0
$\begingroup$

This question has been addressed previously in here but my question is different. It is from Hogg and McKean's "Introduction to Mathematical Statistics".

Theorem $5.2.7.$ Let $\{X_n\}$ be a sequence of random variables bounded in probability and let $\{Y_n\}$ be a sequence of random variables that converges to $0$ in probability. Then $$X_n Y_n \xrightarrow{P} 0 .$$

Proof: Let $\epsilon > 0$ be given. Choose $B_{\epsilon} > 0$ and an integer $N_{\epsilon}$ such that $n \geq N_{\epsilon} \implies P[|X_n| \leq B_{\epsilon}] \geq 1−\epsilon.$

Then $$ \begin{align} \overline{\lim_{n \to \infty}} P[|X_nY_n|\geq \epsilon] &\leq \overline{\lim_{n \to \infty}} P[|X_nY_n|\geq \epsilon,|X_n|≤B_{\epsilon}] \\ &+ \overline{\lim_{n \to \infty}} P[|X_nY_n| \geq \epsilon,|X_n| > B_{\epsilon}] \\ &\leq \overline{\lim_{n \to \infty}} P [|Y_n| ≥ \epsilon/B_{\epsilon}] + \epsilon = \epsilon. \end{align}$$ from which the desired result follows.

How exactly does the desired result simply follow?

In the proof above, $\overline{\lim}$ represents the limit supremum (limsup) of that sequence. My issue with this proof is that they show that $$\overline{\lim_{n \to \infty}} P[|X_nY_n|\geq \epsilon] \leq \epsilon$$ but what we want to show is that there exists $N$ such that $$\sup_{n \geq N} \{P[|X_nY_n|\geq \epsilon]\} \leq \delta$$ for any given $\delta$ unrelated to $\epsilon$, which can be made arbitrarily small. The variable $\epsilon$ appears as part of the condition for boundedness of $X_n$ in probability. I don't understand how the statement that the limsup is less than $\epsilon$ would imply that the limsup actually goes to zero. Please help me understand the proof clearly.

$\endgroup$

2 Answers 2

2
$\begingroup$

The following general result would come handy in deducing what the authors did above.

Result $1.$ Let $(\Omega,\boldsymbol{\mathfrak A},\mu)$ be a measure space. Let $\langle f_n\rangle_{n\in\mathbb N}$ be a sequence of extended real-valued $\boldsymbol{\mathfrak A}$-measurable functions on $D\in\boldsymbol{\mathfrak A};$ let $f$ also be an extended real-valued $\boldsymbol{\mathfrak A}$-measurable function on $D.$ Then $$f_n\overset{\mu}{\longrightarrow} f\iff \forall ~\delta > 0~\exists ~N_\delta\in\mathbb N:\mu\{D:\vert f_n-f\vert\geq\delta\}<\delta~~\forall~ n\geq N_\delta.\tag 1\label 1$$

To see $\eqref 1\implies ~\mu-\textrm{convergence}, $ note that you can take $\delta:=\eta\wedge\varepsilon$ where both $\eta, ~\varepsilon> 0$ and are arbitrarily chosen. By consideration of such $\delta, $ it implies $\{D:\vert f_n-f\vert\geq\varepsilon\} \subset\{D:\vert f_n-f\vert\geq\delta\}$ following which we get $$\mu\{D:\vert f_n-f\vert\geq\varepsilon\} \leq \mu\{D:\vert f_n-f\vert\geq\delta\}<\delta\leq \eta~~\forall~ n\geq N_\delta\implies \mu\{D:\vert f_n-f\vert\geq\varepsilon\}<\eta~~\forall~ n\geq N_\delta,$$ which is nothing but $\lim_{n\to\infty}\mu\{D:\vert f_n-f\vert\geq\varepsilon\}=0, $ the definition of convergence in measure.

$\blacksquare$

You can now formally show how it is related to the author's statement.


Reference:

$\rm [I]$ Real Analysis: Theory of Measure and Integration, J. Yeh, World Scientific, $2014, $ sec. $1\S6\rm[III], $ pp. $111-112.$

$\endgroup$
2
  • $\begingroup$ Thanks for the answer. It is proposition 6.17 in Yeh's book. It took me a while to figure out what $\eta \wedge \varepsilon$ meant. Then I found this link on math SE math.stackexchange.com/q/3454725/145325. So essentially, we apply the logic to minimum of $\eta,\varepsilon$ after which I can say, like the author did, that "the desired result follows." :) :) $\endgroup$ Oct 31, 2023 at 8:30
  • 1
    $\begingroup$ Yeh's book is very much underrated and yet one of the most comprehensive books, and I would recommend it to anyone. $\endgroup$ Oct 31, 2023 at 8:34
1
$\begingroup$

If the $\limsup$ is less than $\epsilon$ for every positive $\epsilon$, then it's zero. It can't be, for example, $1/3, $ because you could take $\epsilon=1/4$. It can't be $0.00042,$ because you could take $\epsilon=0.00001$. It can't be $2^{-10^{100}}$, because you could take $\epsilon=2^{-10^{100}-1}$. And so on.

You don't need a $\delta$ separate from $\epsilon$, it doesn't add any more generality.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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