Question two is about statistical inverse problems, about statistical inversion and

we will talk about a method of computation that will be very important in the exam.

So I hope you have put some work into this exercise and I hope that this video helps

you because you will need those computations.

It's not difficult but it's just a specific type of computational trick that if you haven't

mastered it yet you will quickly if you pay close attention.

The basic mathematical setup is not too difficult so we have an unknown parameter x, we have

some also unknown additive noise and this gives us a measurement y.

So x is something and we have some error and this will give us y.

So you can think of this as being, well you look at for example a thermometer, you're

trying to find out how warm or cold it is and you're just looking at the thermometer

and just by, you know, you can't see exactly where the line is, you will make some additive

measurement either plus some fraction of a degree or minus a fraction of some degree

and a usual model is to assume that this read-off error will be something close to a Gaussian.

It's centered so you're usually not overestimating the temperature systematically, plus you're

either underestimating or overestimating in the same way.

And you also have some prior measure that specifies what you think, in this case the

temperature might be, so you have some belief about the temperature, maybe just by feeling,

right, looks like it could be between 10 and 20 degrees Celsius for example, and it will

make some measurement and this will hopefully contract your posterior and your posterior

belief about this parameter x.

The only formula that you will need is the one for Gaussian densities and if a parameter

is distributed according to a Gaussian measure with mean mu and variance sigma squared, then

this means that it has density 1 over square root of 2 pi sigma squared times e to the

minus x minus mu squared divided by 2 sigma squared.

So this is a very important formula that you definitely need.

I will also give this on the exam as I've done on the exercise exam, on the test exam,

but still this is a very important formula and the most important part of this, well

not the most important, but another important thing is if you integrate over this density

then this will be 1.

So we'll need this very much.

Okay, this is the density of x alone, this is the density of epsilon alone and the joint

density of x and y as a pair of random variables, this is not just a product of two densities

but it has the specific form.

So you could derive this as we have done in the lecture, but you can just use this.

Where does this come from?

It is essentially just a product of those two where you write epsilon as y minus x,

if you bring epsilon to one side you have y minus x on the other side, this is essentially

what you do here.

Okay, so this is the joint density of x and y, this is just a function, nothing else and

the first task is to compute the conditional density y given x is equal to x of y.

And we know the formula for that, this is just the joint density of x and y divided

by the marginal density of x, which we have in this half here.

So we just have to write these down, the numerator is 1 over 2 pi square root of sigma squared

t squared e to the minus x minus mu squared divided by 2 sigma squared minus y minus x

squared divided by 2 tau squared and we divide this by 1 over square root of 2 pi sigma squared

e to the minus x minus mu squared divided by 2 sigma squared.

And you can easily see that this term and this term cancels, also we get a square root

of here if we cancel the square root of 2 pi here and sigma squared cancels.

So what remains is 1 over square root of 2 pi tau squared e to the minus y minus x squared