# Why do we use the Unregularized Cost to plot a Learning Curve?

I'm taking Andrew Ng's Machine Learning Course.
In the section on determining the variance/bias of your model, he suggests the following.

For a given regularization parameter and set of features
Create differently sized subsets of your training data.
For each training data subset, using regularization,
~ train a model,
~ then calculate the error on the subset and the error on the validation set.

Once that's done,
plot the unregularized cost for both training and validation sets as a function of the size of the training data subset.

The idea is that, if the training error and validation error remain very different at large training set sample sizes then the model has high variance.
If training error and validation error converge too quickly then the model has high bias.

My question is...
Since the models we're testing were calculated using a regularization constant, why aren't we plotting regularized cost as a function of the training data size?

• Yeah. That's what he says, right? Jul 7, 2016 at 1:01
• Yeah, right after dinner. Jul 7, 2016 at 1:06
• Q1) Why don't we include regularization when computing J-train, J-cv, and J-test? Regularization is built-in to theta when you train the system. We do not need to include it twice. When we measure J-train, J-cv, and J-test, we want to measure the true error, without any additional penalties. Jul 7, 2016 at 1:24
• That's the FAQ answer I got pointed to when I asked on the Coursera Forum. Jul 7, 2016 at 1:25
• I erased my comments to create space. I think I got the wrong videoclip after all... Oh well,... I think the idea is the same, though: you train the set with regularization, get your estimated parameters, and then use these parameters (no need for regularization to blunt their overfitting effect any more) to check whether the model has high bias or high variance, and whether getting more training data is likely to help. Jul 7, 2016 at 13:55

Background: I believe you are referring to this lecture dealing with Regularization and Bias/Variance in the context of polynomial regression.

The algorithm fmincg produces optimized estimated $\hat \theta$ coefficients (or parameters), based on a gradient descent computation derived from the objective function:

$$J(\theta)=\frac{1}{2m}\left(\displaystyle\sum_{i=1}^m(h_{\theta}(x^{(i)})-y^{(i)})^2\right)+\frac{\lambda}{2m}\left(\sum_{j=i}^n\theta_j^2\right)$$

where $m$ is the number of examples (or subjects/observations), each denoted as $x^{(i)}$; $j$ the number of features; and $\lambda$ the regularization parameter. The optimization gradients include the regularization $\lambda$ for each parameter other than $\theta_0$: it is found in the expression: $\frac{\lambda}{m}\theta_j$ after differentiating the equation above.

The issue at hand is to select the optimal $\lambda$ value to prevent overfitting the data, but also avoiding high bias.

To this end, a vector of possible lambda values is supplied, which in the course exercise is $[0,0.001,0.003,0.01,0.03,0.1,0.3,1,3,10]$, to optimize the coefficients $\Theta$. In this process, and for each iteration through the different lambda values, all other factors (basically the model matrix) remain constant.

Consequently, the differences between the $\Theta$ vectors of parameters that will be obtained are a direct consequence of the different regularization parameters $\lambda$ chosen.

At each iteration and using gradient descent the parameters that minimize the objective function are calculated on the entire training set to eventually plot a validation curve of squared errors over lambda values. This is different than in the case of the learning curves (cost vs. number of examples), where the training set is segmented in increasing numbers of observations as explained right here.

At this point, we have obtained optimal estimated parameters on the training set, and their differences are directly related to the regularization parameter.

Therefore, it makes sense to now set aside the regularization and see what would be the cost or errors, applying each different set of $\Theta$'s to both the training and cross validation sets, looking for a minimum in the crossvalidation set errors. We are not looking to optimize further the parameters $\theta$, we are just checking how the choice of different $\lambda$ values (with its associated coefficients) is reflected in the loss (or cost) function, initially dropping the errors, but eventually, and after having taken care of overfitting, progressively increasing these errors due to bias: This explains why the training error (cost or loss function) is defined as:

$$J_{train}=\frac{1}{2m}\left[\displaystyle\sum_{i=1}^m(h_{\theta}(x^{(i)})-y^{(i)})^2\right]$$

and accordingly, the CV error as:

$$J_{cv}=\frac{1}{2m}\left[\displaystyle\sum_{i=1}^m(h_{\theta}(x^{(i)}_{cv})-y^{(i)}_{cv})^2\right]$$

Basically, the squared errors. In a way the confusion stems from the similarity between the function to minimize by choosing optimal parameters (objective function), and the cost or loss function, meant to assess the errors.

Because if you want to know the actual cost, you need to look at the unregularized cost.

Consider LASSO regression, where we we tell a kind of mathematical white lie in the formula by adding a term that reflects the sum of the parameter values. Adding this term influences the parameter estimates. But at the end of the day when we want to make predictions with this model, we will be evaluating those predictions purely using the sum of squared prediction errors, without the extra term.

For an analogy, think about setting your alarm clock a half hour ahead so that you never wake up late. Sure, you wake up a half hour early, but you don't go through the rest of the day thinking that it's a half hour later than it is. The time on your alarm clock is the regularized cost, and the time on everyone else's clock is the unregularized cost.