Backpropagation to train an evolving radial basis function neural network

Abstract

In this paper, a stable backpropagation algorithm is used to train an online evolving radial basis function neural network. Structure and parameters learning are updated at the same time in our algorithm, we do not make difference in structure learning and parameters learning. It generates groups with an online clustering. The center is updated to achieve the center is near to the incoming data in each iteration, so the algorithm does not need to generate a new neuron in each iteration, i.e., the algorithm does not generate many neurons and it does not need to prune the neurons. We give a time varying learning rate for backpropagation training in the parameters. We prove the stability of the proposed algorithm.

Appendix

Proof of Theorem 1

We select the following Lyapunov function L ₁(k − 1) as:

$$ L_{1}(k-1)=\widetilde{c}_{ij}^{2}(k-1)+\widetilde{\sigma }_{ij}^{2}(k-1)+ \widetilde{v}_{j}^{2}(k-1) $$

(20)

By updating (13), we have:

$$ \begin{aligned} \widetilde{c}_{ij}(k)&=\widetilde{c}_{ij}(k-1)-\eta (k-1)D_{1ij}(k-1)e\left( k-1\right) \\ \widetilde{\sigma }_{ij}(k)&=\widetilde{\sigma }_{ij}(k-1)-\eta (k-1)D_{2ij}(k-1)e\left( k-1\right) \\ \widetilde{v}_{j}(k)&=\widetilde{v}_{j}(k-1)-\eta (k-1)D_{3j}(k-1)e\left( k-1\right)\\ \end{aligned} $$

Now we calculate ΔL ₁(k − 1):

$$ \begin{aligned} \Updelta L_{1}(k-1)&=\left[ \widetilde{c}_{ij}(k-1)-\eta (k-1)D_{1ij}(k-1)e\left( k-1\right) \right] ^{2} \\ &\quad-\widetilde{c}_{ij}^{2}(k-1)+\left[ \widetilde{\sigma }_{ij}(k-1)-\eta (k-1)D_{2ij}(k-1)e\left( k-1\right) \right] ^{2} \\ &\quad-\widetilde{\sigma }_{ij}^{2}(k-1)+\left[ \widetilde{v}_{j}(k-1)-\eta (k-1)D_{3j}(k-1)e\left( k-1\right) \right] ^{2} \\ &\quad-\widetilde{v}_{j}^{2}(k-1) \\ &=\eta ^{2}(k-1)\left\{ D_{1ij}^{2}(k-1)+D_{2ij}^{2}(k-1)+D_{3j}^{2}(k-1)\right\} e^{2}\left( k-1\right) \\ &\quad-2\eta (k-1)\left[ D_{1ij}(k-1)\widetilde{c}_{ij}(k-1)+D_{2ij}(k-1) \widetilde{\sigma }_{ij}(k-1)+D_{3j}(k-1)\widetilde{v}_{j}(k-1)\right] e\left( k-1\right) \end{aligned} $$

(21)

Substituting (12) into the last term of (21) and using (14) gives:

$$ \begin{aligned} \Updelta L_{1}(k-1)&=-2\eta (k-1)\left[ e\left( k-1\right) -\zeta (k-1)\right] e\left( k-1\right) \\ &\quad+\eta ^{2}(k-1)\left\{ D_{1ij}^{2}(k-1)+D_{2ij}^{2}(k-1)+D_{3j}^{2}(k-1)\right\} e^{2}\left( k-1\right) \\ &\leq \eta ^{2}(k-1)\left\{ 1+D_{1ij}^{2}(k-1)+D_{2ij}^{2}(k-1)+D_{3j}^{2}(k-1)\right\} e^{2}\left( k-1\right) \\ &\quad-\eta (k-1)e^{2}\left( k-1\right) +\eta (k-1)\zeta ^{2}(k-1) \\ &\leq \eta ^{2}(k-1)\left\{ 1+q(k-1)\right\} e^{2}\left( k-1\right) \\ &\quad-\eta (k-1)e^{2}\left( k-1\right) +\eta (k-1)\zeta ^{2}(k-1) \\ &\leq -\eta (k-1)\left\{ 1-\eta (k-1)\left[ 1+q(k-1)\right] \right\} e^{2}\left( k-1\right) \\ &\quad+\eta (k-1)\zeta ^{2}(k-1)\\ \end{aligned} $$

Using the case $e^{2}\left( k-1\right) \geq \frac{\overline{\zeta }^{2}}{1-\eta}$ of dead zone (14), then $\eta (k-1)={\frac{\eta _{0}}{1+q(k-1)}}>0:$

$$ \begin{aligned} \Updelta L_{1}(k-1)&\leq -\eta (k-1)\left\{ 1-\frac{\eta _{0}}{1+q(k-1)}\left[ 1+q(k-1)\right] \right\} e^{2}\left( k-1\right) \\ &\quad+\eta (k-1)\zeta ^{2}(k-1) \\ \Updelta L_{1}(k-1)&\leq -\eta (k-1)\left( 1-\eta _{0}\right) e^{2}\left( k-1\right) +\eta (k-1)\zeta ^{2}(k-1)\\ \end{aligned} $$

With $\zeta ^{2}\left( k-1\right) \leq \overline{\zeta }^{2}:$

$$ \Updelta L_{1}(k-1)\leq -\eta (k-1)\left[ \left( 1-\eta _{0}\right) e^{2}\left( k-1\right) -\overline{\zeta }^{2}\right] $$

(22)

From the dead-zone, $e^{2}\left( k-1\right) \geq \frac{\overline{\zeta }^{2}}{1-\eta _{0}}$ and η(k − 1) > 0, ΔL ₁(k − 1) ≤ 0. L ₁(k) is bounded. If $e^{2}\left( k-1\right) <\frac{\overline{\zeta }^{2}}{1-\eta _{0}},$ from (14) we know η(k − 1) = 0, all of weights are not changed, they are bounded, so L ₁(k) is bounded.

When $e^{2}\left( k-1\right) \geq \frac{\overline{\zeta }^{2}}{1-\eta _{0}},$ summarize (22) from 2 to T:

$$ \sum\limits_{k=2}^{T}\eta (k-1)\left[ \left( 1-\eta _{0}\right) e^{2}\left( k-1\right) -\overline{\zeta }^{2}\right] \leq L_{1}(1)-L_{1}(T) $$

(23)

Since L ₁(T) is bounded and using $\eta (k-1)=\frac{\eta _{0}}{1+q(k-1)}>0:$

$$ \underset{T\rightarrow \infty }{\lim}\sum\limits_{k=2}^{T}\left( \frac{\eta _{0}}{1+q(k-1)}\right) \left[ \left( 1-\eta _{0}\right) e^{2}\left( k-1\right) -\overline{\zeta }^{2}\right] <\infty $$

(24)

Because $e^{2}\left( k-1\right) \geq \frac{\overline{\zeta }^{2}}{1-\eta }, \left( \frac{\eta _{0}}{1+q(k-1)}\right) \left[ \left( 1-\eta _{0}\right) e^{2}\left( k-1\right) -\overline{\zeta }^{2}\right] \geq 0,$ so:

$$ \underset{k\rightarrow \infty }{\lim }\left( \frac{\eta _{0}}{1+q(k-1)}\right) \left[ \left( 1-\eta _{0}\right) e^{2}\left( k-1\right) -\overline{\zeta }^{2}\right] =0 $$

(25)

Because L ₁(k − 1) is bounded, so q(k − 1) < ∞, and as $\frac{\eta _{0}}{1+q(k-1)}>0:$

$$ \underset{k\rightarrow \infty }{\lim }\left( 1-\eta _{0}\right) e^{2}\left( k-1\right) =\overline{\zeta }^{2} $$

(26)

That is (15). When $e^{2}\left( k-1\right) <\frac{\overline{\zeta }^{2}}{\left[ 1-\eta _{0}\right]},$ it is already in this zone.