[Machine Learning][BP]The Vectorized Back Propagation Algorithm

　　Reference: https://www.cs.swarthmore.edu/~meeden/cs81/s10/BackPropDeriv.pdf

　　I spent nearly one hour to deduce the vector form of the back propagation. Just in case that I may forget, but need to utilize them, I will write down all the formula here to make a backup.

Structure:

　　Standard BP Network with $\displaystyle \lambda$ hidden layers, one input layer and one output layer.

　　Activation function: sigmoid.

Notations:

$\displaystyle W^{i+1,i}$, denotes the weight matrix connecting from $i$th layer to $i+1$th layer.

$\displaystyle N^i$, denotes the net input of the $i$th layer.

$\displaystyle A^i$, denotes the activation input of the $i$th layer.

$\displaystyle \delta ^i$, denotes the error of the $i$th layer.

$\displaystyle \epsilon$, denotes the learning rate.

*, stands for element by element multiplication.

(omit), stands for matrix multiplication.

　　Specifically,

$\displaystyle X$, denotes the input layer, while equals $\displaystyle A^0$.

$\displaystyle A^{\lambda + 1}$, denotes the output layer.

$\displaystyle Y$, denotes the expected output.

Propagations:

　　Forward:

$\displaystyle N^i = W^{i,i-1}A^{i-1}$.

$\displaystyle A^i = \frac{1}{1+e^{-N^i}}$.

　　Backward:

$\displaystyle \Delta W^{i+1,i} = \epsilon \delta^{i+1}(A^{i})^{T}$.

$\displaystyle \delta ^i = ((\delta^{i+1})^{T}W^{i+1,i})^{T}$.

$\displaystyle \delta ^{\lambda + 1} = (Y - A^{\lambda + 1})*A^{\lambda + 1}*(1-A^{\lambda + 1})$.

Deduction:

　　I am not capable of taking the partial derivative of vector or matrix over vector or matrix, so I derive these formulas by observing the formula for each element in the matrix and extend it to the vector form.

 $\displaystyle \Delta W^{\lambda+1,\lambda}_{i,j} = \epsilon (Y_i - A^{\lambda+1}_i)A^{\lambda+1}_i(1-A^{\lambda +1}_i)A^{\lambda}_j$.

　　Let's assume $\displaystyle \delta ^{\lambda+1}_{i} := (Y_i - A^{\lambda+1}_i)A^{\lambda+1}_i(1-A^{\lambda +1}_i)$.

$\displaystyle \Delta W^{\lambda,\lambda-1}_{i,j}=\epsilon (\delta^{\lambda+1})^{T}W^{\lambda+1,\lambda}_{col(i)}A_j^{\lambda-1}$.

　　Let's assume $\displaystyle \delta ^{\lambda}_{i} := (\delta^{\lambda+1})^{T}W^{\lambda+1,\lambda}_{col(i)}$.

　　The left are reserved for the readers to complete.