MATLAB: Problem with solve and icdf

erroricdfMATLABsolve

Hi, I need to solve equations of the type b = @(x,s) solve(bb==x-s*icdf('Normal',bb,0,1),bb)

 and after the function call

>> b(0.2,0.4)

I get this error. Can anyone help me?

Best,

Christian

Error using symengine Unable to prove 'bb < 0 | 1 < bb' literally. Use 'isAlways' to test the statement mathematically.

Error in sym/subsindex (line 821) X = find(mupadmex('symobj::logical',A.s,9)) – 1;

Error in sym/privsubsasgn (line 1096) L_tilde2 = builtin('subsasgn',L_tilde,struct('type','()','subs',{varargin}),R_tilde);

Error in sym/subsasgn (line 933) C = privsubsasgn(L,R,inds{:});

Error in norminv (line 57) p(p < 0 | 1 < p) = NaN;

Error in icdf (line 91) x = norminv(p,a,b);

Error in @(x,s)solve(bb==x-s*icdf('Normal',bb,0,1),bb)

Best Answer

It is probably best to avoid the Symbolic Math Toolbox for such problems.

Try this:

b = @(x,s,bb) x-s*icdf('Normal',bb,0,1)-bb;
B = fsolve(@(bb)b(x,s,bb), 0.01)

You may need a different initial parameter estimate (I used 0.01 here), depending on what ‘x’ and ‘s’ are. This uses the Optimization Toolbox fsolve (link) function. It is more robust than fzero.

Related Solutions

MATLAB: How to use fsolve to solve this equation

One problem is that you need to add ‘x’ to your syms statement. Also, in your call to fsolve, do not use quotes ('') around the name of the function you are using.

I am not certain what you are doing, but I did not have any problems using this code:

syms t x
aaa=int(3*t,t,0,x)-2
aa = matlabFunction(aaa)

produced:

aaa =
(3*x^2)/2 - 2

Using matlabFunction to create an anonymous function out of it,

aa = @(x) (3.*x.^2)./2-2
aa1 = fsolve(aa,1)

Produced:

aa1 =
       1.1547e+000

Or, if you want to do everything symbolically,

aa2s = solve(aaa == 0, x)
aa2d = double(aa2s)

produces:

aa2s =
    (2*3^(1/2))/3
   -(2*3^(1/2))/3
aa2d =
       1.1547e+000
      -1.1547e+000

MATLAB: Integration of norminv function

The problem is that norminv(x,0,1) goes to -Inf as x goes to zero and Inf as x goes to 1 (and ditto for y and z), so it's hard to integrate accurately. If you can manage with a larger tolerance,

q3 = integral3(fun3,0,1,0,1,0,1,'AbsTol',1e-3);

does not return any warnings (the default tolerance is 1e-10). If you need higher precision, you'll just need to be patient, as the warnings are not fatal. However, I don't know if the precision is actually met under such circumstances.

Best Answer

Related Solutions

MATLAB: How to use fsolve to solve this equation

MATLAB: Integration of norminv function

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