# Cost Functions

Let $$a^L = \sigma(z^L)$$ be the output of the neural network and $$y$$ the label (expected output). Different cost functions come up for different goals, and each has a "canonical" activation associated with it that optimises calculations.

<table><thead><tr><th width="332">Goal</th><th width="225">Cost Function</th><th>Activation</th></tr></thead><tbody><tr><td>Regression (continuous from <span class="math">-\infin</span> to <span class="math">\infin</span>)</td><td>MSE</td><td>Linear - <span class="math">\sigma(z) = z</span></td></tr><tr><td>Binary Classification</td><td>Binary Cross-Entropy</td><td>Sigmoid</td></tr><tr><td>Multi-Class Classification</td><td>Categorical Cross-Entropy</td><td>Softmax</td></tr></tbody></table>

## MSE (Mean Squared Error)

Also known as the **half-squared error**, MSE is simply half the **Euclidean distance** between $$a^L$$ and $$y$$:

$$
C(a^L,y) =\frac{1}{2} \lVert a^L-y \rVert^2 = \frac{1}{2}\sum\_{i=1}^n (a^L\_i - y\_i)^2
$$

The purpose of the half become clear when we note that

$$
\frac{\partial C}{\partial a^L\_i} = a^L\_i - y\_i
$$

So, in terms of vectors,

$$
\frac{\partial C}{\partial a^L} = a^L - y
$$

This makes the calculation of $$\delta^L$$ quite easy:

$$
\delta^L = (a^L - y) \odot \sigma^\prime(z^L)
$$

Note that with **linear activation** $$\sigma(z) = z$$, we would get $$\delta^L = a^L - y$$. This is desirable!

## Binary Cross-Entropy Loss

Given a label $$y$$, which is either $$0$$ or $$1$$ (hence binary), we have

$$
C(a^L, y) = -\[y\ln (a^L) + (1-y)\ln (1- a^L)]
$$

Then we have

$$
\frac{\partial C}{\partial a^L} = -\frac{y}{a^L} + \frac{1-y}{1-a^L} = \frac{a-y}{a(1-a)}
$$

Beautifully, if $$\sigma(z)$$ is the sigmoid function then we have $$\sigma^\prime(z) = \sigma(z)(1-\sigma(z))$$, so

$$
\delta^L = \frac{a-y}{a(1-a)} \cdot a(1-a) = a-y
$$

as we'd like!

## Categorical Cross-Entropy Loss + Softmax

CE loss itself can look like a weird choice:

$$
C(a^L,y) = -\sum\_{i=1}^n y\_i\ln(a^L\_i)
$$

However, we can make it highly efficient by using **softmax** on the final neurons rather than **sigmoid** again. Softmax is simply

$$
a\_i = \frac{e^{z\_i}}{\sum\_j e^{z\_j}}
$$

When combined with **softmax**, we get a very nice derivation for $$\delta$$:

$$
\delta^L = a^L - y
$$

### Proof

This proof is a bit more involved. Using multivariate chain rule:

$$
\delta^L\_i = \frac{\partial C}{\partial z\_i} = \sum\_k \frac{\partial C}{\partial a^L\_k} \frac{\partial a^L\_k}{\partial z\_i}
$$

We can see, from the definition of $$C$$, that

$$
\frac{\partial C}{\partial a^L\_i} = -\frac{y\_i}{a^L\_i}
$$

It is a bit harder to calculate $$\frac{\partial a\_k}{\partial z\_i}$$. If $$k = i$$, then by the quotient rule we get

$$
\begin{align\*}
\frac{\partial a^L\_i}{\partial z\_i} &= \frac{\partial}{\partial z\_i}\left\[\frac{e^{z\_i}}{\sum\_j e^{z\_j}}\right] \\
&= \frac{e^{z\_i}\sum\_j e^{z\_j} - e^{z\_i}e^{z\_i}}{\left(\sum\_j e^{z\_j}\right)^2} \\
&= \frac{e^{z\_i}}{\sum\_j e^{z\_j}} - \left(\frac{e^{z\_i}}{\sum\_j e^{z\_j}}\right)^2 \\
&= a^L\_i - (a^L\_i)^2 \\
&= a^L\_i(1- a^L\_i)
\end{align\*}
$$

If $$k \neq i$$, it's a bit different:

$$
\begin{align\*}
\frac{\partial a^L\_k}{\partial z\_i} &= \frac{\partial}{\partial z\_i}\left\[\frac{e^{z\_k}}{\sum\_j e^{z\_j}}\right] \\
&= \frac{0 \cdot \sum\_j e^{z\_j} - e^{z\_i}e^{z\_k}}{\left(\sum\_j e^{z\_j}\right)^2} \\
&= -\frac{e^{z\_i}}{\sum\_j e^{z\_j}} \cdot \frac{e^{z\_k}}{\sum\_j e^{z\_j}} \\
&= -a^L\_ia^L\_k
\end{align\*}
$$

Therefore

$$
\begin{align\*}
\delta^L\_i &=  \frac{\partial C}{\partial a^L\_i} \frac{\partial a^L\_i}{\partial z\_i} + \sum\_{k\neq i} \frac{\partial C}{\partial a^L\_k} \frac{\partial a^L\_k}{\partial z\_i} \\
&= -\frac{y\_i}{a^L\_i} \cdot a^L\_i(1- a^L\_i) + \sum\_{k \neq i} -\frac{y\_k}{a^L\_k} \cdot -a^L\_ka^L\_i \\
&= -y\_i(1-a^L\_i) + \sum\_{k \neq i} y\_ka^L\_i \\
&= -y\_i + a^L\_iy\_i + \sum\_{k \neq i} y\_ka^L\_i \\
&= -y\_i + a^L\_i\sum\_{k} y\_k \\
&= a^L\_i - y\_i
\end{align\*}
$$


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