Okay, so last part of this lecture will be Bayesian inversion, but we can apply Bayes

theorem to inverse problems. As usual we say y is given by f plus epsilon, we say everything

is one dimensional and one thing to note is here we actually have three random variables.

So of course the parameter is not actually random but we call the random variable, moderate

variable we modulate by a random variable we have X we have epsilon and Y and

because of this relation they're not independent right so they're not

independent but strongly coupled but nevertheless they are three random

variables and usually this assumption that we will also make here X and

epsilon are independent from each other so given just X and epsilon those two

factors will be independent so the distribution of X and epsilon the density

of those two variables those will indeed decouple into the product rho X of X

which is the prior and rho epsilon of epsilon which is the distribution the

density of the probability density of the noise term so this will be an

assumption that we make because Y is equal to f of X plus epsilon this implies

that epsilon is equal to Y minus f of X this means that the distribution of X and

Y which you can also write down right so there is no hidden behind all this

there's a three-dimensional object rho X epsilon Y of X epsilon and Y and we're

looking at the marginalized versions of that so rho of X comma Y of X comma Y

this will be this will be sorry we put this correctly rho of X just wait a

minute okay so the best way to see what this is is so I'm good a bit of hand

waving here but this will not be too disturbing so this is the marginalization

of this let's say three slots distribution X epsilon and Y well we

integrate over all of epsilon this is the definition of this marginalization

here the joint distribution of X and Y and what we see is well this this Y here

is actually given by f of X plus epsilon which means that we can write this

integral as this as X epsilon and f of X plus epsilon this here is f of X plus

epsilon d epsilon what we can now do is we call W we don't have to call it W

you can call it we can just call it Y of course this is f of X plus epsilon so

epsilon is equal to Y minus f of X of course so we have the integral over rho

X epsilon Y over X Y minus f of X and Y d epsilon and this then becomes d sorry

I'm missing my right here again okay let me redo this this thing here it's not

difficult it's just I want to do it less technical than it might look like okay

let's try again so what we do is this is definitely correct this step here and

due to the fact that epsilon is equal to Y minus f of X this integral is actually

the same as just point wise evaluation of X epsilon Y in X Y minus f of X

comma Y so this this step in here well this is due to the fact that this is

kind of a Dirac measure on the manifold where epsilon is equal to Y minus f of X

so for the full technical details we would have to attend a more slightly

more advanced class of calculus but let me just trust me that this is this is

correct and similarly we could also get this from integrating this density X

epsilon and Y over X Y minus f of X comma and let's say W dW so this is

exactly this the same same argument because W has to be equal to Y in order

for this to be nonzero this integral also simplifies to this evaluation here

but this is just a marginalization of rho X epsilon in X and Y minus f of X and

we said that X and epsilon independent so this marginalization here is given

by given by the product rho X of X times rho epsilon but not now epsilon because

we don't know epsilon we only know X and Y so we put Y minus f of X in here

because this is what the argument says here okay so sorry this looked more

complicated than it actually is and now we can actually forget about epsilon we

don't need any point wise measurements of epsilon of course we don't have that