3
$\begingroup$

I am following A first look at rigorous probability theory by Rosenthal, and I am having troubles with limits of random variables.

Specifically proposition 3.1.5. (iii) states that if $Z_1,Z_2...$ are random variables such that $\lim_{n \to \infty}Z_n(w)$ exists for each $w \in \Omega $, and $Z(w)=\lim_{n \to \infty}Z_n(w)$, then Z is also a random variable.

The proof starts out with: For $z \in R$ $$\{Z \le z\} = \bigcap_{m = 1}^\infty \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty \{ Z_k \le z + 1/m \}$$

Why is this true? the book hints to the definition of limit $Z(w)=\lim_{n \to \infty}Z_n(w)$.

Thanks in advance for your explanations

$\endgroup$
2
$\begingroup$

Suppose $w\in\{Z\leq z\}$ then for each $m$, there is $n(m)$ such that for all $k\geq n$, we have $Z_k(w)\leq z+\frac{1}{m}$, which follows because $Z(w)=\lim Z_n(w)$. This means $$ w\in\bigcap_{k=n(m)}^\infty\{Z_k\leq z+1/m\}\subset\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty\{Z_k\leq z+1/m\}. $$ So the rightmost expression contains $w$ for each $m\geq 1$. So we must have $$ w\in\bigcap_{m=1}^\infty\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty\{Z_k\leq z+1/m\}.\tag{*} $$ This proves one direction.

For the other direction, suppose $w\in A$ where $A$ is the right hand side of (*). Suppose that $Z(w)>z$ then there is some integer $M$ such that $Z(w)-\frac{2}{M}>z$. But this means for $k$ sufficiently large, $Z_k(\omega)\geq z+\frac{2}{M}>z+\frac{1}{M}$. Consequently, $$ w\not\in\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty\{Z_k\leq z+1/M\}. $$ In particular, $w$ cannot be in $A$, giving us the desired contradiction.

$\endgroup$
  • $\begingroup$ could we not omit the $1/m$ and still have a true inequality in the forward direction of the proof. Or is it because we do not know if the $Z_n(w)$ approach $Z(w)$ from right or left? $\endgroup$ – Monolite Sep 11 '14 at 10:26
  • 1
    $\begingroup$ I think if you omit $\frac{1}{m}$, even the left-to-right inclusion wouldn't work. Suppose there is $w$ such that $z=Z(w)$. Then it's possible that $Z_n(w)$ can get arbitrarily close to $z$ while alternating both above and below $z$. $\endgroup$ – Kim Jong Un Sep 11 '14 at 12:06
  • $\begingroup$ Did you mean $w\in\{Z^{-1 }\mid Z\leq z\}$ instead of $w\in\{Z\leq z\}$ ? $\endgroup$ – Monolite Dec 23 '14 at 18:40

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.