not manage to do so.
              $endgroup$
              – Api
              Dec 9 '18 at 17:41












              $begingroup$
              What did you try? Why could you not manage? Can you come up with a simple example of the barrier you found?
              $endgroup$
              – John Doty
              Dec 9 '18 at 21:42




              $begingroup$
              What did you try? Why could you not manage? Can you come up with a simple example of the barrier you found?
              $endgroup$
              – John Doty
              Dec 9 '18 at 21:42











              3












              $begingroup$

              Here is an example in which I use Manipulate to plot two graphs for different parameter values. This example may help you to adapt your code in a similar manner.



              Clear[alfa, newalfa, a1, a2, x, s, chn];
              Manipulate[
              SeedRandom[s];
              alfa = RandomReal[1, 20];
              newalfa = alfa*(1 + chn);
              Manipulate[
              Plot[{Sin[a1 x], Sin[a2 x]}, {x, 0, 10}],
              Row[{Control[{a1, alfa, Animator, AnimationRunning -> False}]}],
              Row[{Control[{a2, newalfa, Animator, AnimationRunning -> False}]}]
              ],
              {{s, 1, "s"}, 1, 100, 1},
              {{chn, 0, "change"}, -0.2, 0.2, 0.02}
              ]


              This code yields the following graphs.



              enter image description here






              share|improve this answer









              $endgroup$


















                3












                $begingroup$

                Here is an example in which I use Manipulate to plot two graphs for different parameter values. This example may help you to adapt your code in a similar manner.



                Clear[alfa, newalfa, a1, a2, x, s, chn];
                Manipulate[
                SeedRandom[s];
                alfa = RandomReal[1, 20];
                newalfa = alfa*(1 + chn);
                Manipulate[
                Plot[{Sin[a1 x], Sin[a2 x]}, {x, 0, 10}],
                Row[{Control[{a1, alfa, Animator, AnimationRunning -> False}]}],
                Row[{Control[{a2, newalfa, Animator, AnimationRunning -> False}]}]
                ],
                {{s, 1, "s"}, 1, 100, 1},
                {{chn, 0, "change"}, -0.2, 0.2, 0.02}
                ]


                This code yields the following graphs.



                enter image description here






                share|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Here is an example in which I use Manipulate to plot two graphs for different parameter values. This example may help you to adapt your code in a similar manner.



                  Clear[alfa, newalfa, a1, a2, x, s, chn];
                  Manipulate[
                  SeedRandom[s];
                  alfa = RandomReal[1, 20];
                  newalfa = alfa*(1 + chn);
                  Manipulate[
                  Plot[{Sin[a1 x], Sin[a2 x]}, {x, 0, 10}],
                  Row[{Control[{a1, alfa, Animator, AnimationRunning -> False}]}],
                  Row[{Control[{a2, newalfa, Animator, AnimationRunning -> False}]}]
                  ],
                  {{s, 1, "s"}, 1, 100, 1},
                  {{chn, 0, "change"}, -0.2, 0.2, 0.02}
                  ]


                  This code yields the following graphs.



                  enter image description here






                  share|improve this answer









                  $endgroup$



                  Here is an example in which I use Manipulate to plot two graphs for different parameter values. This example may help you to adapt your code in a similar manner.



                  Clear[alfa, newalfa, a1, a2, x, s, chn];
                  Manipulate[
                  SeedRandom[s];
                  alfa = RandomReal[1, 20];
                  newalfa = alfa*(1 + chn);
                  Manipulate[
                  Plot[{Sin[a1 x], Sin[a2 x]}, {x, 0, 10}],
                  Row[{Control[{a1, alfa, Animator, AnimationRunning -> False}]}],
                  Row[{Control[{a2, newalfa, Animator, AnimationRunning -> False}]}]
                  ],
                  {{s, 1, "s"}, 1, 100, 1},
                  {{chn, 0, "change"}, -0.2, 0.2, 0.02}
                  ]


                  This code yields the following graphs.



                  enter image description here







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Dec 8 '18 at 17:07









                  Tugrul TemelTugrul Temel

                  841213




                  841213























                      3












                      $begingroup$

                      In all honesty this does nothing more than @John Doty suggested, but it illustrates my point in an earlier comment.



                      Define



                      h = Range[1, 10, 0.5]
                      a = 1
                      b = 1
                      f[a_, b_, h_, x_] := PDF[NormalDistribution[h, b + a], x]
                      g[a_, b_, h_, x_] := PDF[InverseGaussianDistribution[a + h, b], x]


                      Then



                      Do[{hh = h[[j]], Print[Plot[{f[a, b, hh, x], g[a, b, hh, x]}, {x, -5, 5}, 
                      PlotLegends -> "Expressions"]]}, {j, 1, Length[h]}]


                      It will print 20 PDF plots given a and b evaluated over range x, for each value of h. From what I understand in the OP, the part of the code below the definition of f4 can go into the brackets above, with the plot put inside a Print. Also, it looks like the brackets around amin, amax are reduntant, but I may be wrong. The operations inside the Do loop can be as complex as needed, so the calculation of the tables can be put in directly. Unfortunately I was not able to run the code in the OP and see what I was supposed to get.






                      share|improve this answer









                      $endgroup$













                      • $begingroup$
                        Thank you for answering, I was not able to implement the procedure though
                        $endgroup$
                        – Api
                        Dec 9 '18 at 17:37
















                      3












                      $begingroup$

                      In all honesty this does nothing more than @John Doty suggested, but it illustrates my point in an earlier comment.



                      Define



                      h = Range[1, 10, 0.5]
                      a = 1
                      b = 1
                      f[a_, b_, h_, x_] := PDF[NormalDistribution[h, b + a], x]
                      g[a_, b_, h_, x_] := PDF[InverseGaussianDistribution[a + h, b], x]


                      Then



                      Do[{hh = h[[j]], Print[Plot[{f[a, b, hh, x], g[a, b, hh, x]}, {x, -5, 5}, 
                      PlotLegends -> "Expressions"]]}, {j, 1, Length[h]}]


                      It will print 20 PDF plots given a and b evaluated over range x, for each value of h. From what I understand in the OP, the part of the code below the definition of f4 can go into the brackets above, with the plot put inside a Print. Also, it looks like the brackets around amin, amax are reduntant, but I may be wrong. The operations inside the Do loop can be as complex as needed, so the calculation of the tables can be put in directly. Unfortunately I was not able to run the code in the OP and see what I was supposed to get.






                      share|improve this answer









                      $endgroup$













                      • $begingroup$
                        Thank you for answering, I was not able to implement the procedure though
                        $endgroup$
                        – Api
                        Dec 9 '18 at 17:37














                      3












                      3








                      3





                      $begingroup$

                      In all honesty this does nothing more than @John Doty suggested, but it illustrates my point in an earlier comment.



                      Define



                      h = Range[1, 10, 0.5]
                      a = 1
                      b = 1
                      f[a_, b_, h_, x_] := PDF[NormalDistribution[h, b + a], x]
                      g[a_, b_, h_, x_] := PDF[InverseGaussianDistribution[a + h, b], x]


                      Then



                      Do[{hh = h[[j]], Print[Plot[{f[a, b, hh, x], g[a, b, hh, x]}, {x, -5, 5}, 
                      PlotLegends -> "Expressions"]]}, {j, 1, Length[h]}]


                      It will print 20 PDF plots given a and b evaluated over range x, for each value of h. From what I understand in the OP, the part of the code below the definition of f4 can go into the brackets above, with the plot put inside a Print. Also, it looks like the brackets around amin, amax are reduntant, but I may be wrong. The operations inside the Do loop can be as complex as needed, so the calculation of the tables can be put in directly. Unfortunately I was not able to run the code in the OP and see what I was supposed to get.






                      share|improve this answer









                      $endgroup$



                      In all honesty this does nothing more than @John Doty suggested, but it illustrates my point in an earlier comment.



                      Define



                      h = Range[1, 10, 0.5]
                      a = 1
                      b = 1
                      f[a_, b_, h_, x_] := PDF[NormalDistribution[h, b + a], x]
                      g[a_, b_, h_, x_] := PDF[InverseGaussianDistribution[a + h, b], x]


                      Then



                      Do[{hh = h[[j]], Print[Plot[{f[a, b, hh, x], g[a, b, hh, x]}, {x, -5, 5}, 
                      PlotLegends -> "Expressions"]]}, {j, 1, Length[h]}]


                      It will print 20 PDF plots given a and b evaluated over range x, for each value of h. From what I understand in the OP, the part of the code below the definition of f4 can go into the brackets above, with the plot put inside a Print. Also, it looks like the brackets around amin, amax are reduntant, but I may be wrong. The operations inside the Do loop can be as complex as needed, so the calculation of the tables can be put in directly. Unfortunately I was not able to run the code in the OP and see what I was supposed to get.







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered Dec 8 '18 at 19:15









                      TitusTitus

                      645417




                      645417












                      • $begingroup$
                        Thank you for answering, I was not able to implement the procedure though
                        $endgroup$
                        – Api
                        Dec 9 '18 at 17:37


















                      • $begingroup$
                        Thank you for answering, I was not able to implement the procedure though
                        $endgroup$
                        – Api
                        Dec 9 '18 at 17:37
















                      $begingroup$
                      Thank you for answering, I was not able to implement the procedure though
                      $endgroup$
                      – Api
                      Dec 9 '18 at 17:37




                      $begingroup$
                      Thank you for answering, I was not able to implement the procedure though
                      $endgroup$
                      – Api
                      Dec 9 '18 at 17:37


















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