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.
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)