Given a regression setting with covariates $X_{n \times m}$ and response $Y_{n \times p}$ where $p>1$, i.e the responses are vector-valued or multivariate, is there a Nadaraya-Watson estimator for kernel regression in this setting?

This boils down to how the following can be computed with this form of $Y$ :


But since above, $y_i$ is now multivariate as well, what happens to this multiplication operation in the numerator, in this generalization to multivariate responses?


1 Answer 1


The mathematical operation here allows also the use of a vector instead of a scalar. Think of it as a weighted sum of vectors:

$$\frac{\sum_i w_i \mathbf y_i}{\sum_k w_k} = \sum_i \left(\frac{w_i}{\sum_k w_k} \right) \mathbf y_i = \sum_i \tilde w_i \mathbf y_i$$

where the coefficients are given in terms of kernel functions

$$ w_i = K_h(\mathbf x-\mathbf x_i)\\[1em] \Rightarrow \quad\tilde w_i = \frac{K_h(\mathbf x-\mathbf x_i)}{\sum_{k=1}^{n}K_h(\mathbf x-\mathbf x_k)}$$

With this, multivariate Nadaraya-Watson kernel regression simply boils down to a one-dimensional regression in each dimension.

  • $\begingroup$ Awesome. i previously thought it would be a contradiction if we did multiple 1-d regressions as all dimensions may not have equally useful information. But it's good to know, that this is not a detrimental answer-and the right answer indeed. Thanks! $\endgroup$
    – hearse
    Nov 26, 2014 at 3:26
  • $\begingroup$ @PraneethVepakomma: still, it was slightly misleading (--whereas I think you got it right). I made a mistake the definition of the coefficients. Actually it was correct as the normalization factor cancels out, but now its more convenient. $\endgroup$
    – davidhigh
    Nov 26, 2014 at 8:47
  • $\begingroup$ how is the bandwidth h estimated here, in this case? Do you know of R-packages that implement this for multivariate Y? $\endgroup$
    – hearse
    Nov 26, 2014 at 18:10

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