Linear Regression Gradient Descent: Feature Normalization/Scaling for Prediction

[Ackbach]

Cross-posted on SE.DS Beta.

I'm just doing a simple linear regression with gradient descent in the multivariate case. Feature normalization/scaling is a standard pre-processing step in this situation, so I take my original feature matrix $X$, organized with features in columns and samples in rows, and transform to $\tilde{X}$, where, on a column-by-column basis,
$$\tilde{X}=\frac{X-\bar{X}}{s_{X}}.$$
Here, $\bar{X}$ is the mean of a column, and $s_{X}$ is the sample standard deviation of a column. Once I've done this, I prepend a column of $1$'s to allow for a constant offset in the $\theta$ vector. So far, so good.

If I did not do feature normalization, then my prediction, once I found my $\theta$ vector, would simply be $x\cdot\theta$, where $x$ is the location at which I want to predict the outcome. But now, if I am doing feature normalization, what does the prediction look like? I suppose I could take my location $x$ and transform it according to the above equation on an element-by-element basis. But then what? The outcome of $\tilde{x}\cdot\theta$ would not be in my desired engineering units. Moreover, how do I know that the $\theta$ vector I've generated via gradient descent is correct for the un-transformed locations? I realize all of this is a moot point if I'm using the normal equation, since feature scaling is unnecessary in that case. However, as gradient descent typically works better for very large feature sets ($> 10k$ features), this would seem to be an important step. Thank you for your time!

 

[I like Serena]

If I'm not mistaken, you're trying to find $\theta$, such that $X\theta$ is as close as possible to some $y$.
That is, find the $\theta$ that minimizes $\|X\theta - y\|$.
Afterwards, the prediction is $\hat y = X\theta$.
Is that correct?

If so, then after normalization, we find $\tilde \theta$, such that $\tilde X\tilde\theta$ is as close as possible to the also normalized $\tilde y = \frac{y-\bar y}{s_y}$, yes?
In that case the prediction becomes:
$$\hat y = \bar y + \tilde X\tilde\theta s_y$$
which is in engineering units.

 

[Ackbach]

Actually, I just learned the answer (see the SE.DS Beta link): you transform the prediction location $x$ in precisely the same way you did for the columns of $X$, but component-wise. So you do this:

1. To each element of $x$, you add the mean of the corresponding column of $X$.
2. Divide the result by the standard deviation of the corresponding column of $X$.
3. Prepend a $1$ to the $x$ vector to allow for the bias. Call the result $\tilde{x}$.
4. Perform $\tilde{x}\cdot\theta$, which is the prediction value.

As it turns out, if you do feature normalization, then the $\theta$ vector DOES contain all the scaling and engineering units you need. And that actually makes sense, because you're not doing any transform to the outputs of the training data.