How can I show that $Y_n(t)$ is iid again?

Question

Let $(X_n) $ be a sequence of iid random variables where $F$ is the distribution function of $X_1$. For $t \in \mathbb R$, how can I show that the random variables $Y_n(t):=1_{X_n \leq t}$, $n \in \mathbb N$, is iid again?

Answer 1

$f_t(x)=I_{(-\infty , t]}$ is a measurable function and $Y_n(t)=f_t(X_n)$. For any measurable function $f$ the variables $f(X_1),f(X_2),...$ are i.i.d..

Answer 2

If $X,Y$ are independent random variables then so are $f(X),g(Y)$ if $f,g$ are measurable functions.

This because:$$P(f(X)\in A\wedge g(Y)\in B)=P(X\in f^{-1}(A)\wedge Y\in g^{-1}(B))=$$$$P(X\in f^{-1}(A))P(Y\in g^{-1}(B))=P(f(X)\in A)P(g(Y)\in B)$$

The second equality on base of independence.

The same principle can be used to proved for a finite number of random variables.

This can be applied that the $Y_n$ are independent.

Next to that it is evident that $\mathsf1_{X_n\leq t}$ and $\mathsf1_{X_m\leq t}$ have equal distribution whenever $X_n$ and $X_m$ have equal distributions.