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I have the following equation:

p = 1/(1 + E^(-51.179 + 0.849 x1 + 2.3649 x2 + 1.155 x3 + 1.46 x4 + 1.31 x5))

and I want to find the set of values for x1 to x5 such that the fraction be equal to 0.5.In other words, I want x1-x5 values for which p==0.5. I tried NSolve and Reduce but those didn't work. All x1-x5 are between 0 and 10.

I searched the list but I couldn't come up with a solution. Any help is appreciated.

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1 Answer 1

up vote 2 down vote accepted
FindInstance[p == 0.5, {x1, x2, x3, x4, x5}]

Gives:

{{x1 -> -3.5 + 4.9 I, x2 -> -0.6 - 1.1 I, x3 -> -3.2 + 0.5 I, 
  x4 -> -0.2 - 2.5 I, x5 -> 45.4637 + 44.3225 I}}

You can also limit the domain and ask to find n instances

   sol = FindInstance[p == 0.5, {x1, x2, x3, x4, x5}, Reals, 3]
{{x1 -> -7.5, x2 -> -3.7, x3 -> -0.5, x4 -> 14.4, x5 -> 35.0001},
 {x1 -> 6.6, x2 -> -11.2, x3 -> -5., x4 -> 5.9,  x5 -> 52.8424}, 
 {x1 -> 11.4, x2 -> 14.9, x3 -> 5.2, x4 -> 11.3, x5 -> -12.3974}}

Check

p /. sol    

{0.5, 0.5, 0.5}

To specifically select values that meet your criteria do the following:

FindInstance[{p == 0.5, x1 > 0 && x1 <= 10, x2 > 0 && x2 <= 10, 
  x3 > 0 && x3 <= 10, x4 > 0 && x4 <= 10, x5 > 0 && x5 <= 10}, {x1, 
  x2, x3, x4, x5}, Reals]

Gives

{{x1 -> 1.06349, x2 -> 4.66239, x3 -> 10., x4 -> 10., x5 -> 10.}}

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thanks for your respone.Is it possible to find a unique set? I found that equation using LogitModelFit given data on five predictors and an outcome variable. –  Amin Oct 25 '13 at 4:46
    
@Amin. If I understand Mathematics well, having one equation with more than one unknowns will not give you a unique solution. –  RunnyKine Oct 25 '13 at 4:51
    
You're right.Indeed, I'm talking in the context of statistics where we try to find best solutions.For instance, those coefficients are obtained based on least squares. I think I should restrict values of x1-x5 in more reasonable limits then I would be able to find some acceptable results. –  Amin Oct 25 '13 at 4:55
    
@Amin. I can't really help you with that as I am no statistician. But that may be a way to go. But at least my post answered your question right? –  RunnyKine Oct 25 '13 at 4:58
1  
p == 0.5 when the argument of the exponential is zero -- i.e. $1/(1+\exp(0))=0.5$. So your solution set is the hyperplane -51.179 + 0.849 x1 + 2.3649 x2 + 1.155 x3 + 1.46 x4 + 1.31 x5 == 0. –  Stephen Luttrell Oct 25 '13 at 9:52

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