2
$\begingroup$

I have this following problem on my Probability problem set.

Let $(X_{n})$ be a sequence of random variables and $X$ another random variable in $(\Omega, \mathcal{F}, P)$ such that $P(\{\omega \in \Omega: \limsup X_{n}(\omega) \leq X(\omega)\}) = 1$. Show that for any $\epsilon > 0$, there is an event $A$ with $P(A) < \epsilon$ and $N \in \mathbb{N}$ large enough so $X_{n}(\omega) < X(\omega) + \epsilon$ for all $n \geq N$ and for all $\omega \in A^{c}$.

Here's my work.

Given $\omega \in \Omega$, $(X_{n} (\omega))$ is a sequence of real numbers and I will omit $\omega$ but I have already chosen one in $\Omega$. Let $k \in \mathbb{N}$. Then, $\limsup X_{n} \leq X + 1/k$ if and only if there is some $n_{0}$ such that $X_{n} \leq X + 1/k, \forall n>n_{0}$. If this is right, this translates to:

$\{\omega \in \Omega: \limsup X_{n}(\omega) \leq X(\omega)\} = \bigcap_{k=1}^{\infty}\bigcup_{n=1}^{\infty}\bigcap_{z=n}^{\infty}\{\omega \in \Omega: X_{z} < X + 1/k\}$

Then, I think that the result follows because we can take $A$ as the complement of the set in the LHS, which will have measure zero, less then any given $\epsilon > 0$ and if we let $\omega$ in the RHS set, we will have that for all $n > N$, with $N$ large enough, the desired inequality.

My question is if the translation from limsup properties to the set language is right and if the whole argument is sound. Thanks a lot for the support!

$\endgroup$
1
$\begingroup$

Let $\Omega'=\{\limsup_n X_n\le X\}$. For fixed $n\ge 1$, let $$ A_{n}=\cup_{m\ge n}\{X_m\ge X+\epsilon\}\cap\Omega'. $$

$\{A_{n}\}_{n\ge 1}$ is a sequence of decreasing sets s.t. $\cap_{n\ge 1}A_{n}=\emptyset$. Hence, $\mathsf{P}\{A_{n}\}\to 0$ so that we may choose $n_\epsilon$ s.t. $\mathsf{P}\{A_{n_\epsilon}\}<\epsilon$. Take $A=\Omega'^{c}\cup A_{n_\epsilon}$. Then $\mathsf{P}(A)\le \mathsf{P}(\Omega'^{c})+\mathsf{P}(A_{n_\epsilon})<\epsilon$ and $X_n<X+\epsilon$ on $A^{c}$ for all $n\ge n_\epsilon$.

$\endgroup$

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.