Gradient descent and normal equation method for solving linear regression gives different solutions

Question

I'm working on machine learning problem and want to use linear regression as learning algorithm. I have implemented 2 different methods to find parameters theta of linear regression model: Gradient (steepest) descent and Normal equation. On the same data they should both give approximately equal theta vector. However they do not.

Both theta vectors are very similar on all elements but the first one. That is the one used to multiply vector of all 1 added to the data.

Here is how the thetas look like (fist column is output of Gradient descent, second output of Normal equation):

Grad desc Norm eq
-237.7752 -4.6736
-5.8471   -5.8467
9.9174    9.9178
2.1135    2.1134
-1.5001   -1.5003
-37.8558  -37.8505
-1.1024   -1.1116
-19.2969  -19.2956
66.6423   66.6447
297.3666  296.7604
-741.9281 -744.1541
296.4649  296.3494
146.0304  144.4158
-2.9978   -2.9976
-0.8190   -0.8189

What can cause the difference in theta(1, 1) returned by gradient descent compared to theta(1, 1) returned by normal equation? Do I have bug in my code?

Here is my implementation of normal equation in Matlab:

function theta = normalEque(X, y)
    [m, n] = size(X);
    X = [ones(m, 1), X];
    theta = pinv(X'*X)*X'*y;
end

Here is code for gradient descent:

function theta = gradientDesc(X, y)
    options = optimset('GradObj', 'on', 'MaxIter',  9999);
    [theta, ~, ~] = fminunc(@(t)(cost(t, X, y)),...
                    zeros(size(X, 2), 1), options);
end

function [J, grad] = cost(theta, X, y)
    m = size(X, 1);
    X = [ones(m, 1), X];
    J = sum((X * theta - y) .^ 2) ./ (2*m);
    for i = 1:size(theta, 1)
        grad(i, 1) = sum((X * theta - y) .* X(:, i)) ./ m;
    end
end

I pass exactly the same data X and y to both functions (I do not normalize X).

Edit 1:

Based on answers and comments I checked few my code and run some tests.

First I want to check if the problem can be cause by X beeing near singular as suggested by @user1489497's answer. So I replaced pinv by inv - and when run it I really got warning Matrix is close to singular or badly scaled.. To be sure that that is not the problem I obtained much larger dataset and run tests with this new dataset. This time inv(X) did not display the warning and using pinv and inv gave same results. So I hope that X is not close to singular any more.

Then I changed normalEque code as suggested by woodchips so now it looks like:

function theta = normalEque(X, y)
    X = [ones(size(X, 1), 1), X];
    theta = pinv(X)*y;
end

However the problem is still there. New normalEque function on new data that are not close to singular gives different theta as gradientDesc.

To find out which algorithm is buggy I have run linear regression algorithm of data mining software Weka on the same data. Weka computed theta very similar to output of normalEque but different to the output of gradientDesc. So I guess that normalEque is correct and there is a bug in gradientDesc.

Here is comparison of thetas computed by Weka, normalEque and GradientDesc:

Weka(correct) normalEque    gradientDesc
779.8229      779.8163      302.7994
  1.6571        1.6571        1.7064
  1.8430        1.8431        2.3809
 -1.5945       -1.5945       -1.5964
  3.8190        3.8195        5.7486
 -4.8265       -4.8284      -11.1071
 -6.9000       -6.9006      -11.8924
-15.6956      -15.6958      -13.5411
 43.5561       43.5571       31.5036
-44.5380      -44.5386      -26.5137
  0.9935        0.9926        1.2153
 -3.1556       -3.1576       -1.8517
 -0.1927       -0.1919       -0.6583
  2.9207        2.9227        1.5632
  1.1713        1.1710        1.1622
  0.1091        0.1093        0.0084
  1.5768        1.5762        1.6318
 -1.3968       -1.3958       -2.1131
  0.6966        0.6963        0.5630
  0.1990        0.1990       -0.2521
  0.4624        0.4624        0.2921
-12.6013      -12.6014      -12.2014
 -0.1328       -0.1328       -0.1359

I also computed errors as suggested by Justin Peel's answer. Output of normalEque gives slightly lesser squared error but the difference is marginal. What is more when I compute gradient of cost of theta using function cost (the same as the one used by gradientDesc) I got gradient near zero. Same done on output of gradientDesc does not give gradient near zero. Here is what I mean:

>> [J_gd, grad_gd] = cost(theta_gd, X, y, size(X, 1));
>> [J_ne, grad_ne] = cost(theta_ne, X, y, size(X, 1));
>> disp([J_gd, J_ne])
  120.9932  119.1469
>> disp([grad_gd, grad_ne])
  -0.005172856743846  -0.000000000908598
  -0.026126463200876  -0.000000135414602
  -0.008365136595272  -0.000000140327001
  -0.094516503056041  -0.000000169627717
  -0.028805977931093  -0.000000045136985
  -0.004761477661464  -0.000000005065103
  -0.007389474786628  -0.000000005010731
   0.065544198835505  -0.000000046847073
   0.044205371015018  -0.000000046169012
   0.089237705611538  -0.000000046081288
  -0.042549228192766  -0.000000051458654
   0.016339232547159  -0.000000037654965
  -0.043200042729041  -0.000000051748545
   0.013669010209370  -0.000000037399261
  -0.036586854750176  -0.000000027931617
  -0.004761447097231  -0.000000027168798
   0.017311225027280  -0.000000039099380
   0.005650124339593  -0.000000037005759
   0.016225097484138  -0.000000039060168
  -0.009176443862037  -0.000000012831350
   0.055653840638386  -0.000000020855391
  -0.002834810081935  -0.000000006540702
   0.002794661393905  -0.000000032878097

This would suggest that gradient descent simply did not converge to global minimum... But that is hardly the case as I run it for thousands of iterations. So where is the bug?

Solution 1

As others mentioned, an ill-conditioned hessian matrix is likely the cause of your problem.

The number of steps that standard gradient descent algorithms take to reach a local optimum is a function of the largest eigenvalue of the hessian divided by the smallest (this is known as the condition number of the Hessian). So, if your matrix is ill-conditioned, then it could take an extremely large number of iterations for gradient descent to converge to an optimum. (For the singular case, it could converge to many points, of course.)

I would suggest trying three different things to verify that an unconstrained optimization algorithm works for your problem (which it should): 1) Generate some synthetic data by computing the result of a known linear function for random inputs and adding a small amount of gaussian noise. Make sure that you have many more data points than dimensions. This should produce a non-singular hessian. 2) Add a regularization terms to your error function to increase the condition number of the hessian. 3) Use a second order method like conjugate gradient or L-BFGS rather than gradient descent to reduce the number of steps needed for the algorithm to converge. (You will probably need to do this in conjunction with #2).

Solution 2

Could you post a little more about what you X looks like? You're using pinv() which is Moore-Penrose pseudo inverse. If the matrix is ill-conditioned this could cause problems with obtaining the inverse. I would bet that the gradient-descent method is closer to the mark.