2
$\begingroup$

I have a sequence $(X_n: \Omega \to \mathbb{R})_{n=1}^\infty$ of pairwise independent random variables.

Define for $n \geq 1: X_n' := X_n I_{\{X_n \leq n\}}$ where $I_A$ is the indicator function on $A$. Is it true that $(X_n')_{n=1}^\infty$ is a sequence of pairwise independent random variables?

Intuitively, this seems true. I know that pairwise independence is preserved under a Borel transformation $g: \mathbb{R} \to \mathbb{R}$ so I tried to write

$$X_n' = g \circ X_n$$

for some suitable $g$ but did not come up with anything useful. Any hints?

$\endgroup$
  • $\begingroup$ It should follow from the definition of independent random variables. About your third paragraph, are you sure that this result holds? I think $g$ needs to preserve measure or something like that. $\endgroup$ – Yanko Dec 15 '18 at 15:13
  • $\begingroup$ $g$ must be measurable, which is what I meant with "Borel". $\endgroup$ – user370967 Dec 15 '18 at 15:14
  • $\begingroup$ If $g$ is a constant you sure the claim holds? $\endgroup$ – Yanko Dec 15 '18 at 15:16
  • $\begingroup$ I'm thinking about it. Give me a second please. Thanks for your useful comment. $\endgroup$ – user370967 Dec 15 '18 at 15:16
  • $\begingroup$ Fix Borel sets $A,B$. We have to prove that $P(g\circ X \in A, g \circ Y \in B) = P(g \circ X \in A) P(g \circ Y \in B)$ where $g=1$ is the constant 1 function. If $1 \in A \cap B$, then we have $1 = 1$. Otherwise, we have $0=0$, so I don't think there is a contradiction. What do you think? $\endgroup$ – user370967 Dec 15 '18 at 15:23
0
$\begingroup$

Try $$ g_n\colon x\mapsto x\mathbf 1_{\left[-\infty,n\right]}(x). $$

$\endgroup$
  • $\begingroup$ Don't you mean $x\mapsto x\mathbf1_{(-\infty,n]}(x)$? $\endgroup$ – drhab Dec 15 '18 at 15:36
  • $\begingroup$ Yeah I noticed it too but the idea remains the same. $\endgroup$ – user370967 Dec 15 '18 at 15:36
0
$\begingroup$

It is immediate that $X\mathbf1_{X_n\leq n}$ is measurable wrt $\sigma(X_n)$ because it is the product of two random variables that are both measurable wrt $\sigma(X_n)$.

That is enough to conclude that also the $X_n'$ are pairwise disjoint.

There is indeed a measurable function $g:\mathbb R\to\mathbb R$ such that $X_n'=g\circ X_n$.

Let $h:\mathbb R^2\to\mathbb R$ denote the function prescribed by $(x,y)\mapsto xy$ and let $k_n:\mathbb R\to\mathbb R^2$ denote the function that is prescribed by $x\mapsto(x,\mathbf1_{(-\infty,n]}(x))$.

Both functions can be shown to be measurable so that also their composition is measurable.

Then function $g=h\circ k_n$ will do the job.

$\endgroup$
  • $\begingroup$ Thank you for your answer. Gives more insight. $\endgroup$ – user370967 Dec 15 '18 at 15:37
  • $\begingroup$ You are very welcome. $\endgroup$ – drhab Dec 15 '18 at 15:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy