f(X) and f(Y) are identically distributed

Question

Suppose that $X$ is a random variable defined on a probability space ( Ω , $\mathcal F$ , P )

and $P_X$ is the probability measure on ( $\mathbb{R} $, $\mathcal R$)

and $Y$ is a random variable defined on ( Ω' , $\mathcal F'$ , P' )

IF $ X $and $Y $ are identically distributed

If $f:$ $\mathbb{R} $ $->$ $\mathbb{R} $ is Borel Measurable, And $X$ and $Y$ are identically distributed, then if we define $U = f(X)$ and $V= f(Y)$.

$U$ and $V$ are also identically distributed

(Tried Solution):

After many hints and many hours all I manage to do is the below :

We know $P_x$ $=$ $P_Y$

and $P_X(B)$ , $B \in \mathcal R$ $= $$P_Y(B)$ , $B \in \mathcal R$

$P(X^{-1}(B)) = P(Y^{-1}(B))$

$ P(\omega\in Ω : X(ω) \in B) = P(\omega\in Ω' : Y(ω) \in B)$

Somehow we have to show that the above is equal to the below:. $$......$$ $$......$$

$ P(\omega\in Ω : f o X(ω) \in B) = P(\omega\in Ω' : foY(ω) \in B)$

Hence, $ P(\omega\in Ω : U(ω) \in B) = P(\omega\in Ω' :V(ω) \in B)$

Thus, $U = f(X)$ and $V= f(Y)$ are identically distributed .

Could you please explain the solution thoroughly, because I really want to understand the solution.

Answer 1

To show that $f(X)$ and $f(Y)$ are identically distributed, we need to show that for any Borel set $B$, we have $P(f(X) \in B) = P(f(Y) \in B)$.

We have : $$ P(f(X) \in B) = P(\{w : f(X(w)) \in B\}) = P(\{w : X(w) \in f^{-1}(B)\}) $$

where $f^{-1}(B) = \{t \in \mathbb R : f(t) \in B\}$. Note that this set is also Borel if $B$ is, since $f$ is Borel measurable. Therefore, all the above probabilities are well defined.

Finally, note that as $X$ and $Y$ are identically distributed, $$ P(\{w : X(w) \in f^{-1}(B)\}) = P_X(f^{-1}(B))= P_Y(f^{-1}(B)) = P(\{w : Y(w) \in f^{-1}(B)\}) $$

and now go back using the logic of the first line to $P(f(Y) \in B)$ and conclude.

Now ask questions freely on this answer.

Answer 2

For every Borel set $B$:

$$P_U(B)=P_{f(X)}(B)=P(f(X)\in B)=P(X\in f^{-1}(B))=P_X(f^{-1}(B))=$$$$P_Y(f^{-1}(B))=P(Y\in f^{-1}(B))=P(f(Y)\in B)=P_{f(Y)}(B)=P_V(B)$$

The fifth equality rests on the fact that $X$ and $Y$ have identical distribution.

In the notation above e.g. $P(X\in A)$ actually abbreviates $P(\{X\in A\})$ where $\{X\in A\}$ is on its turn an abbreviation of $\{\omega\in\Omega\mid X(\omega)\in A\}$.