# Backpropagation through a layer norm

Posted on 18 May, 2022

Derivation of the backpropagation equations for layer normalization.

## Layer normalization

Layer normalization is a normalization over each layer. In practice it is implemented as a normalization over columns (column major languages like Julia) or rows (row major languages like Python). There are other kinds of normalization like batch normalization, which is a normalization across batches. Interestingly layer norm was only popularised after batch normalization in this 2016 paper.

The function used for layer norm is:

$Z^l = a^{l}\frac{X^{l}-\mu^{l}}{\sigma^{l}+\epsilon} + b^{l}$

Where $\mu^l$ and $\sigma^{l}$ are the mean and standard deviation for each layer $l$ respectively, $a$ and $b$ are trainable parameters and $\epsilon$ is a small value used for numerical stability.

The mean and standard deviation are calculated as follows:

\begin{align} \mu^l &= \frac{1}{n}\sum_r^n x_r^l \\ \sigma^l &= \sqrt{\frac{1}{n}\sum^n_r (x_r^l - \mu^l)^2} \end{align}

### Backpropagation

Let us consider a single layer so that we can drop the $l$ superscript. We will ignore the trainable parameters because they will just scale the result. This is the same as $a=1$ and $b=0$. For a scalar loss function $L$, we need to find out the affect the vector $x$ will have the final loss. That is:

$\frac{\partial L}{\partial x} = \frac{\partial z}{\partial x} \frac{\partial L}{\partial z}$

Since $z$ and $x$ are vectors the Jacobian $\frac{\partial z}{\partial x}$ is a matrix. We take the partial derivative of each component of $z$ with respect to every component of $x$. That is:

$\frac{\partial z}{\partial x} = \begin{bmatrix} \frac{\partial z_1}{\partial x_1} & \frac{\partial z_2}{\partial x_1} & \dots & \frac{\partial z_n}{\partial x_1} \\ \frac{\partial z_1}{\partial x_2} & \frac{\partial z_2}{\partial x_2} & \dots & \frac{\partial z_n}{\partial x_2} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial z_1}{\partial x_n} & \frac{\partial z_2}{\partial x_n} & \dots & \frac{\partial z_n}{\partial x_n} \end{bmatrix}$

Consider $z_1$:

$z_1 = \frac{x_1-\mu}{\sigma+\epsilon}$

It explicitly depends on $x_1$. However it is also indirectly dependent on all the other components of $x$ because $\sigma$ and $\mu$ are both calculated with them. Hence we have a whole column of derivatives instead of one value.

The derivative with respect to $x_1$ will look slightly different to the derivative with respect to the other components because of this explicit dependency. The difference is:

\begin{align} \frac{\partial }{\partial x_i} x_1 &= \begin{cases} 1 & i=1 \\ 0 & i\neq 1 \end{cases} \\ &= \delta_{i1} \end{align}

In general we can use the Kronecker delta symbol $\delta_{ik}$ for convenience.

We can now calculate the derivate for $z_k$ with respect to a component $x_i$. This will require repeated applications of the product rule and the chain rule:

\begin{align} \frac{\partial z_k}{\partial x_i} &= \frac{\partial }{\partial x_i}(x_k -\mu)(\sigma + \epsilon)^{-1} \\ &= (\sigma + \epsilon)^{-1}\frac{\partial }{\partial x_i}(x_k -\mu) + (x_k -\mu)\frac{\partial }{\partial x_i}(\sigma + \epsilon)^{-1} \\ &= \frac{1}{\sigma + \epsilon}\left(\delta_{ik} - \frac{\partial \mu}{\partial x_i}\right) - (x_k -\mu)(\sigma + \epsilon)^{-2}\left(\frac{\partial \sigma}{\partial x_i} + 0\right) \\ &= \frac{1}{\sigma + \epsilon}\left(\delta_{ik} - \frac{\partial \mu}{\partial x_i}\right) - \frac{x_k -\mu}{(\sigma + \epsilon)^{2}}\frac{\partial \sigma}{\partial x_i} \end{align}

This is dependent on the derivative of the mean:

\begin{align} \frac{\partial \mu}{\partial x_i} &= \frac{\partial}{\partial x_i} \frac{1}{n}\sum_r^n x_r \\ &= \frac{1}{n}(0 + ... + 0 + 1 + 0 + ... + 0) \\ &= \frac{1}{n} \end{align}

Intuitively if a component increases by $\Delta x$ it increases the whole mean by $\frac{\Delta x}{n}$.

It is also dependent on the derivative of the standard deviation:

\begin{align} \frac{\partial \sigma}{\partial x_i} &= \frac{\partial}{\partial x_i} \left( \frac{1}{n}\sum^n_r (x_r - \mu)^2 \right)^{\frac{1}{2}} \\ &= \frac{1}{2}\left( \frac{1}{n}\sum^n_r (x_r - \mu)^2 \right)^{-\frac{1}{2}} \frac{\partial}{\partial x_i} \left( \frac{1}{n}\sum^n_r (x_r - \mu)^2 \right) \\ &= \frac{1}{2\sigma}\left(\frac{1}{n}\sum^n_{r}2(x_r - \mu) \left(\delta_{ir} - \frac{\partial \mu}{\partial x_i}\right) \right) \\ &= \frac{1}{n\sigma}\left((x_i -\mu) - \sum^n_r (x_r - \mu)\frac{\partial \mu}{\partial x_i} \right)\\ &= \frac{x_i -\mu}{n\sigma} \end{align}

Because $\sum^n_r (x_r - \mu) = \sum^n_r x_r - \mu \sum^n_r 1 = (n\mu) - \mu(n) = 0$

In summary (including the trainable parameters $a$ and $b$):

\begin{align} \frac{\partial z_k}{\partial x_i} &= a\left(\frac{1}{\sigma + \epsilon}\left(\delta_{ik} - \frac{\partial \mu}{\partial x_i}\right) - \frac{x_k -\mu}{(\sigma + \epsilon)^{2}}\frac{\partial \sigma}{\partial x_i} \right) \\ \frac{\partial \mu}{\partial x_i} &= \frac{1}{n} \\ \frac{\partial \sigma}{\partial x_i} &= \frac{x_i -\mu}{n\sigma} \\ \frac{\partial z_k}{\partial a} &= \frac{x_k -\mu}{\sigma + \epsilon} \\ \frac{\partial z_k}{\partial b} &= 1 \end{align}

### Julia implementation

We can get the gradients directly from Flux:

Flux uses ChainRulesCore to define rrule for backpropagation. However it doesn’t use the final rules derived above; instead it breaks the equations down into pieces much like was done in deriving them. One disadvantage of this is that you don’t get performance gains from terms cancelling out as happened with $\frac{\partial \sigma}{\partial x_i}$.

We can check our final equations match with the following:

The values in grads[1][:, k] and grads2 should differ by an order of $10^{-15}$ or less.