Solving a function with fsolve

Hello everyone! As im new in matlab i would like your help to solve an issue that i have with fsolve .

I have created a function in a new matlab file and i named the matlab file as the name of the function. I named it NEcdf.

function fx = NEcdf(x,hetta_RL,omega_RL,delta_RL,alpha_VaR)
alpha_VaR=0.99;
hetta_RL=-0.0047;
omega_RL=0.0077;
delta_RL=0.0041;
fx = ((cdf('Normal',(x-hetta_RL)./omega_RL))-((exp(omega_RL^2)./2.*(delta_RL).^2 - (x-hetta_RL)./delta_RL)).*(cdf('Normal',(x-hetta_RL)./omega_RL - omega_RL./delta_RL))-alpha_VaR);
end

Then i have the following codes:

alpha_VaR=0.99;

hetta_RL=-0.0047;

omega_RL=0.0077;

delta_RL=0.0041;

f=fsolve(@(x) NEcdf(hetta_RL,omega_RL,delta_RL,alpha_VaR)
VaR_NE=fsolve(f,0)

And i dont get the result i expect from this solver.

Thank you in advance!!!

Answer 1

[Stephan]

VaR_NE = fsolve(@NEcdf,0)

function fx = NEcdf(x)
alpha_VaR=0.99;
hetta_RL=-0.0047;
omega_RL=0.0077;
delta_RL=0.0041;
fx = ((cdf('Normal',(x-hetta_RL)./omega_RL))-((exp(omega_RL^2)./2.*(delta_RL).^2-...
    (x-hetta_RL)./delta_RL)).*(cdf('Normal',(x-hetta_RL)./omega_RL-...
    omega_RL./delta_RL))-alpha_VaR);
end

[Andreas S]

I'm running your suggested VaR_NE and the problem still remains. When I run the VaR_NE the result is the number I use at fsolve brackets. For example when I use zero at fsolve it gives me zero as e result. When I use 1 the result is zero again. When I use number two or bigger, the result I get is hetta_RL. What's is the problem?

[Stephan]

I guess it could be a problem how you use the cdf function. Calling this function with two arguments is used when you have a distribution object as input parameter. But you only have numbers. So my guess is, that you have to give at least 3 input parameters to the cdf function. The first one is 'normal', the second should be the input vector for that you want to calculate, number 3 (and maybe 4) are mu (and sigma) of your normal distribution. The fact that you don't do this, let's me guess this could be the problem.