How to replace a for loop with something faster

Hi,

I have to speed up a portion of my code as much as possible, since the associated computational time is way too high (especially because this part of the code belongs to a nested function, which is called in an optimization process). I 'd like to avoid using a for loop for my vector recursion since its pretty time consuming

I have heard about using the "filter" function or a Cmex file to accelerate the process in a similar discussion (<http://www.mathworks.com/matlabcentral/answers/28396-speed-up-recursive-loop)>, but I dont have a clue of how to apply them to my problem. Therefore, I would really appreciate any kind of help :)

            phi=phi'; % the input has to be a row vector

            % recursion for calculating A(t,T,Phi)=A_ and B(t,T,Phi)=B_ (see
            % p.592 in Heston and Nandi article)

            %A=zeros(phi.*r);
            row=size(phi,1);
            A=zeros(row,T-1);
            B=zeros(row,T-1);
            A(:,T)=0;
            B(:,T)=0; 
            *for i=1:T-1
                A(:,T-i)=A(:,T-i+1)+phi.*r+B(:,T-i+1).*w-.5*log(1-2*a.*B(:,T-i+1));
                B(:,T-i)=phi.*(lam_+g_)-.5*g_^2+b.*B(:,T-i+1)+.5.*(phi-g_).^2./(1-2.*a.*B(:,T-i+...
                    1));
            end*

            A_=A(:,1)+phi.*r+B(:,1).*w-.5*log(1-2.*a.*B(:,1)); % A(t;T,phi)
            B_=phi.*(lam_+g_)-.5*g_^2+b.*B(:,1)+.5*(phi-g_).^2./(1-2.*a.*B(:,1)); % B(t;T,phi)

Thanks a lot!!!

Notes: r w a b lam_ g_ S_0 Sig_ are scalars, phi is a vector that might contain complex numbers

Basically the backward recursion is as described in the for loop, from final conditions A(T,T)=B(T,T)=0

[Simon]

Hi!

Can you give example values for the parameters so that we can execute the code?

[Maël]

Sure:

a= 1.32e-6; b= 0.589; lam_= -0.5; g_= 422.0950; w=5.02e-6 T= 60;S_0= 1132.60;Sig_=1.4416e-04;r=6.6048e-05;

phi= [-79.6315i; -93.2105i]

Thanks a lot for your help Simon!

Answer 1

[Andrei Bobrov]

Please try is it:

p1 = phi.*r;
p2 = phi.*(lam_+g_)-.5*g_^2;
p3 = .5.*(phi-g_).^2;
a2 = 2*a;
for ii = 1:T-1
    k = T-ii+1;
    p4 = 1-a2.*B(:,k);
    A(:,k-1) = A(:,k) + p1 + B(:,k).*w - .5*log(p4);
    B(:,k-1) = p2 + b.*B(:,k) + p3./p4;
end

[Maël]

Thanks a lot!!

Answer 2

[Simon]

Hi!

If you don't need the results of every loop, you can write:

p1 = phi.*r;
p2 = phi.*(lam_+g_)-.5*g_^2;
p3 = .5.*(phi-g_).^2;
a2 = 2*a;

A=zeros(row,1);
B=zeros(row,1);

for ii = 1:T-1
    p4 = 1-a2*B;
    A = A + p1 + B*w - .5*log(p4);
    B = p2 + b*B + p3./p4;
end

It saves some more time.

[Maël]

Thanks a lot to both of you, both your answers help to reduce computational time significantly. I don't need the results of every loop (just the last one), so Simon's solution is a bit faster. Is there any way to avoid using a for loop altogether, I heard the filter function could be used for recursion on vectors? Do you know anything aboutt that?

[Simon]

Hi!

I'm not perfectly sure but I think it is not possible using filter, because you have two dependent variables A and B.

[Maël]

Uhm :) even if we used two different first order filters (one for A and one for B)?

[Simon]

No, this will give you the independent calculation of A and B, but this is not possible since A depends on B.

[Maël]

You're absolutely right, it missed that.. Thank you for your help!