Question

Let $\{X_{n}\}$ be a sequence of $r \cdot v$'s and $Y_{n}=\left(\frac{S_{n}-E(S_{n})}{n}\right)$ where $S_{n}=\sum_{i=1}^{n} X_{i}$ then the necessary and sufficient condition for the sequence $\{X_{n}\}$ to satisfy W.L.L.N is

• A. $E\left(\frac{Y_{n}}{1+Y_{n}}\right) \rightarrow 0$ as $n \rightarrow \infty$
• B. $E\left(\frac{Y_{n}^{2}}{1+Y_{n}^{2}}\right) \rightarrow 0$ as $n \rightarrow \infty$
• C. $E\left(\frac{Y_{n}}{1+Y_{n}^{2}}\right) \rightarrow 0$ as $n \rightarrow \infty$
• D. $E\left(\frac{Y_{n}^{2}}{1+Y_{n}}\right) \rightarrow 0$ as $n \rightarrow \infty$

Hint:

Convergence in probability can always be expressed as the expectation of a bounded function. The form $\frac{x^2}{1+x^2}$ is the most common "distance" function used to prove WLLN for variables without finite second moments.

Answer 1

The Correct Option is B

We examine the definition of convergence in probability and its relationship to bounded expectations.

Step 1: \color{redDefine WLLN
The Weak Law of Large Numbers (WLLN) holds if $Y_n = \frac{S_n - E(S_n)}{n} \xrightarrow{P} 0$ as $n \to \infty$.
Convergence in probability $Y_n \xrightarrow{P} 0$ means for any $\epsilon > 0$, $P(|Y_n| > \epsilon) \to 0$.

Step 2: \color{redLink to bounded functions
A sequence $Y_n \xrightarrow{P} 0$ if and only if $E[g(|Y_n|)] \to 0$ for any continuous, bounded, non-negative function $g$ such that $g(0) = 0$ and $g(x) > 0$ for $x > 0$.

Step 3: \color{redEvaluate the options
The function $g(x) = \frac{x^2}{1+x^2}$ is a standard choice in analysis to metrize convergence in probability.
It is bounded (between 0 and 1), continuous, and $g(0)=0$.
Therefore, $Y_n \xrightarrow{P} 0 \iff E\left(\frac{Y_n^2}{1+Y_n^2}\right) \to 0$.

Step 4: \color{redVerify sufficiency
By Markov's inequality:
$P(|Y_n| > \epsilon) = P\left(\frac{Y_n^2}{1+Y_n^2} > \frac{\epsilon^2}{1+\epsilon^2}\right) \le \frac{1+ \epsilon^2}{\epsilon^2} E\left(\frac{Y_n^2}{1+Y_n^2}\right)$.
If the expectation goes to 0, the probability must also go to 0.
Thus, the condition is $E\left(\frac{Y_{n}^{2}}{1+Y_{n}^{2}}\right) \rightarrow 0$.