Noisy Targets as deterministic target function

[davidrcoelho]

Hi everyone,

As I understood, noisy target is when we observe that for a same value of x we get different values of y.

Then, we model this function as a distribution P(y|x) (instead of a deterministic one)

But later on in the book, Yaser says:

"This view suggests that a deterministic target function can be considered a special case of a noisy target, just with zero noise. Indeed, we can formally express any function f as a distribution P(y|x) by choosing P(y|x) to be zero for all y except y = f(x)".

My question is:

How we can consider a value of y is equal to f(x), since the function f is a distribution.
Suppose we have two different values of y for the same input x, what of these two y's is different to f(x)?

[yaser]

Originally Posted by davidrcoelho

Hi everyone,

As I understood, noisy target is when we observe that for a same value of x we get different values of y.

Then, we model this function as a distribution P(y|x) (instead of a deterministic one)

But later on in the book, Yaser says:

"This view suggests that a deterministic target function can be considered a special case of a noisy target, just with zero noise. Indeed, we can formally express any function f as a distribution P(y|x) by choosing P(y|x) to be zero for all y except y = f(x)".

My question is:

How we can consider a value of y is equal to f(x), since the function f is a distribution.
Suppose we have two different values of y for the same input x, what of these two y's is different to f(x)?

Just to clarify, the statement above from the book is talking about viewing a deterministic target as (a special case of) a noisy target, not the other way around as the title of this thread "Noisy Targets as deterministic target function" may suggest. A noisy target can be viewed as a deterministic target plus noise, but not as a stand-alone deterministic target.

The case you mention, where there are two possible 's for the same is a case of a noisy target, not a deterministic target.

[davidrcoelho]

It is clear now. Thanks a lot!