[944] General Question About Measurability

[David Itkin]

Just to add to this, I just realized that when I replied to Richard I hit
"reply" rather than "reply all". Here was my response:

Hi Richard,

Z will also be G-measurable. In general if [image: X_1,...,X_n] are random
variables and [image: f: \mathbb{R}^n \rightarrow \mathbb{R}] is a Borel
function (i.e. [image: f^{-1}(-\infty,a) \in \mathcal{B}(\mathbb{R}^n)] for
all [image: a \in \mathbb{R}] where [image: \mathcal{B}(\mathbb{R}^n)] is
the Borel sigma algebra in [image: \mathbb{R}^n]) then [image:
f(X_1,...,X_n)] will be a random variable (i.e. G-measurable). So in
particular this will hold for all continuous functions f.

In the above example we have [image: f:\mathbb{R}^2 \rightarrow
\mathbb{R}] given
by [image: f(x,y) = x + y], which is Borel so X + Y is a random variable.

Thanks,

David

On Tue, Nov 7, 2017 at 10:37 AM, Gautam Iyer <gi1242+944 at cmu.edu> wrote:

> On Tue, Nov 07, 2017 at 08:37:50AM -0500, Richard Rosenbaum wrote:
>
> > Say X, Y, and Z are RVs. G is a sigma algebra.
> >
> > Z = X + Y
> >
> > If X and Y are both G-measurable, does this imply that Z is also
> > G-measurable?
>
> Yes! Try it, it isn't too hard.
>
> > I guess more generally - say a random variable is some function of
> > other RVs. does the measurability of the two RVs that the function
> > takes as inputs imply that the function itself is measurable with
> > respect to that same sigma algebra?
>
> Almost always. To be 100% correct:
>
>     Suppose f is a function from R^2 to R (i.e. takes two real numbers
>     as inputs), AND is measurable with respect to the Borel sigma
>     algebra. If X and Y are any two random variables, then f(X, Y) is
>     also a random variable.
>
> It's really really hard to write down non-Borel measurable functions. So
> almost every function you encouter will be Borel measurable, and so
> almost every operation you do to random variables will yield a random
> variable.
>
> Best,
>
> Gautam
>
> --
> The system requirements said 'Requires Windows 95 or better', so I
> bought a Mac.
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