# Why is this sequence of random variables pairwise independent?

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?

• 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. – Yanko Dec 15 '18 at 15:13
• $g$ must be measurable, which is what I meant with "Borel". – user370967 Dec 15 '18 at 15:14
• If $g$ is a constant you sure the claim holds? – Yanko Dec 15 '18 at 15:16
• I'm thinking about it. Give me a second please. Thanks for your useful comment. – user370967 Dec 15 '18 at 15:16
• 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? – user370967 Dec 15 '18 at 15:23

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

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

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.

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