Help: Speed up for loop with two functions

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Kai Koslowsky 2021-10-12

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https://ww2.mathworks.cn/matlabcentral/answers/1562341-help-speed-up-for-loop-with-two-functions

评论： Kai Koslowsky 2021-11-3

Hello everyone, i am currently running this code. But unfortunately it takes for ever to finsish. A 12x10 matrix took nearly 30 minutes and i need one for 1415596x10. Is there any way i can speed up the process? Is there something that i am missing?

My goal is to optimize my estimation parameters of fun with a particle swarm algorithm. For each day in my sample (6009 days), i have multiple data. I want to minimize the error between observed data (implied_volatility) and my estimation data (via fun) with my 10 parameters.

A = [log_moneyness maturity]; xdata = A; ydata = implied_volatility; 
fun = @(x,xdata) (x(1)+x(2).*xdata(:,2)) + (x(3)+x(4).*xdata(:,2)).*((x(5)+x(6).*xdata(:,2)).*...
    (xdata(:,1) -(x(7)+x(8).*xdata(:,2))) + sqrt((xdata(:,1) - (x(7)+x(8).*xdata(:,2)).^2 ...
    + (x(9)+x(10).*xdata(:,2)).^2))); %original function
parameters_forall = table2array(T_param (:,2:11)); %1415596x10 matrix
%implied_volatility is a 1415596x1 matrix with original values from market
%now i want to minimize the error of my estimate by:   
tDays = size(parameters_forall(:,1));
%%
for i = 1:tDays
    x = parameters_forall(i,:);
    residue = @(x) (sum(fun(x,xdata)-implied_volatility(i)).^2);
    y = particleswarm(residue, 10);
    swarm_parameters(i,:) = y;  
end

Grateful for any help, all the best to you!

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Kai Koslowsky 2021-10-14

在 MATLAB Online 中打开

You are absolutely right, about having to adjust N. What i therefore did was:

E = grp2idx(SPX.date); %1415596x1 dates
t = unique (E); %6009 unique dates
Nt = histc(E(:),t); % gives me 6009x1 with number of occourences each day.
null = 0;
N = [null;N];% N now has the size 6010x1
nDays = size(Nt);%6009
for i=1:nDays
    kk = N(i)+1;
    ii = N (i+1);
    Xdata = xdata(kk:ii,:);
    Ydata = ydata(kk:ii,:);
    residue = @(x) sqrt((1/ii).*(sum(fun(x,Xdata)-Ydata).^2));
    y = particleswarm(residue, 10,lb,ub,[]);
    swarm_parameters(i,:) = y; 
end

So instead of subtracting from "ii", i add +1 to "kk". This does the trick as well.

My option data set goes from 1996-2020. In the early years, the average number of options ranges from 10-40 a day, in the later years it increases to 900-2000 per day (each with different maturity and strike price).

I do not however understand the meaning of your second part. When the residue is 0, does that mean it is a optimal estimation parameter?

Maybe a quick background explanation to what i am doing overall. I observed Call/Put data on the SP500 from the market, evaluated implied volatilities of these quotes, Via my function i try to fit an interpolant to them (via SVI Method by Gatheral), with adjustments for maturity and log-moneyness. Now i need to evaluate the interpolant at maturity of 30 calender days and a fine grid of strike prices (No idea yet on how to do so). Then i will interpolated my estimated implied volatilites back to option prices and will try to calculate the risk-neutral density via the work of Breeden/Litzenberger. In the end i will compare the risk neutral PDF to the PDF i observed at the market.

Walter Roberson 2021-10-14

在 MATLAB Online 中打开

residue = @(x) sqrt((1/ii).*(sum(fun(x,Xdata)-Ydata).^2));

At the moment, I do not undertand why the residue should be proportional to the index of the last entry? The implication is that if you were to have exactly the same data in (say) January 2018 and (say) March 2018, that the residue should be smaller for the second case because it would have a greater ii and so 1/ii would be smaller ??

On the other hand, the 1/ii factor acts as a constant multiple, and for any two proposed x vectors for the same ii, the relative residue is what is important for determining which is smaller residue, and the relative order is not affected by 1/ii . Likewise, unless you had imaginary components, sum of squared would always be positive, and the relative order of sqrt() of two positive numbers is the same as the relative order of the two numbers itself.

What you need to know for minimization is the relative successes. So you might as well use

residue = @(x) sum((fun(x,Xdata)-Ydata).^2);

Please note that this has an important change to your formula. Watch the location of the () !!

      fun(x,Xdata)           %invoke function to obtain prediction
                - Ydata      %subtract actual value
    (                  ).^2  %square the difference
sum(                       ) %sum the squares

whereas your formula had

        fun(x,XData)             %invoke function to obtain prediciton
                    - Ydata      %subtract actual value
    sum(                    )    %sum the differences
                             .^2 %square the sum

Note: You never store the residues, so you do not care about the absolute values, only about the relative values.

Kai Koslowsky 2021-10-15

编辑：Kai Koslowsky 2021-10-15

So do i understand you correctly, that the resulting parameters do not change with or without divisor and sqrt? I thought, that when particleswarm is trying to minimize the residue, it will minimize the result via the best parameters and not only the value of the function, so that it fits to ydata. How can i make sure, that my given formula is regarded in each point (sqrt and divisor)?

Edit

For the divisor i agree. I am rather lost on how to make sure, that my parameters are minimized by the function i sent (see Comment above).

Kai Koslowsky 2021-11-3

Dear Mr. Robinson,

would you like to post something as a answer below, so i can accept your answer? This is the least that i can do. I am truely grateful for your help. I have looked over all your comments many times, and it helped me a lot.

All the best to you!

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