Backpropagation

The goal of any supervised learning algorithm is to find a function that best maps a set of inputs to their correct output. The motivation for backpropagation is to train a multi-layered neural network such that it can learn the appropriate internal representations to allow it to learn any arbitrary mapping of input to output.[8]

To understand the mathematical derivation of the backpropagation algorithm, it helps to first develop some intuition about the relationship between the actual output of a neuron and the correct output for a particular training example. Consider a simple neural network with two input units, one output unit and no hidden units, and in which each neuron uses a linear output (unlike most work on neural networks, in which mapping from inputs to outputs is non-linear)[lower-alpha 7] that is the weighted sum of its input.

A simple neural network with two input units (each with a single input) and one output unit (with two inputs)

Initially, before training, the weights will be set randomly. Then the neuron learns from training examples, which in this case consist of a set of tuples ${\displaystyle (x_{1},x_{2},t)}$ where ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ are the inputs to the network and t is the correct output (the output the network should produce given those inputs, when it has been trained). The initial network, given ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$, will compute an output y that likely differs from t (given random weights). A loss function ${\displaystyle L(t,y)}$ is used for measuring the discrepancy between the target output t and the computed output y. For regression analysis problems the squared error can be used as a loss function, for classification the categorical crossentropy can be used.

As an example consider a regression problem using the square error as a loss:

${\displaystyle L(t,y)=(t-y)^{2}=E,}$

where E is the discrepancy or error.

Consider the network on a single training case: ${\displaystyle (1,1,0)}$. Thus, the input ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ are 1 and 1 respectively and the correct output, t is 0. Now if the relation is plotted between the network's output y on the horizontal axis and the error E on the vertical axis, the result is a parabola. The minimum of the parabola corresponds to the output y which minimizes the error E. For a single training case, the minimum also touches the horizontal axis, which means the error will be zero and the network can produce an output y that exactly matches the target output t. Therefore, the problem of mapping inputs to outputs can be reduced to an optimization problem of finding a function that will produce the minimal error.

Error surface of a linear neuron for a single training case

However, the output of a neuron depends on the weighted sum of all its inputs:

${\displaystyle y=x_{1}w_{1}+x_{2}w_{2},}$

where ${\displaystyle w_{1}}$ and ${\displaystyle w_{2}}$ are the weights on the connection from the input units to the output unit. Therefore, the error also depends on the incoming weights to the neuron, which is ultimately what needs to be changed in the network to enable learning.

In this example, upon injecting the training data the loss function becomes

${\displaystyle E=(t-y)^{2}=y^{2}=(x_{1}w_{1}+x_{2}w_{2})^{2}=(w_{1}+w_{2})^{2}.}$

Then, the loss function ${\displaystyle E}$ takes the form of a parabolic cylinder with its base directed along ${\displaystyle w_{1}=-w_{2}}$. Since all sets of weights that satisfy ${\displaystyle w_{1}=-w_{2}}$ minimize the loss function, in this case additional constraints are required to converge to a unique solution. Additional constraints could either be generated by setting specific conditions to the weights, or by injecting additional training data.

One commonly used algorithm to find the set of weights that minimizes the error is gradient descent. By backpropagation, the steepest descent direction is calculated of the loss function versus the present synaptic weights. Then, the weights can be modified along the steepest descent direction, and the error is minimized in an efficient way.

Diagram of an artificial neural network to illustrate the notation used here

Calculating the partial derivative of the error with respect to a weight ${\displaystyle w_{ij}}$ is done using the chain rule twice:

${\displaystyle {\frac {\partial E}{\partial w_{ij}}}={\frac {\partial E}{\partial o_{j}}}{\frac {\partial o_{j}}{\partial w_{ij}}}={\frac {\partial E}{\partial o_{j}}}{\frac {\partial o_{j}}{\partial {\text{net}}_{j}}}{\frac {\partial {\text{net}}_{j}}{\partial w_{ij}}}}$

(Eq. 1)

In the last factor of the right-hand side of the above, only one term in the sum ${\displaystyle {\text{net}}_{j}}$ depends on ${\displaystyle w_{ij}}$, so that

${\displaystyle {\frac {\partial {\text{net}}_{j}}{\partial w_{ij}}}={\frac {\partial }{\partial w_{ij}}}\left(\sum _{k=1}^{n}w_{kj}o_{k}\right)={\frac {\partial }{\partial w_{ij}}}w_{ij}o_{i}=o_{i}.}$

(Eq. 2)

If the neuron is in the first layer after the input layer, ${\displaystyle o_{i}}$ is just ${\displaystyle x_{i}}$.

The derivative of the output of neuron ${\displaystyle j}$ with respect to its input is simply the partial derivative of the activation function:

${\displaystyle {\frac {\partial o_{j}}{\partial {\text{net}}_{j}}}={\frac {\partial \varphi ({\text{net}}_{j})}{\partial {\text{net}}_{j}}}}$

(Eq. 3)

which for the logistic activation function case is:

${\displaystyle {\frac {\partial o_{j}}{\partial {\text{net}}_{j}}}={\frac {\partial }{\partial {\text{net}}_{j}}}\varphi ({\text{net}}_{j})=\varphi ({\text{net}}_{j})(1-\varphi ({\text{net}}_{j}))=o_{j}(1-o_{j})}$

This is the reason why backpropagation requires the activation function to be differentiable. (Nevertheless, the ReLU activation function, which is non-differentiable at 0, has become quite popular, e.g. in AlexNet)

The first factor is straightforward to evaluate if the neuron is in the output layer, because then ${\displaystyle o_{j}=y}$ and

${\displaystyle {\frac {\partial E}{\partial o_{j}}}={\frac {\partial E}{\partial y}}}$

(Eq. 4)

If half of the square error is used as loss function we can rewrite it as

${\displaystyle {\frac {\partial E}{\partial o_{j}}}={\frac {\partial E}{\partial y}}={\frac {\partial }{\partial y}}{\frac {1}{2}}(t-y)^{2}=y-t}$

However, if ${\displaystyle j}$ is in an arbitrary inner layer of the network, finding the derivative ${\displaystyle E}$ with respect to ${\displaystyle o_{j}}$ is less obvious.

Considering ${\displaystyle E}$ as a function with the inputs being all neurons ${\displaystyle L=\{u,v,\dots ,w\}}$ receiving input from neuron ${\displaystyle j}$,

${\displaystyle {\frac {\partial E(o_{j})}{\partial o_{j}}}={\frac {\partial E(\mathrm {net} _{u},{\text{net}}_{v},\dots ,\mathrm {net} _{w})}{\partial o_{j}}}}$

and taking the total derivative with respect to ${\displaystyle o_{j}}$, a recursive expression for the derivative is obtained:

${\displaystyle {\frac {\partial E}{\partial o_{j}}}=\sum _{\ell \in L}\left({\frac {\partial E}{\partial {\text{net}}_{\ell }}}{\frac {\partial {\text{net}}_{\ell }}{\partial o_{j}}}\right)=\sum _{\ell \in L}\left({\frac {\partial E}{\partial o_{\ell }}}{\frac {\partial o_{\ell }}{\partial {\text{net}}_{\ell }}}{\frac {\partial {\text{net}}_{\ell }}{\partial o_{j}}}\right)=\sum _{\ell \in L}\left({\frac {\partial E}{\partial o_{\ell }}}{\frac {\partial o_{\ell }}{\partial {\text{net}}_{\ell }}}w_{j\ell }\right)}$

(Eq. 5)

Therefore, the derivative with respect to ${\displaystyle o_{j}}$ can be calculated if all the derivatives with respect to the outputs ${\displaystyle o_{\ell }}$ of the next layer – the ones closer to the output neuron – are known. [Note, if any of the neurons in set ${\displaystyle L}$ were not connected to neuron ${\displaystyle j}$, they would be independent of ${\displaystyle w_{ij}}$ and the corresponding partial derivative under the summation would vanish to 0.]

Substituting Eq. 2, Eq. 3 Eq.4 and Eq. 5 in Eq. 1 we obtain:

${\displaystyle {\frac {\partial E}{\partial w_{ij}}}={\frac {\partial E}{\partial o_{j}}}{\frac {\partial o_{j}}{\partial {\text{net}}_{j}}}{\frac {\partial {\text{net}}_{j}}{\partial w_{ij}}}={\frac {\partial E}{\partial o_{j}}}{\frac {\partial o_{j}}{\partial {\text{net}}_{j}}}o_{i}}$

${\displaystyle {\frac {\partial E}{\partial w_{ij}}}=o_{i}\delta _{j}}$

with

${\displaystyle \delta _{j}={\frac {\partial E}{\partial o_{j}}}{\frac {\partial o_{j}}{\partial {\text{net}}_{j}}}={\begin{cases}{\frac {\partial L(o_{j},t)}{\partial o_{j}}}{\frac {d\varphi ({\text{net}}_{j})}{d{\text{net}}_{j}}}&{\text{if }}j{\text{ is an output neuron,}}\\(\sum _{\ell \in L}w_{j\ell }\delta _{\ell }){\frac {d\varphi ({\text{net}}_{j})}{d{\text{net}}_{j}}}&{\text{if }}j{\text{ is an inner neuron.}}\end{cases}}}$

if ${\displaystyle \varphi }$ is the logistic function, and the error is the square error:

${\displaystyle \delta _{j}={\frac {\partial E}{\partial o_{j}}}{\frac {\partial o_{j}}{\partial {\text{net}}_{j}}}={\begin{cases}(o_{j}-t_{j})o_{j}(1-o_{j})&{\text{if }}j{\text{ is an output neuron,}}\\(\sum _{\ell \in L}w_{j\ell }\delta _{\ell })o_{j}(1-o_{j})&{\text{if }}j{\text{ is an inner neuron.}}\end{cases}}}$

To update the weight ${\displaystyle w_{ij}}$ using gradient descent, one must choose a learning rate, ${\displaystyle \eta >0}$. The change in weight needs to reflect the impact on ${\displaystyle E}$ of an increase or decrease in ${\displaystyle w_{ij}}$. If ${\displaystyle {\frac {\partial E}{\partial w_{ij}}}>0}$, an increase in ${\displaystyle w_{ij}}$ increases ${\displaystyle E}$; conversely, if ${\displaystyle {\frac {\partial E}{\partial w_{ij}}}<0}$, an increase in ${\displaystyle w_{ij}}$ decreases ${\displaystyle E}$. The new ${\displaystyle \Delta w_{ij}}$ is added to the old weight, and the product of the learning rate and the gradient, multiplied by ${\displaystyle -1}$ guarantees that ${\displaystyle w_{ij}}$ changes in a way that always decreases ${\displaystyle E}$. In other words, in the equation immediately below, ${\displaystyle -\eta {\frac {\partial E}{\partial w_{ij}}}}$ always changes ${\displaystyle w_{ij}}$ in such a way that ${\displaystyle E}$ is decreased:

${\displaystyle \Delta w_{ij}=-\eta {\frac {\partial E}{\partial w_{ij}}}=-\eta o_{i}\delta _{j}}$

The mathematical expression of the loss function must fulfill two conditions in order for it to be possibly used in backpropagation.[9] The first is that it can be written as an average ${\textstyle E={\frac {1}{n}}\sum _{x}E_{x}}$ over error functions ${\textstyle E_{x}}$, for ${\textstyle n}$ individual training examples, ${\textstyle x}$. The reason for this assumption is that the backpropagation algorithm calculates the gradient of the error function for a single training example, which needs to be generalized to the overall error function. The second assumption is that it can be written as a function of the outputs from the neural network.

Let ${\displaystyle y,y'}$ be vectors in ${\displaystyle \mathbb {R} ^{n}}$.

Select an error function ${\displaystyle E(y,y')}$ measuring the difference between two outputs. The standard choice is the square of the Euclidean distance between the vectors ${\displaystyle y}$ and ${\displaystyle y'}$:

$${\displaystyle E(y,y')={\tfrac {1}{2}}\lVert y-y'\rVert ^{2}}$$

The error function over ${\textstyle n}$ training examples can then be written as an average of losses over individual examples:

$${\displaystyle E={\frac {1}{2n}}\sum _{x}\lVert (y(x)-y'(x))\rVert ^{2}}$$