Analyzing Convergence in Probability

We are given a sequence of independent and identically distributed (i.i.d.) random variables $\{X_i\}_{i \geq 1}$ with $E(X_i) = 0$ and $V(X_i) = 1$. We need to identify the statements that hold true regarding convergence in probability.

Convergence of Sums: $\sum X_i$

Let $S_n = \sum_{i=1}^n X_i$. We know $E(S_n) = n E(X_i) = n \times 0 = 0$ and $V(S_n) = n V(X_i) = n \times 1 = n$.

• Option 3: Consider $\frac{1}{n^{1/2}} \sum_{i=1}^n X_i = \frac{S_n}{n^{1/2}}$. The variance is $V\left(\frac{S_n}{n^{1/2}}\right) = \frac{V(S_n)}{(n^{1/2})^2} = \frac{n}{n} = 1$. Since the variance is constant and non-zero, and by the Central Limit Theorem (CLT), $\frac{S_n}{n^{1/2}}$ converges in distribution to $N(0, 1)$, it does not converge to 0 in probability. Thus, option 3 is false.
• Option 2 (B): Consider $\frac{1}{n^{3/4}} \sum_{i=1}^n X_i = \frac{S_n}{n^{3/4}}$. The variance is $V\left(\frac{S_n}{n^{3/4}}\right) = \frac{V(S_n)}{(n^{3/4})^2} = \frac{n}{n^{3/2}} = \frac{1}{n^{1/2}}$. As $n \to \infty$, the variance $V\left(\frac{S_n}{n^{3/4}}\right) \to 0$. Since the mean is 0 and the variance converges to 0, the variable $\frac{S_n}{n^{3/4}}$ converges to 0 in probability. Thus, option 2 (B) is true.

Convergence of Sums of Squares: $\sum X_i^2$

Let $Y_i = X_i^2$. The variables $Y_i$ are also i.i.d.

First, we find the expected value of $Y_i$: $E(Y_i) = E(X_i^2)$. Using the variance formula $V(X_i) = E(X_i^2) - (E(X_i))^2$, we have: $1 = E(X_i^2) - (0)^2$ $E(X_i^2) = 1$. So, $E(Y_i) = 1$.

• Option 4 (D): Consider $\frac{1}{n} \sum_{i=1}^n X_i^2 = \frac{1}{n} \sum_{i=1}^n Y_i$. By the Law of Large Numbers (LLN), the sample mean of $Y_i$ converges in probability to its expected value: $ \frac{1}{n} \sum_{i=1}^n Y_i \to E(Y_i) \quad \text{in probability} $ $ \frac{1}{n} \sum_{i=1}^n X_i^2 \to 1 \quad \text{in probability} $ Thus, option 4 (D) is true.
• Option 1: This option claims $\frac{1}{n} \sum_{i=1}^n X_i^2 \to 0$ in probability, which contradicts our finding that it converges to 1. Thus, option 1 is false.

Conclusion

Based on the analysis using the Law of Large Numbers and properties of variance, the true statements are:

• Option 2 (B): $\frac{1}{n^{3/4}} \sum_{i=1}^n X_i \to 0$ in probability.
• Option 4 (D): $\frac{1}{n} \sum_{i=1}^n X_i^2 \to 1$ in probability.