# How to make predictions using smoothing splines

In ordinary least squares regression, for outcome vector $$y$$ and design matrix $$X$$ (full rank), the estimated coefficient values are $$\hat{\beta} = (X^TX)^{-1}X^TY$$. Given a new set of covariates $$X_{new}$$, the predicted values are $$y_{new} = X_{new}\hat{\beta}.$$

For smoothing splines,

My question: is $$\hat{\theta}$$ analogous to $$\hat{\beta}_{OLS}$$? That is, are the predicted values given by a smoothing spline $$y_{new} = X_{new}\hat{\theta}$$?

Smoothing splines are a basis expansion method and for all of these methods prediction goes the same way. We have a model like $$f(x) = \sum_{j=1}^m \beta_j h_j(x)$$ where the $$h_j$$ are completely known and the only thing to be estimated is $$\beta$$ which determines the relative weights of the basis functions. For a point $$x_0 \in \mathbb R^p$$, we first represent it in terms of the basis functions as $$(h_1(x_0), \dots, h_m(x_0))\in\mathbb R^m$$ and then it's like a linear regression on this new representation of $$x_0$$ so our prediction is $$\langle (h_1(x_0), \dots, h_m(x_0)), \beta\rangle = \sum_{j=1}^m \beta_j h_j(x_0) = f(x_0).$$ The mapping from $$\mathbb R^p$$ to $$\mathbb R^m$$ given by $$x_0 \mapsto (h_1(x_0), \dots, h_m(x_0))$$ is completely known (when we choose to use a cubic regression spline, say, we've implicitly chosen our basis functions [up to changes of basis]), so once $$\hat\beta$$ is obtained from the training data we have everything we need to predict for any new point $$x_0$$.
In your case the form of $$\hat\theta$$ represents the fact that a smoothing spline leads to a generalized ridge regression, so $$\theta$$ is estimated with shrinkage, but for prediction we just do $$x_0 \mapsto \sum_j \hat\theta_j N_j(x_0)$$ which is exactly because this is a basis expansion method.