# How to combine soft-max value from two or more neural networks

Consider a scenario where we have some input variable $$X$$ (maybe an image), which goes through several noisy path and we generate $$K$$ noisy observations:

\begin{align} Y_i = X_i +Z_i, i \in \{1,\ldots, K\} \end{align} where noises are independent. Here I am assuming additive structure but it doesn't have to be.

Suppose that I use the same neural network on $$Y_i$$'s to classify. Also, suppose that I have access to the soft-max values In particular, let $$p_i$$ be the soft-max value for input $$Y_i$$,

Question: How do I combine these values in the best way to determine my class? One option is \begin{align} p_{new} = \frac{1}{K} \sum_{i=1}^K p_i \end{align}

One can also just threshold at $$0.5$$ and find the majority.

I am looking for an answer that would have some theoretical backing.

Enter ensemble classifier fusion. Under this approach, classifiers are trained with the same feature sets and training/testing objects, and then classifier votes are combined using the ensemble majority voting (EMV) and ensemble weighted majority voting (EWMV) ensemble combination techniques [1]. Let $$d_{l,\omega}({\bf x}) \in \{0,1\}$$ be the decision rule for an object $$\mathbf{x}$$ by the $$l$$th classifier $$(l=1,2,\ldots,L)$$ for class $$\omega$$ $$(\omega=1,2,\ldots,\Omega)$$. The support for EMV and EWMV, respectively, is functionally composed as $$$$\boldsymbol{\mu}_\omega({\bf x}) = \sum _{l=1}^L d_{l,\omega}({\bf x}), \quad \quad d_{l,\omega}({\bf x}) \in \{0,1\},$$$$ $$$$\boldsymbol{\mu}_\omega({\bf x}) = \sum _{l=1}^L w_l d_{l,\omega}({\bf x}), \quad \quad d_{l,\omega}({\bf x}) \in \{0,1\},$$$$ where $$w_l$$ is the normalized weight reflecting the accuracy of the $$l$$th classifier. Here, accuracy is based on the proportion of classified test objects assigned to the diagonal of the confusion matrix divided by the number of test objects. Let the set of class labels be $$\omega=1,2,\ldots,\Omega$$ and the ensemble decision for object $$\mathbf{x}$$ be $$\cal{E}({\bf x}\rightsquigarrow \omega)$$. The decision rule for test object $$\mathbf{x}$$ is $$$${\cal{E}}({\bf x}\rightsquigarrow \omega) \equiv \arg \underset{c} \max \{ \boldsymbol{\mu}_c({\bf x})\}.$$$$