In the Wikipedia article on radial basis function network, I didn't understand what was meant by "center vector for neuron i", in other words "center of the RBF units called also prototype".


In RBF networks each neuron in the hidden layer applies a computation that is related to its "center vector".

First consider the set of input neurons as a vector $\mathbf{x}$, and that each hidden layer neuron receives the complete input vector as its own input.

Second, each hidden layer neuron $i$ is parametrised through a vector (center vector) $\mathbf{c_i}$ of equal dimension than $\mathbf{x}$.

The computation of each hidden layer neuron $i$ consists of:

  1. Evaluate the distance (according to a metric, which may or may not be Euclidean) between the input $\mathbf{x}$ and the center vector $\mathbf{c_i}$. When the input is equal to the center vector, the output for that neuron will be maximal (see 2).

  2. Evaluate a Gaussian function that decays with increasing distance between the vectors. That's the output of each hidden layer neuron. In the following equation, $\beta$ is a parameter that specifies the decay rate of the Gaussian (converse to the standard deviation in a regular Gaussian distribution).

\begin{equation} \rho_i(\mathbf{x})=\exp{[-\beta\Vert\mathbf{x}-\mathbf{c_i}\Vert^2]} \end{equation}

Now combine linearly the output of all hidden layer neurons to obtain the total neural network output. For that, you need an additional set of parameters or weights $a_i$ (one $a_i$ for each hidden layer neuron, but the $a_i$ values are scalars while each $\mathbf{c_i}$ is a vector).

The final equation, that yields the total network output (a scalar value), is:

\begin{equation} \varphi(\mathbf{x}) = \sum_{i=1}^N a_i \rho(||\mathbf{x}-\mathbf{c}_i||) \end{equation}

And that's it.

| improve this answer | |
  • $\begingroup$ I have all the honor if you can response me to the question "ventral stream architecture" asked now in this site :):) $\endgroup$ – Christina Nov 2 '13 at 17:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.