案例学习：预测房价

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來自 University of Washington 的課程

机器学习：回归

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案例学习：预测房价

從本節課中

Ridge Regression

You have examined how the performance of a model varies with increasing model complexity, and can describe the potential pitfall of complex models becoming overfit to the training data. In this module, you will explore a very simple, but extremely effective technique for automatically coping with this issue. This method is called "ridge regression". You start out with a complex model, but now fit the model in a manner that not only incorporates a measure of fit to the training data, but also a term that biases the solution away from overfitted functions. To this end, you will explore symptoms of overfitted functions and use this to define a quantitative measure to use in your revised optimization objective. You will derive both a closed-form and gradient descent algorithm for fitting the ridge regression objective; these forms are small modifications from the original algorithms you derived for multiple regression. To select the strength of the bias away from overfitting, you will explore a general-purpose method called "cross validation". <p>You will implement both cross-validation and gradient descent to fit a ridge regression model and select the regularization constant.

- Emily FoxAmazon Professor of Machine Learning

Statistics - Carlos GuestrinAmazon Professor of Machine Learning

Computer Science and Engineering

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Okay.

So now let's think a little bit about the solution.

So this is our w hat ridge.

And what happens if I set lambda equal to 0?

Well, I get w hat ridge is equal to

H transpose H inverse, H transpose y.

And that might look very familiar to you.

That was exactly equal to our w hat least squares, our old solution,

before we introduced this notion of ridge regression.

And what if set lambda all the way to infinity?

Well then i get w hat ridge equals zero.

Because it's like dividing by infinity.

When we have this infinity appearing in this inverse.

Remember the inverse was like our matrix analog of division, so

that's intuition for why w hat ridge is exactly equal to zero.

Okay.

So this is a little sanity check.

That when lambda is equal to zero, this

closed form solution we have is exactly equal to our least square solution.

That's what we had discussed at the very beginning of this module.

And likewise, when we crank lambda all the way up to infinity,

our solution is equal to zero.

But now what we have is we have a closed form for

what the solution is for some lambda in between zero and infinity.

Let's also recall the discussion we had about our previous solution,

w hat least squares, where we said this H

transpose H looks exactly like the following where

this is our little cartoon of our H matrix, where the number for H transpose.

Let me actually write it here, it'll be clearer.

The number of rows is equivalent to the number of observations.

That's N and the number of columns

is equal to the number of features,

which we denote as D.

And what we said not last module, but the module before that,

we said H transpose H, the multiplication of these two matrices is invertible.

In general, if the number of observations is greater than the number of features,

but really it's the number of linearly independent observations being greater

than the number of features, and

we said the complexity of this inverse is cubic in the number of features D.

Well now let's think about similar properties, but

of our ridge regression solution.

Where this H transpose H is exactly like we had it before.

But now before we take our inverse, we're adding lambda times the identity matrix.

And when you take a scaler and multiply by the identity matrix,

you just get that value along the diagonal, so we get a whole bunch of

lambdas along the diagonal and zero everywhere else in this matrix.

So what ends up happening now is the result, H transpose H plus lambda

times identity, is invertible always when lambda is greater than zero.

Even if the number of observations or number of

linearly independent observations is less than the number of features.

So this is really important.

When you have lots of features.

So for large D, which has lots of features, and remember,

that's how we motivated using ridge regression.

We're in these really complicated models where you have lots and

lots of features, a lot of flexibility and the potential to over fit.

Now we see something very explicit about how it helps us.

And just to return to the discussion on the naming

of ridge regression being called a regularization technique.

If you remember,

I said that we're regularizing our standardly square solution.

Well we can see that here, because lambda

times the identity is making H transpose

H plus lambda identity more regular.

That's what's allowing us to do this inverse even in this

other situation, this harder situation, and

because this result is more regular, we call it regularized.

Okay.

But, the complexity of the inverse is still cubic in the number of features we

have and often when we're thinking about ridge regression, like I said we're

thinking about cases where you have lots and lots of features, so doing this close

form solution that we've shown here can be computationally prohibitive.

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