Let $X$ and $Y$ be random random variables and let $A \in \mathcal{B}$. Prove that the function $Z$ defined by $$Z(\omega)=\begin{cases} X(\omega),& \text{ if } \omega \in A \\ Y(\omega),& \text{ if } \omega \in A^{c} \end{cases}$$ is a random variable

Proof so far: $$Z^{-1}((-\infty,a])=\{\omega:Z(\omega)\geq a\}=\{\omega: Z(\omega)\geq a, \omega \in A\}\cup\{\omega: Z(\omega)\leq a, \omega \in A^{c}\}=Y^{-1}[a,\infty) \cup X^{-1}([a,\infty))$$ So $Z$ is measurable

  • 5
    $\begingroup$ Every single equality is wrong, please take your time and rework the details of your proof. $\endgroup$ – Dominik Sep 23 '15 at 16:39
  • $\begingroup$ It might be worth mentioning once again, since the accepted answer wrongly claims that the proof in the question is correct modulo some small typos, that the identity $$Z^{-1}((-\infty,a])=Y^{-1}[a,\infty) \cup X^{-1}([a,\infty))$$ is actually squarely wrong, as would be the identity $$Z^{-1}([a,\infty))=Y^{-1}([a,\infty)) \cup X^{-1}([a,\infty))$$ $\endgroup$ – Did Nov 27 '15 at 10:15

Let $X, Y$ be random variables in $(\Omega, \mathcal B, \mathbb P)$.

If $A \in \mathcal B$, then $1_A$ and $1_{A^C}$ are random variables.

Note that

$$Z = X1_A + Y1_{A^C}$$

Since sums or products of random variables in $(\Omega, \mathcal B, \mathbb P)$ are random variables in $(\Omega, \mathcal B, \mathbb P)$, $Z$ is a random variable in $(\Omega, \mathcal B, \mathbb P)$.

As for your proof, I think you should say:

  1. $\forall a \in \mathbb R$

  2. have $Z \ge a$ instead of $Z \le a$

  3. $Z$ is $\mathcal B$-measurable


Let $B$ be a borel set then $$Z^{-1}(B)=[Z^{-1}(B)\cap A] \cup [Z^{-1}(B)\cap A^{c}]=[X^{-1}(B)\cap A] \cup [Y^{-1}(B)\cap A^{c}] \\ = X^{-1}(B) \cup Y^{-1}(B)$$ By the property of $\sigma$ field we see that $Z$ is a random variable

  • $\begingroup$ @Dominik check out my solution $\endgroup$ – Josh Sep 24 '15 at 13:34
  • $\begingroup$ Josh, I don't think Dominik was notified because he didn't edit or comment on this post. Looks okay. Pretty much the same as the one you gave in OP (by uniqueness lemma) I guess. I think you can omit the A's at the end but should replace 1 of the X's with a Y $\endgroup$ – BCLC Nov 26 '15 at 23:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.