In Andrew Ng's lectures he introduces Locally Weighted Linear Regression (page 15) for which he describes the weight as:

$$ \begin{align} w^{(i)} = exp\Big(-\frac{(x^{(i)} - x)^2}{2 \tau^2}\Big) \end{align} $$

However in the problem assignment for Locally Weighted Logistic Regression, the weight is described as:

$$ \begin{align} w^{(i)} = exp\Big(-\frac{||x^{(i)} - x||^2}{2 \tau^2}\Big) \end{align} $$

where $x^{(i)}$ had dimensions $1 \times n$ and $x$ has dimensions $m \times n$.

Looking at the solution set, it seems that they first get the difference between $x$ and $x^{(i)}$ (giving an $m \times n$ matrix) followed by element-by-element squaring (using the Matlab/Octave .^2 operator), and then sum along the rows to give a $m \times 1$ vector. In the linear algebra review notes (pages 7-8), multiple types of norms are described, but the sequence of operations performed doesn't seem to be consistent with any of the descriptions of norms. What is going on here?


1 Answer 1


First off, because I am not sure that this wasn't a part of your question, the norm they are using is defined as $|| \mathbf{x} || = \sqrt{\sum_{i=1}^n x_i^2}$, so if we take the norm of $\mathbf{x} - \mu$ (where $\mu$ is our mean or center) and square it, we get $|| \mathbf{x} - \mu ||^2 = \sum_{i=1}^n (x_i - \mu)^2$. The two expressions are equivalent.

It sounds like they are assigning a different center, $\mu_{i,j}$ for each $x_i$ and each hidden node $h_j$ for $j = 1, 2, \ldots, m$. In that case, for each input value $x_i$ and hidden node $h_j$, you would have:

$$ w_{i, j} = \exp \bigg(\frac{(x_i - \mu_{i,j})^2}{2\tau_j^2} \bigg) $$

The input to node $h_j$ is then the sum of these values over all inputs $x_i$ for that node: $\sum_{i = 1}^n w_{i, j}$. I think what they are doing in the code is just the vectorized form of this.

  • $\begingroup$ Ah feel so stupid! Didn't write it out and so didn't see that the outer square was cancelling out the square root. Thank you! $\endgroup$
    – oort
    Mar 11, 2017 at 18:58

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