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How to make it isolate the Beta parameter?

Posted 1 year ago
6 Replies
2 Total Likes


I'm trying to isolate the beta parameter in the inequality belowenter image description here and that Mathematica give me the specified conditions such that the inequality is holding. I'm beginning in Mathematica so I'm not sure this is feasible as my computer is just stopping on Running for unlimited time (I've stopped after 1h30 minutes). I put the Mathematica file below also, Thanks in advance!

6 Replies
Posted 1 year ago

Reduce might work better than Solve for this problem.

I started this without using any of your additional inequalities about domains and added those, one at a time, after seeing from the result that Reduce needed additional information. In the very last steps I even introduced a couple of inequalities which seem should have been deduced from what I gave it earlier, but Reduce didn't seem to be able to find those on its own. Reduce completes this calculation in a few seconds.

Reduce[{(-a+c β+t θ1)^2(6b^2+4b λ+λ^2)/(8b^3)>(a-c-t θ3)^3/4b,
  0<c<a, 0<b<1, 0<λ<1, 0<t<(a-c α)/θ2, t<(a-c β)/θ1, 0<θ1<θ2<θ3<1, 0<θ3, β c+t θ1<a, α c+t θ2<a}, β]

but the result returned from that is complicated and includes lots of things which you know to be true and which clutter up what you are trying to see.

Simplify can be used to discard things which you know to be true if you include that information in the optional third argument, which in this case is just repeating your additional inequalities.

  Reduce[{(-a+c β+t θ1)^2(6b^2+4b λ+λ^2)/(8b^3)>(a-c-t θ3)^3/4b,
    0<c<a,0<b<1,0<λ<1,0<t<(a-c α)/θ2,t<(a-c β)/θ1, 0<θ1<θ2<θ3<1,0<θ3,β c+t θ1 <a,α c+t θ2<a},β],
  {0<c<a,0<b<1,0<λ<1,0<t<(a-c α)/θ2,t<(a-c β)/θ1, 0<θ1<θ2<θ3<1,0<θ3,β c+t θ1 <a,α c+t θ2<a}]

which returns

β*c + t*θ1 + Sqrt[2]*b^2*Sqrt[(a - c - t*θ3)^3/(6*b^2 + 4*b*λ + λ^2)] < a ||
c + t*θ3 > a

That doesn't quite completely isolate β, but sometimes results are close enough that the extra work needed to format the result doesn't seem like it is worth the effort to me. That output also includes the alternative c + t*θ3 > a which might be eliminated with the addition of another domain inequality, but hopefully this is close enough and gives you an idea how you can do similar things in the future.

Please check all this very carefully to try to make certain I haven't made any mistakes in reformulating your problem.

Thanks a lot I will try that ! You just put a power 3 instead of power 2 on the right side of the inequality but correcting this it worked !

Crossposted here.

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We will look at the issue from another perspective. Assume there is the one set of life data that is fitted using a Weibull distribution. This data set has a shape parameter β. If we separate the data into two groups and fit each group with a separate Weibull distribution, we get β1 and β2 as the shape parameters of the two groups. If we suppose that β1 < β2, will β1 < β < β2 always be true, or is there some other relationship between these beta values?

Posted 1 year ago

Thank you for catching that. I apologize for my mistake.

With the correct equation it doesn't do as well isolating β. Not using Simplify and manually extracting the desired relation for β is perhaps better.

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