find intervals for both B and C which make the exponential function F converge to some points.

I have series Y with 258 observations.

I define new series by deleting first and last 50 observations

newY = DlnY(51:end-50);

I also define the series Y lagged 2.

logY = log(Y);
DlnY = diff(logY);
Y2 = [NaN; NaN; DlnY(1:end-2)];

I have exponential function as follows:

F = 1- exp((-B/0.009)*(Y2-C)^2)

F_func = @(B, C, Y2) 1 - exp((-B / 0.009) * (Y2 - C).^2);

The parameter c is in the range of series newY.

C = [min(newY), max(newY)];

The parameter B in the interval from 1 to infinity.

B = [1, Inf];

I) I want to find intervals for both B and C which make the function F converge to 0.

II) I want to find intervals for both B and C which make the function F converge to 0.5.

III) I want to find intervals for both B and C which make the function F converge to 1.

How can I write a MATLAB code find these intervals?

Can you please help me to write these codes. Thank you.

[Torsten]

What do you mean by "converge" in this context ?

[BURCU]

"Equal to" and "close to"

[BURCU]

Edit:

I) I want to find intervals for both B and C which make the function F equal to (close to) 0.

II) I want to find intervals for both B and C which make the function F equal to (close to) 0.5.

III) I want to find intervals for both B and C which make the function F equal to (close to) 1.

[Walter Roberson]

I have exponential function as follows:

F = 1- exp((-B/0.009)*(Y2-C)^2)

Your Y2 is a vector, so your F is a vector. It is not really reasonable to talk about a vector converging to a value -- not unless you mean that the sequence F(end-3) F(end-2) F(end-1) F(end) is a curve with asymptope equal to the given value.

[BURCU]

Yes, you're right. Actually, I wanted it that way, but I couldn't express it properly. How can I calculate this by using MATLAB codes?

[Matt J]

Actually, I wanted it that way

Which way? Edit your original question to articulate what you really mean.

Answer 1

[Walter Roberson]

There is no point in calculating most of the F values since we are only interested in convergence. So we only test the last few values. The exact number we test is arbitrary.

F4_func = @(BC) 1 - exp((-BC(1) / 0.009) * (Y2(end-3:end) - BC(2)).^2);

targets = [0, .5, 1];
numtargets = length(targets);

bestBC = cell(numtargets,1);

for Tidx = 1 : numtargets
    target = targets(Tidx);
    objfun = @(BC) sum((F4_func(BC) - target).^2);
    A = []; b = []; Aeq = []; beq = [];
    lb = [1 min(newY)];
    ub = [inf max(newY)];
    nonlfunc = [];
    BC0 = [2 mean(newY)];
    bestBC{Tidx} = fmincon(objfun, BC0, A, b, Aeq, beq, lb, ub, nonlfunc);
end

[BURCU]

Thank you! In order to print the results, I will use the code

 disp(bestBC); 

Am I right?

[Walter Roberson]

celldisp(bestBC)

would be more like it.

You could probably instead use

bestBC = zeros(numtargets, 2);
%...
bestBC(Tidx,:) = fmincon(...)

I just wasn't absolutely positive that fmincon() would return vectors of values even in the case where fmincon essentially failed.

[BURCU]

Thsnk you so much