We'll start if I find out how to make it in presentation mode.

Yeah, like this.

Let me just check.

Yeah, it works.

OK, so as you can clearly see, Vincent is not here today.

He's currently traveling outside of Germany.

That's why he cannot give you this lecture.

So you stopped last week from what

he said in slide number, I think, set 14 on statistical

classifiers, slide 39.

And you had here an example of classification

if you have a Gaussian classifier

on the two Gaussian distributions which

have the same covariance matrix.

So if we come back here to the formula, what we have is

this term here is the same for both

since we have the same covariance matrix sigma.

So when you look for the highest value, it kind of drops.

So you can simplify it out.

And that makes a linear boundary between the classes.

So that's what we have here.

So the decision function is linear in the component of C.

So that's why we have here this linear boundary.

However, if you don't have identical covariance matrices,

so if you have different matrices for the different classes,

so you have a matrix sigma k for each class k,

then you cannot simplify this out anymore.

So what you get is a quadratic function.

So a quadratic function with regard to the feature vector.

So this term here does not have C.

And here you have two terms which contain C.

Each contains simply C. But this one here,

it contains C to the square.

So that means in case of different covariance matrices

on your different classes, you will get quadratic class

separation.

So here you have an example of what it can look like.

OK.

OK, so if you want to use a Gaussian classifier,

that means you will have to go through your data.

And for each class omega k, you have

to compute the prior probability, the center

of gravity of the class, and the covariance matrix.

So you can estimate them from the training data.

It's typically not possible to know them

with an infinite precision.

But if you have enough data, that should be totally fine.

And then you use as decision the argmax,

so the k for which you get the highest

value using this formula.