Weak Law of Large Numbers (W.L.L.N.) Condition

The question asks for the necessary and sufficient condition for a sequence of random variables $\{X_n\}$ to satisfy the Weak Law of Large Numbers (W.L.L.N.). We are given $S_n = \sum_{i=1}^n X_i$ and $Y_n = \left( \frac{S_n - E(S_n)}{n} \right)$.

Analysis of Variable $Y_n$

• $Y_n$ represents the standardized difference between the sample mean $\frac{S_n}{n}$ and its expectation $E(\frac{S_n}{n})$.
• It follows that $E(Y_n) = 0$.

Evaluating the Convergence Criterion

The core condition is expressed using the expectation of a function involving $Y_n$. Let's examine the criterion:

$ E\left( \frac{Y_n^2}{1 + Y_n^2} \right) \to 0 \text{ as } n \to \infty $

• This expectation condition implies convergence in probability. Specifically, it means the term $\frac{Y_n^2}{1 + Y_n^2}$ converges to 0 in probability as $n \to \infty$.
• Since $0 \le \frac{Y_n^2}{1 + Y_n^2} < 1$, the convergence of $\frac{Y_n^2}{1 + Y_n^2}$ to 0 implies $Y_n^2 \to 0$ in probability.
• $Y_n^2 \to 0$ in probability is equivalent to $Y_n \to 0$ in probability.

Establishing the W.L.L.N. Link

• $Y_n \to 0$ in probability means $\frac{S_n - E(S_n)}{n} \to 0$ in probability.
• This can be rewritten as $\frac{S_n}{n} - E\left(\frac{S_n}{n}\right) \to 0$ in probability.
• Assuming the expectation $E(X_1)$ exists (let it be $\mu$), then $E\left(\frac{S_n}{n}\right) = \mu$.
• The convergence becomes $\frac{S_n}{n} - \mu \to 0$ in probability.
• This is precisely the statement of the Weak Law of Large Numbers.

Therefore, the condition $E\left( \frac{Y_n^2}{1 + Y_n^2} \right) \to 0$ is the necessary and sufficient criterion for the sequence $\{X_n\}$ to satisfy the W.L.L.N.