# Does L2-normalization of ridge regression punish intercept? If not, how to solve its derivative?

I am new to ML. I was informed that the L2-normalization of ridge regression does not punish the intercept $\theta_{0}$. As in the cost function: $$\nabla_{\theta}J(\theta)=\frac{1}{2}\sum_{i=1}^{m}(h_{\vec \theta}(x^{(i)})-y^{(i)})^2+\lambda\sum_{j=1}^{n}{\theta_{j}^{2}}$$ The L2-normalization term $\lambda\sum_{j=1}^{n}{\theta_{j}^{2}}$ only sums from $j=1$ to $n$, not from $j=0$ to $n$. I also read that:

in most cases (all cases?), you're better off not regularizing $\theta_{0}$, since its unlikely to reduce overfitting and shrinks the space of representable functions

which comes from the last answer of user48956 of Why is a zero-intercept linear regression model predicts better than a model with an intercept?

I am confused about how to solve the derivative of the cost function, since: $$\nabla_{\theta}J(\theta)=\frac{1}{2}（X\theta-Y）^{T}（X\theta-Y）+\lambda(\theta^{'})^{T}\theta^{'},$$ where $\theta^{'}=\left[ \begin{matrix} \theta_{1} \\ \theta_{2} \\ ...\\ \theta_{n} \end{matrix} \right]$ , $\theta=\left[ \begin{matrix} \theta_{0} \\ \theta_{1} \\ ...\\ \theta_{n} \end{matrix} \right]$ and $X=\left[ \begin{matrix} 1 & X_{1}^{(1)} & X_{2}^{(1)} & ...& X_{n}^{(1)} \\ 1 & X_{1}^{(2)} & X_{2}^{(2)} & ...& X_{n}^{(2)} \\ ...\\ 1 & X_{1}^{(m)} & X_{2}^{(m)} & ...& X_{n}^{(m)} \end{matrix} \right]$.

$\theta^{'}$ and $\theta$ are different. Hence they cannot be mixed from my point of view. And the derivative is about $\theta$，which contains $\theta^{'}$. After googling and viewing the questions on this forum, there is still no way for me to get the solution: $$\theta=(X^TX+\lambda*I)^{-1}X^TY$$ Can anybody give me a clue? Thanks in advance for your help!

However, I think there are two quick fixes to this problem:

First of all, we do not add the all 1 column to $X$. Namely $X=\left[ \begin{matrix} X_{1}^{(1)} & X_{2}^{(1)} & ...& X_{n}^{(1)} \\ X_{1}^{(2)} & X_{2}^{(2)} & ...& X_{n}^{(2)} \\ ...\\ X_{1}^{(m)} & X_{2}^{(m)} & ...& X_{n}^{(m)} \end{matrix} \right]$. That is to say we do not include the intercept at all in the model:$$y=\theta_{1}X_{1}+\theta_{2}X_{2}+...+\theta_{n}X_{n}.$$ I believe this method is adopted in the classic book Machine Learning in Action by Peter Harrington which I am currently reading. In its implementation of ridge regression (P166 and P177 if you also have the book), all the $X$ passed to ridge regression does not have the all 1 column. So no intercept is fitted at all.

Secondly, the intercept is also being punished in reality.

scikit's logistic regression regularizes the intercept by default.

which once again comes from the last answer of user48956 of Why is a zero-intercept linear regression model predicts better than a model with an intercept?

Both of the two quick fixes lead to the solution $$\theta=(X^TX+\lambda*I)^{-1}X^TY.$$

So can the derivative of L2-normalization of ridge regression actually being solved or are just solved by quick fixes?

• See also stats.stackexchange.com/questions/86991. The answer to your Q is that everything can be solved: simply decompose the first term into theta_0 and theta_prime. You can immediately solve for optimal theta_0: it's the overall mean. Leading to your "quick fix #1" as the proper solution for theta_prime on the mean-subtracted data. Jan 8 '18 at 15:12
• The second solution is pretty universally considered incorrect. The penalization of the intercept is one of a few design errors in sklearn. Jan 8 '18 at 15:17
• @MatthewDrury I think I mentioned a hack to "fix" this and made you sad. But I want to tell OP that the hack is make a huge intercept, such as 1e9. Then the regularization takes little effect on it. Jan 8 '18 at 16:27
• Yah, you remeber correctly, everytime i do that i die inside a little. Jan 8 '18 at 16:44
• @amoeba Thanks a lot! I work out the solution in my answer under your guidance. Can you help me check it? What's more, I discover that Machine Learning in Action subtracts mean from $Y$ and not adds the all 1 column to $X$ which are both right way to go. However, when doing feature standardizing, it divides $X-\mu$ by variance not by standard deviation! I asked another question for this: not by standard deviation Can you help me again on this question? Thanks! Jan 9 '18 at 10:41

The Elements of Statistical Learning by Hastie et al. points out in P63 that:

the intercept $\theta_{0}$ has been left out of the penalty term

Furthermore, it says:

The ridge solutions are not equivariant under scaling of the inputs, and so one normally standardizes the inputs before solving (3.41) (3.41 is the cost function). It can be shown (Exercise 3.5) that the solution to (3.41) can be separated into two parts, after reparametrization using centered inputs: each $X_{j}^{(i)}$ gets replaced by $X_{j}^{(i)}-\overline{x_{j}}.$ We estimate $\theta_{0}$ by $\overline{y}=\frac{1}{m}\sum_{i=1}^{m}y^{(i)}$ The remaining coefficients get estimated by a ridge regression without intercept, using the centered $X_{j}^{(i)}$. Henceforth we assume that this centering has been done, so that the input matrix $X$ has $n$ (rather than $n + 1$) columns.

Although I wonder why The Elements of Statistical Learning first suggests feature standardization and then only feature centering is conducted. Maybe to agree with Exercise 3.5 which only uses feature centering.

Anyway, I believe it's right to apply z-score standardization to features. So I now try to solve the derivative of the cost function of ridge regression following the suggestion of the above commenter amoeba. Thanks him or her a lot!

First, the cost function: $$\nabla_{ \theta}J(\theta)=\frac{1}{2}\sum_{i=1}^{m}(y_{i}-\theta_{0}-\frac{X_{1}^{(i)}-\overline{X_1}}{\sigma_1}\theta_1-\frac{X_{2}^{(i)}-\overline{X_2}}{\sigma_2}\theta_2-...-\frac{X_{n}^{(i)}-\overline{X_n}}{\sigma_n}\theta_n)^2+\lambda\sum_{j=1}^{n}{\theta_{j}^{2}},$$ where $\overline{X_j}$ is the mean of attribute $X_{j}$ and $\sigma_j$ is the standard deviation of $X_{j}$. To make it shorter: $$\nabla_{ \theta}J(\theta)=\frac{1}{2}\sum_{i=1}^{m}(y_{i}-\theta_{0}-\sum_{j=1}^{n}\frac{X_j^{(i)}-\overline{X_j}}{\sigma_{j}}\theta_j)^2+\lambda\sum_{j=1}^{n}{\theta_{j}^{2}}$$ Now we first compute the value of $\theta_0$ in the above expression by setting the derivative with respect to $\theta_0$ equal to zero. Since $\lambda\sum_{j=1}^{n}{\theta_{j}^{2}}$ does not have $\theta_{0}$, we get: $$\nabla_{ \theta_0}J(\theta)=-\sum_{i=1}^{m}(y_{i}-\theta_{0}-\sum_{j=1}^{n}\frac{X_j^{(i)}-\overline{X_j}}{\sigma_{j}}\theta_j)=0$$ That is: $$\sum_{i=1}^{m}(y_{i}-\theta_{0})-\sum_{i=1}^{m}\sum_{j=1}^{n}\frac{X_j^{(i)}-\overline{X_j}}{\sigma_{j}}\theta_j=0$$ As $$\sum_{i=1}^{m}\sum_{j=1}^{n}\frac{X_j^{(i)}-\overline{X_j}}{\sigma_{j}}\theta_j=0$$ (because $\overline{X_j}$ is the mean of attribute $X_{j}$ ), so now we have $$\sum_{i=1}^{m}(y_{i}-\theta_{0})=0,$$ obviously: $$\theta_0=\overline{y}=\frac{1}{m}\sum_{i=1}^{m}y^{(i)}$$

So the intercept of feature-standardized ridge regression is always $\overline{y}$. Hence if we first centralize $Y$ by subtracting its mean (get $(y_i)^{'}$ for data example $i$), not include all 1 column in $X$, and then do feature standardization on $X$ (get $(X_j^{(i)})^{'}$ for $X_{j}$ of data example $i$), the cost function will simply be $$\nabla_{ \theta}J(\theta)=\frac{1}{2}\sum_{i=1}^{m}((y_{i})^{'}-\sum_{j=1}^{n}(X_j^{(i)})^{'}\theta_j)^2+\lambda\sum_{j=1}^{n}{\theta_{j}^{2}}$$ That is $$\nabla_{\theta}J(\theta)=\frac{1}{2}（X^{'}\theta-Y^{'}）^{T}（X^{'}\theta-Y^{'}）+\lambda(\theta)^{T}\theta,$$ where $\theta=\left[ \begin{matrix} \theta_1 \\ \theta_2 \\ ...\\ \theta_n \end{matrix} \right]$, $X^{'}$ does not have all 1 column and standardized of $X$, $Y^{'}$ is centered with respect to $Y$. Now $\theta$ (without $\theta_0$) can be solved with: $$\theta=((X^{'})^TX^{'}+\lambda*I)^{-1}(X^{'})^TY^{'}$$ For standardized features, the linear model will be $$y=\overline{y}+\theta{_1}X_1^{'}+\theta{_2}X_2^{'}+...+\theta{_n}X_n^{'}---(1),$$ where $$X_i^{'}=\frac{X_{i}-\overline{X_i}}{\sigma_i}---(2)$$ If we use (2) in (1) as suggested in the answer of Plasty Grove. So for origin input data, the linear model will be
$$y=\overline{y}+\frac{X_{1}-\overline{X_1}}{\sigma_1}\theta_1+\frac{X_{2}-\overline{X_2}}{\sigma_2}\theta_2+...+\frac{X_{n}-\overline{X_n}}{\sigma_n}\theta_n$$ That is $$y=\frac{\theta_1}{\sigma_1}X_1+\frac{\theta_2}{\sigma_2}X_2+...+\frac{\theta_n}{\sigma_n}X_n+\overline{y}-\frac{\overline{X_1}}{\sigma_1}\theta_1-\frac{\overline{X_2}}{\sigma_2}\theta_2-...-\frac{\overline{X_n}}{\sigma_n}\theta_n$$ That's why after we solve coefficients of standardized features, to return coefficients of origin input data (unstandardized features), we must return $\theta_i/\sigma_i$

• Nice detailed example. A couple of comments: you comment on the effect of centering $Y$, but to omit an intercept term and obtain correct estimates, one must center all $X$ features as well. However I agree this example agrees with the rationale for not penalizing the intercept term (to obtain consistent inference). Second, intuition should serve for something here. We all accept that predicting $Y$ by its mean is akin to a 0 parameter model, so to get the additive effect of $X$ in the model, we mustn't penalize the term which merely catches the mean-Y effect in the presence of $X$. Mar 15 '18 at 19:05