For linear regression, we have previously worked out two learning algorithms. One based on gradient descent and one based on the normal equation. In this video, we'll take those two algorithms and generalize them to the case of regularized linear regression. Here's the optimization objective that we came up with last time for regularized linear regression. This first part is our usual objective for linear regression. And we now have this additional regularization term, where lambda is our regularization parameter, and we like to find parameters theta that minimizes this cost function, this regularized cost function, J of theta. Previously, we were using gradient descent for the original cost function without the regularization term. And we had the following algorithm, for regular linear regression, without regularization, we would repeatedly update the parameters theta J as follows for J equals 0, 1, 2, up through n. Let me take this and just write the case for theta 0 separately. So I'm just going to write the update for theta 0 separately than for the update for the parameters 1, 2, 3, and so on up to n. And so this is, I haven't changed anything yet, right. This is just writing the update for theta 0 separately from the updates for theta 1, theta 2, theta 3, up to theta n. And the reason I want to do this is you may remember that for our regularized linear regression, we penalize the parameters theta 1, theta 2, and so on up to theta n. But we don't penalize theta 0. So, when we modify this algorithm for regularized linear regression, we're going to end up treating theta zero slightly differently. Concretely, if we want to take this algorithm and modify it to use the regular rise objective, all we need to do is take this term at the bottom and modify it as follows. We'll take this term and add minus lambda over m times theta j. And if you implement this, then you have gradient descent for trying to minimize the regularized cost function, j of theta. And concretely, I'm not gonna do the calculus to prove it, but concretely if you look at this term, this term hat I've written in square brackets, if you know calculus it's possible to prove that that term is the partial derivative with respect to J of theta using the new definition of J of theta with the regularization term. And similarly, this term up on top which I'm drawing the cyan box, that's still the partial derivative respect of theta zero of J of theta. If you look at the update for theta j, it's possible to show something very interesting. Concretely, theta j gets updated as theta j minus alpha times, and then you have this other term here that depends on theta J. So if you group all the terms together that depend on theta j, you can show that this update can be written equivalently as follows. And all I did was add theta j here is, so theta j times 1. And this term is, right, lambda over m, there's also an alpha here, so you end up with alpha lambda over m multiplied into theta j. And this term here, 1 minus alpha times lambda m, is a pretty interesting term. It has a pretty interesting effect. Concretely this term, 1 minus alpha times lambda over m, is going to be a number that is usually a little bit less than one, because alpha times lambda over m is going to be positive, and usually if your learning rate is small and if m is large, this is usually pretty small. So this term here is gonna be a number that's usually a little bit less than 1, so think of it as a number like 0.99, let's say. And so the effect of our update to theta j is, we're going to say that theta j gets replaced by theta j times 0.99, right? So theta j times 0.99 has the effect of shrinking theta j a little bit towards zero. So this makes theta j a bit smaller. And more formally, this makes the square norm of theta j a little bit smaller. And then after that, the second term here, that's actually exactly the same as the original gradient descent update that we had, before we added all this regularization stuff. So, hopefully this gradient descent, hopefully this update makes sense. When we're using a regularized linear regression and what we're doing is on every iteration we're multiplying theta j by a number that's a little bit less then one, so its shrinking the parameter a little bit, and then we're performing a similar update as before. Of course that's just the intuition behind what this particular update is doing. Mathematically what it's doing is it's exactly gradient descent on the cost function J of theta that we defined on the previous slide that uses the regularization term. Gradient descent was just one of our two algorithms for fitting a linear regression model. The second algorithm was the one based on the normal equation, where what we did was we created the design matrix X where each row corresponded to a separate training example. And we created a vector y, so this is a vector, that's an m dimensional vector. And that contained the labels from my training set. So whereas X is an m by (n+1) dimensional matrix, y is an m dimensional vector. And in order to minimize the cost function J, we found that one way to do so is to set theta to be equal to this. Right, you have X transpose X, inverse, X transpose Y. I'm leaving room here to fill in stuff of course. And what this value for theta does is this minimizes the cost function J of theta, when we were not using regularization. Now that we are using regularization, if you were to derive what the minimum is, and just to give you a sense of how to derive the minimum, the way you derive it is you take partial derivatives with respect to each parameter. Set this to zero, and then do a bunch of math and you can then show that it's a formula like this that minimizes the cost function. And concretely, if you are using regularization, then this formula changes as follows. Inside this parenthesis, you end up with a matrix like this. 0, 1, 1, 1, and so on, 1, until the bottom. So this thing over here is a matrix whose upper left-most entry is 0. There are ones on the diagonals, and then zeros everywhere else in this matrix. Because I'm drawing this rather sloppily. But as a example, if n = 2, then this matrix is going to be a three by three matrix. More generally, this matrix is an (n+1) by (n+1) dimensional matrix. So if n = 2, then that matrix becomes something that looks like this. It would be 0, and then 1s on the diagonals, and then 0s on the rest of the diagonals. And once again, I'm not going to show this derivation, which is frankly somewhat long and involved, but it is possible to prove that if you are using the new definition of J of theta, with the regularization objective, then this new formula for theta is the one that we give you, the global minimum of J of theta. So finally I want to just quickly describe the issue of non-invertibility. This is relatively advanced material, so you should consider this as optional. And feel free to skip it, or if you listen to it and positive it doesn't really make sense, don't worry about it either. But earlier when I talked about the normal equation method, we also had an optional video on the non-invertibility issue. So this is another optional part to this, sort of an add-on to that earlier optional video on non-invertibility. Now, consider a setting where m, the number of examples, is less than or equal to n, the number of features. If you have fewer examples than features, than this matrix, X transpose X will be non-invertible, or singular. Or the other term for this is the matrix will be degenerate. And if you implement this in Octave anyway and you use the pinv function to take the pseudo inverse, it will kind of do the right thing, but it's not clear that it would give you a very good hypothesis, even though numerically the Octave pinv function will give you a result that kinda makes sense. But if you were doing this in a different language, and if you were taking just the regular inverse, which in Octave denoted with the function inv, we're trying to take the regular inverse of X transpose X. Then in this setting, you find that X transpose X is singular, is non-invertible, and if you're doing this in different program language and using some linear algebra library to try to take the inverse of this matrix, it just might not work because that matrix is non-invertible or singular. Fortunately, regularization also takes care of this for us. And concretely, so long as the regularization parameter lambda is strictly greater than 0, it is actually possible to prove that this matrix, X transpose X plus lambda times this funny matrix here, it is possible to prove that this matrix will not be singular and that this matrix will be invertible. So using regularization also takes care of any non-invertibility issues of the X transpose X matrix as well. So you now know how to implement regularized linear regression. Using this you'll be able to avoid overfitting even if you have lots of features in a relatively small training set. And this should let you get linear regression to work much better for many problems. In the next video we'll take this regularization idea and apply it to logistic regression. So that you'd be able to get logistic regression to avoid overfitting and perform much better as well.
