How to use scipy's least_squares

Question

I am trying to implement a simple model estimation in Python. I have an ARCH model:

logR_t = u + theta_1 * logR_t + \epsilon_t

where logR_t are my log-returns vector, u and theta_1 are the two parameters to be estimated and \epsilon_t are my residuals.

In Matlab, I have the following lines to call the optimiser on the function Error_ARCH. The initial guess for the parameters is 1, their lower bounds are -10 and upper bounds are 10.

ARCH.param = lsqnonlin(  @(param) Error_ARCH(param, logR), [1 1], [-10 -10], [10 10]);
[ARCH.Error, ARCH.Residuals] = Error_ARCH( ARCH.param, logR);

Where the error to minimise is given as:

function [error, residuals] = Error_ARCH(param, logreturns)

         % Initialisation
         y_hat = zeros(length(logreturns), 1 );

         % Parameters
         u      = param(1);
         theta1 = param(2);

         % Define model
         ARCH =@(z)  u + theta1.*z;

         for i = 2:length(logreturns)
             y_hat(i) = ARCH( logreturns(i-1) );    
         end

         error = abs( logreturns - y_hat );

         residuals = logreturns - y_hat;

end

I would like a similar thing in Python but I am stuck since I do not know where to specify the arguments to the least_squares function in SciPy. So far I have:

from scipy.optimize import least_squares

def model(param, z):
    """This is the model we try to estimate equation"""
    u = param[0]
    theta1 = param[1]

    return u + theta1*z


def residuals_ARCH(param, z):
    return z - model(param, z)

When I call the lsq optimisizer, I get an error: residuals_ARCH() missing 1 required positional argument: 'z'

guess = [1, 1]
result = least_squares(residuals_ARCH, x0=guess, verbose=1, bounds=(-10, 10))

Thank you for all your help

Answer 1

The least_squares method expects a function with signature fun(x, *args, **kwargs). Hence, you can use a lambda expression similar to your Matlab function handle:

# logR = your log-returns vector
result = least_squares(lambda param: residuals_ARCH(param, logR), x0=guess, verbose=1, bounds=(-10, 10))