# Prove random variables are independent via strict definition

The question:

Suppose $\{X_i\}_{i\in \mathbb N}$ is an $i.i.d.$ sequence. Define $Y_i=X_i1_{(X_i\le i)},\,i\in \mathbb N$.

Prove: $\{Y_i\}_{i\in \mathbb N}$ are independent as well.

Below is my try:

$\{X_i\}_{i\in \mathbb N}$ is an $i.i.d.$ sequence, thus $\forall$ finite index set $F\subset \mathbb N$, we have: $$\forall A_k \in\sigma(X_k)\,for\,k\in F, \,\mathbb P(\cap_{k\in F}A_k)=\Pi_{k\in F}\mathbb P(A_k)$$

Then I need to prove that: $$\forall B_k \in\sigma(Y_k)\,for\,k\in F, \,\mathbb P(\cap_{k\in F}B_k)=\Pi_{k\in F}\mathbb P(B_k)$$

My intuition is that $\sigma(Y_k) \subseteq \sigma(X_k)$ ($\require{enclose}\begin{array}{rl}\enclose{updiagonalstrike,downdiagonalstrike}{Actually\,I\,feel\,that\,\sigma(Y_k) = \sigma(X_k)}\end{array}$ - this equality is NOT true, see comments of @spaceisdarkgreen ). If this is achieved, the proof is done.

We know that, by the definition of σ-algebra generated by random variable $$\sigma(X_i)=\{X_i^{-1}(B):\,B\in \mathscr B(\mathbb R)\}$$ $$\sigma(Y_i)=\{Y_i^{-1}(B):\,B\in \mathscr B((-\infty,k]\}$$

I want to show that every set belonging to $\sigma(Y_i)$ also belongs to $\sigma(X_i)$, but then I found complexity of Borel set plus the inverse mapping makes me a bit lost.

$\forall M \in \mathscr B((-\infty,k]) \subset \mathscr B(\mathbb R)$, then actually $X_i^{-1}(M) \subset Y_i^{-1}(M))$, because of the effect of $1_{(X_k\le k)}\, in\, Y_k$, and $X_i^{-1}(M)\in\sigma(X_i), \,Y_i^{-1}(M)\in \sigma(Y_i)$. Let N = $Y_k^{-1}(M)\setminus X_k^{-1}(M)$, then it should be the case that $N\subset\{\omega:X_i(\omega)>k\}\in \sigma(X_i)$. Thus if we could prove that $N \in \sigma(X_i)$, the proof is done - but this is where I get stuck.

Thus I want to know:

$(1)$ Is the direction of my proof on the right track?

$(2)$ If so, how to prove that $N \in \sigma(X_i)$? If not so, how to prove this via strict definition?

$(3)$ If $(1)$ and $(2)$ are addressed, I'm happy to see any alternative/smart proofs.

• It is true that $\sigma(Y_i)\subset \sigma(X_i)$ but not that $\sigma(Y_i)=\sigma(X_i)$ (when you know $X_i$ you know $Y_i$ but not the other way around). Commented Mar 12, 2017 at 4:55
• This is a situation where moving into a more abstract or general setting may actually make things clearer. Can you establish that for any measurable function $f$, $\sigma(f(X))\subset\sigma(X)$? Then specialize to $f(x)=x1\{x<i\}$. Commented Mar 12, 2017 at 5:05
• Thanks both! That intuitive explanation is very helpful in understanding the inequality relationship of the two sigma algebra's. And I'll try to check with the generic conclusion of $f(x)$ tomorrow. Thanks again! Commented Mar 12, 2017 at 5:22
• Can you please explain your notation : $Y_i=X_i1_{(X_i\le i)}$ ?
– A.G.
Commented Mar 12, 2017 at 13:22
• $Y_i=X_i*1_{(X_i\le i)}$, where $1_{(X_i \le i)}$ equals to 1 if $X_i \le i$, and equals to 0 otherwise. @A.G. Commented Mar 12, 2017 at 14:17

Thanks for @snarfblaat's comments. I think now I could piece together the last part of the puzzle. The following is a proof about $σ(Y_k)⊆σ(X_k)$

Let $X:\Omega \mapsto A$, and $f:A\mapsto B$. $X, \,f,\, f\circ X$ are measurable w.r.t. Borel $\sigma$-algebra, and $\Omega,\,A,\,B$ are some topological spaces. Then $\sigma(f\circ X)\subseteq \sigma(X)$

Proof:

Notice that $\forall$ Borel-set $M \subseteq B,\,$ s.t. $(f\circ X)^{-1}(M) \in \sigma(f\circ X)$

Since, $f^{-1}(M)\subseteq A$ is a Borel-set, and $(f\circ X)^{-1}(M)=X^{-1}(f^{-1}(M))$

We have $(f\circ X)^{-1}(M)=X^{-1}(f^{-1}(M)) \in \sigma(X)$ □

In terms of applying to the concrete case in the question above:

Let $f(x)=x1_{(x\le i)}$. Here $A=\mathbb R$, and $B=(-\infty, i]$, with the natural topology. And it is easy to see that $f$ is measurable w.r.t. Borel $\sigma$-algebra on $B$, by noticing that

$q(x)=1_{(x\le i)}:\mathbb R\mapsto \mathbb R$ is Borel measurable since q is monotone, and so is $p(x)=x:\mathbb R\mapsto \mathbb R$. Thus $f(x)=p(x)q(x)$ is also Borel measurable. (Here I applied Prop 5.7, Prop 5.10, and Prop 5.11 of Real Analysis for Graduate Students by Richard F. Bass)