# Regularization on weights without bias

I was learning neural network using the book "Deep Learning" by Ian Goodfellow, Yoshua Bengio and Aaron Courville. In section 7.1, it says:

...we typically choose to use a parameter norm penalty Ω that penalizes only the weights of the affine transformation at each layer and leaves the biases unregularized. The biases typically require less data to fit accurately than the weights. Each weight specifies how two variables interact. Fitting the weight well requires observing both variables in a variety of conditions. Each bias controls only a single variable...

I don't understand why it says

Each weight speciﬁes how two variables interact

Could someone please explain it please? It would be perfect with some examples.

• I hate that they use the term "bias" for the layer-wise intercept. Regularization always induces bias! – generic_user Feb 22 '18 at 20:31

Here's my understanding of this quote. This is sort of a hand-wavy argument, but still gives some intuition. Let's consider a simple linear layer:

$$y = Wx + b$$

... or equivalently:

$$y_i = x_{1}W_{i,1} + ... + x_{n}W_{i,n} + b_i$$

If we focus on one weight $W_{i,j}$, its value is determined by observing two variables $(x_j, y_i)$. If the training data has $N$ rows, there're only $N$ pairs $(x_j, y_i)$, out of which $W_{i,j}$ is going to learn the correct value. That is a lot of flexibility, which the authors summarize in this phrase:

Fitting the weight well requires observing both variables in a variety of conditions.

In other words, the number of training rows $N$ must be really big in order to capture the correct slope without regularization. On the other hand, $b_i$ affects just $y_i$, which basically means its value can be better estimated from the same number of examples $N$. The authors put it this way:

This means that we do not induce too much variance by leaving the biases unregularized.

In the end, we'd like to regularize the weights that have "more freedom", that's why regularizing $W_{i,j}$ makes more sense than $b_i$.

• Can I understand it this way: $y = w_1x_1 + w_2x_2 + w_3x_3 + ...... + w_nx_n + b$ \\ $y$ and $x_1, x_2, ..... x_n$ are all variables. So we have $Var(y) = w_1^2Var(x_1) + w_2^2Var(x_2) + w_3^2Var(x_3) + .... + w_n^2Var(x_n)$ In this case, I consider $w_2$ and $b$ is constant – roy Feb 23 '18 at 0:02
• If they're considered constant, yes, it's right – Maxim Feb 23 '18 at 13:15

In ML lingo a weight is a coefficient of a bona fide regression variable and bias is the intercept. Also, in regression language interaction has a specific meaning, not the same as used in the text quoted.

In your text how variables interact means simply that there is a function that into translates inputs $x$ into output: $$a=f(b+wx)$$ So, the weight $w$ specifies how the variables $a$ and $x$ interact, using the language of the book. Now, the bias $b$ only controls an output $a$. In other words, they're saying that $x$ via $w$ impacts $a$, while $b$ impact $a$ directly, it doesn't need another variable.

I wouldn't pay too much attention to this argument. It's too wobbly to me. What you need to understand is that if you set all weights $w=0$, the model will still somewhat work, because $b\ne 0$, and it will cause your layer to still produce the value that is around the mean. It will not be a very intelligent forecast, since it doesn't accept any inputs, but it will be showing some kind of an average output. If you set all $w=b=0$ this will not work well, since it'll be producing zero all the time.