Solution Outline

1. Generating Function of $X_n$ and $E[X_n^2]$

Generating Function: The generating function $G_{X_n}(t)$ of a discrete random variable $X_n$ is defined as: $G_{X_n}(t) = \mathbb{E}[t^{X_n}] = \sum_{x \in \text{Supp}(X_n)} P(X_n = x) t^x.$ For $X_n$, the values of $x$ are $\{0, n\}$. Therefore: $G_{X_n}(t) = P(X_n = 0) \cdot t^0 + P(X_n = n) \cdot t^n = (1 - \frac{1}{n^2}) \cdot 1 + \frac{1}{n^2} \cdot t^n.$ Simplify: $G_{X_n}(t) = 1 - \frac{1}{n^2} + \frac{1}{n^2} t^n.$

Second Moment $E[X_n^2]$: The second moment $E[X_n^2]$ is computed as: $E[X_n^2] = \sum_{x \in \text{Supp}(X_n)} x^2 P(X_n = x).$ For $X_n$, this becomes: $E[X_n^2] = (0^2) \cdot P(X_n = 0) + (n^2) \cdot P(X_n = n).$ Substitute $P(X_n = 0) = 1 - \frac{1}{n^2}$ and $P(X_n = n) = \frac{1}{n^2}$: $E[X_n^2] = (n^2) \cdot \frac{1}{n^2} = 1.$

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2. Convergence of $X_n$ When $\alpha = 3$

When $\alpha = 3$, we analyze convergence types:

• Quadratic Convergence: Quadratic convergence means $E[X_n^2] \to 0$ as $n \to \infty$. Here, $E[X_n^2] = 1$, so quadratic convergence does not occur.

• Convergence in Probability: Convergence in probability means for any $\epsilon > 0$: $P(|X_n - 0| > \epsilon) \to 0 \text{ as } n \to \infty.$ Here: $P(|X_n - 0| > \epsilon) = P(X_n = n) = \frac{1}{n^2}.$ As $n \to \infty$, $\frac{1}{n^2} \to 0$. Thus, $X_n$ converges to 0 in probability.

• Convergence in Law: Convergence in law means the distribution of $X_n$ converges to the distribution of 0 (a degenerate random variable). As $P(X_n = 0) \to 1$ and $P(X_n = n) \to 0$ as $n \to \infty$, $X_n$ converges in law to 0.

Conclusion for $\alpha = 3$: $X_n$ converges to 0 in probability and in law, but not quadratically.

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3. Convergence of $X_n$ When $\alpha = 2$

For $\alpha = 2$, the distribution of $X_n$ and its properties remain unchanged because $E[X_n^2] = 1$, regardless of $\alpha$. The reasoning is identical to part (2).

Conclusion for $\alpha = 2$: $X_n$ converges to 0 in probability and in law, but not quadratically.

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Summary of Results

1. Generating Function: $G_{X_n}(t) = 1 - \frac{1}{n^2} + \frac{1}{n^2} t^n.$ Second Moment: $E[X_n^2] = 1.$

2. For $\alpha = 3$:

   • Converges in probability: Yes.
   • Converges in law: Yes.
   • Quadratic convergence: No.

3. For $\alpha = 2$:

   • Converges in probability: Yes.
   • Converges in law: Yes.
   • Quadratic convergence: No.

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Follow-Up Questions

1. Can you compute $E[X_n]$ and verify if it converges to 0 as $n \to \infty$?
2. How does the variance $\text{Var}(X_n)$ behave as $n \to \infty$?
3. What happens if $P(X_n = n)$ is changed to $\frac{1}{n^\alpha}$ for general $\alpha > 0$?
4. How does the convergence of $X_n$ depend on $\alpha$?
5. Can you prove convergence in law using the characteristic function of $X_n$?

Tip: Always check the definitions of convergence types to determine how they apply to a sequence.