# Why is regularization interpreted as a gaussian prior on my weights?

Going through this online class, the notes specify the following. (In the highlights):

I understand how we get a maximum likelihood interpretation. What I do not get is:

Why does using the Frobenious norm of the weight matrix $R(W)$ as a regularizer, lead to the interpretation of the weight matrix $W$ having a prior as being sampled from a gaussian? How is this arrived at? And what if the regularization term was something other than the frobenius norm?

(And to that last point, does this mean that every element in the matrix $W$ is sampled from a gaussian?)

Thank you.

Since we're using MAP we are trying to maximize the probability of the parameters given the data.

$$P(W|x,y) = \frac {P(x,y|W) P(W)} {P(x,y)}$$

$P(x,y)$ can be ignored since it's fixed for our data. So we are trying to maximize $log(P(x,y|W)) + log(P(W))$. Let's look at $P(W)$.

If each element of $W$ is drawn independently from a unit Gaussian, the probability of the matrix is

$$P(W) = \frac {1} {\sqrt{2 \pi}} \prod_{ij} exp\Big( -\frac {w_{ij}^2} 2 \Big)$$

so $log(P(W))$ is $$-\frac 1 2 \sum_{ij}{w_{ij}}^2$$

plus some constant terms. The log-likelihood is then

$$log(P(x,y|W)) - \frac 1 2 \sum_{ij} {{w_{ij}}^2}$$

Since we are thinking in terms of losses we negate the log-likelihood and minimize. Now the first term is cross-entropy and the second term is $\frac 1 2 R(W)$.

If you have a (reasonable in some sense that I don't know how to define) more or less arbitrary penalty function you can obtain a density for it by integrating over its domain and then normalizing to obtain a density. In this case it's just easy to see that it's a Gaussian.

This does not mean that the elements of $W$ actually are sampled from a Gaussian. What it means is that you believe that's what $W$ looks like before you have any evidence to the contrary. In other words the prior on the elements of $W$ is a Gaussian.

• Thanks testuser, however I think you lost me on the last part(s). I guess what I am trying to understand is, if we think that a matrix $W$ "looks" like a gaussian... what does that mean? Doesn't that mean that its elements are all iid sampled from a gaussian? Secondly, I understand the formulas you put up, but how can we tie them directly to the cost function in this particular case? Expansion on this would be appreciated. Thanks! – Creatron Aug 11 '16 at 21:40
• If you ask the person running an experiment, before they've seen any data, "What's your best guess for $W$"? they'll say: "I think each element was sampled independently from a Gaussian with mean zero and variance 1". – testuser Aug 11 '16 at 21:45
• I think it's a little clearer - Ill digest it tonight and let you know if I have any remaining questions. Thanks again. – Creatron Aug 11 '16 at 22:24
• I think I get it testuser. One question though is - why WOULD we ever assume that the weights come from a gaussian, (in this machine learning context)? Why is that a thing to start off with? By what right do we make this assumption? Thanks. – Creatron Aug 13 '16 at 19:52
• In this case I don't think that anyone really thinks that the weights are generated this way. It's just something that people know restricts the set of fitted models and has been observed to lead to less overfitting. – testuser Aug 13 '16 at 21:00