Hi Shar,

> If x and y are gaussian,

> x ~ N(0, R)

> y ~ N(0, V)

>

> x and y are uncorrelated.

>

> and z = x + y, then

> (z | x) ~ N(x, V)

Strictly speaking, this is not true (choose x to be N(0,1), and y=ex

where e is a bernouilli variable with values in {-1,1} independent of

x). For the statement to be true, you *either* need to replace the

word "uncorrelated" by "independent", *or*, "x and y are gaussian"

by "x and y are jointly gaussian". At the end of the day, if you want

your statement to be true, you need the independence between x and y,

and being "uncorrelated" will imply independence, only in the case

when x and y are *jointly* gaussian...

So we assume that x and y are jointly normal and uncorrelated...

> Now if z = f(x) + y

> where f(x) is some complicated nonlinear function.

> Then can I still say,

> (z | x) ~ N(f(x), V) ??

Yes you can (as long as f is a borel measurable map), and the proof

is as follows: we want to know the conditional distribution of z

knowing x, and for this purpose we compute the conditional

characteristic function of z knowing x, i.e. for u in R:

E[exp(iuz)|x]=E[exp(iu(f(x)+y))|x]

=exp(iuf(x))E[exp(iuy)|x] (*)

=exp(iuf(x))E[exp(iuy)] (**)

=exp(iuf(x))exp(-u^2V/2) (***)

Equality (*) stems from the measurability of exp[(iuf(x))] with

respect to s(x) [the sigma-algebra generated by x]. Equality (**)

stems from the independence of exp(iuy) and x. Equality (***) stems

from the assumption y~N(0,V).

We conclude that (see e.g. Lemma 6.13 p.85 of Karatzas and Shreve's

book, brownian motion and stochastic calculus) that conditionally on

x, the law of z is normal with mean f(x) and variance V.

Regards. Noel.