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equation solving - FindFit with a sophisticated function (integral)


I am trying to find a fit to the distribution function (empiricial data) in terms of a function which is itself an integral of a product of two simpler functions (two polynomials), that is the model. In particular, I observe T(x) and the model is that $$T(x) \approx \int_0^xF(\frac{x-y}{1-y})g(y)dy $$


My assumption is that $F(.)$ and $g(.)$ are polynomial functions. And so my problem would be to find the best fit polynomial functions, i.e. to assume that $F(.)= k_1+a_1 (\frac{x-y}{1-y})+ b_1 (\frac{x-y}{1-y})^2$ and similarly $g(.)=k_2+a_2 y + b_2 y^2$ and let FindFit run.


Here is the data (just a small sample)


data={{0.002, 4}, {0.01, 5}, {0.02, 1}, {0.025, 1}, {0.0333, 1}, {0.05, 

1}, {0.0905, 1}, {0.09995, 1}, {0.105, 1}, {0.114, 1}, {0.2, 
5}, {0.222, 2}, {0.25, 1}, {0.3, 1}, {0.35, 1}, {0.4, 7}, {0.5, 
29}, {0.501, 2}, {0.505, 2}, {0.51, 1}, {0.52, 1}, {0.55, 
1}, {0.55555, 1}, {0.6, 12}, {0.64, 2}, {0.65, 5}, {0.666, 1}, {0.7,
18}, {0.73, 1}, {0.74, 1}, {0.75, 30}, {0.76, 3}, {0.77266, 
2}, {0.775, 1}, {0.8, 57}, {0.801, 2}, {0.8018, 1}, {0.802, 
1}, {0.81, 1}, {0.81554, 1}, {0.82, 3}, {0.825, 1}, {0.82888, 
1}, {0.83, 1}, {0.84, 4}, {0.85, 30}, {0.859, 1}, {0.86, 3}, {0.861,
1}, {0.862, 1}, {0.875, 7}, {0.88, 8}, {0.888, 2}, {0.9, 
46}, {0.901, 3}, {0.9018, 1}, {0.902, 1}, {0.9022, 1}, {0.9026, 

1}, {0.9027, 1}, {0.904, 1}, {0.9094, 1}, {0.91, 2}, {0.9202, 
1}, {0.925, 1}, {0.926, 1}, {0.93, 1}, {0.94, 2}, {0.95, 5}, {0.96, 
3}, {0.976, 1}, {0.98, 1}, {0.995, 1}, {1., 11}};

I have been trying to solve this pretty naively as follows:


D=SmoothKernelDistribution[data, 0.02];    
FindFit[CDF[D, x], 
Integrate[(a1 ((x - y)/(1 - y))^2 + b1 ((x - y)/(1 - y)) + 
   k1) (a2 y^2 + b2 y + k2), {y, 0, x}], {a1, b1, k1, a2, b2, 
k2}, {(x - y)/(1 - y), y}]


However I got the error: "(-y+x)/(1-y) is not a valid variable". I am sure that I need more sophisticated approach, but which steps would be there I am pretty lost...


(I did not specify in my code the domains (both x and y belong to [0,1]), but I doubt that this is of any relevance at this stage.)


I would appreciate any hints how to proceed!



Answer



Your datatest has one too many parenthesis.


Dont use func, replace it with the evaluated integral.


Integrate[(a ((x - y)/(1 - y))^2 + b ((x - y)/(1 - y)) + 
    c) (d y^2 + e y + f), {y, 0, x}, Assumptions -> {0 < x < 1}]
(* 

1/6 x (6 (c f + b (d + e + f) + a (4 d + 3 e + 2 f)) - 
    3 (-c e + b (d + e) + a (8 d + 5 e + 2 f)) x + (2 a - b + 
       2 c) d x^2) + (-b (d + e + f) + 
    a (-2 f + e (-3 + x) + 2 d (-2 + x))) (-1 + x) Log[1 - x] *)

Now run FindFit


sol = FindFit[datatest,
  1/6 x (6 (c f + b (d + e + f) + a (4 d + 3 e + 2 f)) - 
      3 (-c e + b (d + e) + a (8 d + 5 e + 2 f)) x + (2 a - b + 
         2 c) d x^2) + (-b (d + e + f) + 

      a (-2 f + e (-3 + x) + 2 d (-2 + x))) (-1 + x) Log[1 - x], {a, 
   b, c, d, e, f}, x, Method -> NMinimize, 
  NormFunction -> (Norm[#, Infinity] &)]

(* {a -> -2.70862, b -> 0.555548, c -> -0.243013, 
 d -> -0.124391, e -> -0.310328, f -> -0.516147} *)

Plot the results


Show[
 ListPlot[datatest],

 Plot[Evaluate[func[a, b, c, d, e, f, x] /. sol], {x, 0, 1}, 
  PlotStyle -> Red]
 ]

Mathematica graphics


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