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## function I do not really understand, could somebody give me a hint?

Asked by Locks

### Locks (view profile)

on 19 May 2013
Accepted Answer by Wayne King

### Wayne King (view profile)

Hi,
I have received a function which consists of several furnctions, but I have some problems to see what is going on here. In input parameters are given in line two: x0 = [1.1768 0.082259 0.83746 -0.54589]; %parameters: psi m xi rho
That's clear so mean that I can change them in order to get the function running. But I am not sure what is happending with those parameters. In the last function, the following is happening

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## 2 Answers

Answer by Wayne King

### Wayne King (view profile)

on 19 May 2013
Accepted Answer

u is just the variable for integration, so I could define a function
function y = myfun(x)
y = x.^2;
end
and then integrate that function like:
quadl(@(u) myfun(u),0,2)
That is just the integral of x^2 from 0 to 2. So in LaTex
\int_{0}^{2} x^2 dx or \int_{0}^{2} u^2 du or whatever you variable you like.
Thank you for accepting my answer if I have helped you.

Locks

### Locks (view profile)

on 19 May 2013
you did, thanks

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Answer by Wayne King

### Wayne King (view profile)

on 19 May 2013

This input
x0 = [1.1768 0.082259 0.83746 -0.54589];
is fed into CallPriceF() as the input argument, parameters.
Then, it is fed into the function, CallPriceIntegralF(), as the input argument, x.
So the code:
kappa = x(1); theta = x(2);
sigma = x(3); rho = x(4);
inside of CallPriceIntegralF() is really
kappa = x0(1); theta = x0(2);
sigma = x0(3); rho = x0(4);

Locks

### Locks (view profile)

on 19 May 2013
thanks, do you know where the u in the last function comes from?
%Help vectors
o = sqrt(-1)*u; b = sigma*rho*o-kappa;
a = o.*(1-o); g = sqrt(b.^2+a*sigma^2); %gamma
B = - (a.*(1-exp(-g*T))) ./ (2*g - (g+b).*(1-exp(-g*T)));
A = -kappa*theta*( (g+b)/sigma^2 * T + 2/sigma^2 * log(1- (g+b)./(2*g).*(1-exp(-g*T))));
CHF = exp(A + B*V0 + o * log(F));

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