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Let X1, X2, . . . be a sequence of iid random variables with common characterist...
May 9, 2024
Let X1, X2, . . . be a sequence of iid random variables with common characteristic function ϕ(t) = exp(−|t|^α)), where α > 0. Determine the distribution of the normalised sum Sn = (n^−1)*(X1 +· · · + Xn) as n → ∞.
Solution by Steps
step 1
Identify the characteristic function of the sum SnS_n
step 2
Use the property that the characteristic function of the sum of independent random variables is the product of their individual characteristic functions
step 3
Calculate the characteristic function of SnS_n using the given characteristic function ϕ(t)=exp⁡(−∣t∣α)\phi(t) = \exp(-|t|^\alpha)
step 4
Apply the Central Limit Theorem (CLT) for iid random variables with a common characteristic function
step 5
Determine the limiting distribution of SnS_n as n→∞n \to \infty by analyzing the behavior of the characteristic function of SnS_n
Answer
The normalized sum SnS_n converges in distribution to a stable distribution with characteristic exponent α\alpha.
Key Concept
Central Limit Theorem for iid random variables and stable distributions
Explanation
The Central Limit Theorem states that the normalized sum of iid random variables with a common characteristic function converges in distribution to a stable distribution, provided that the characteristic function satisfies certain conditions. In this case, the given characteristic function leads to a stable distribution with characteristic exponent α\alpha.
Let (Ω, F, P) be a probability space, and A1, A2, . . . be an increasing sequence of events; that is, A1 ⊆ A2 ⊆ · · · such that P(A1) = P(A2) = · · · = p > 0. Does the sequence of events converge in probability to the event A = U (from n=1 to inf) An? Prove or disprove this. Hint: For a sequence of events, convergence in probability can be written as limn→∞ P(A4An) = 0, where 4 is the symmetric difference.
Solution by Steps
step 1
Consider the definition of convergence in probability for events
step 2
By definition, a sequence of events AnA_n converges in probability to an event AA if lim⁡n→∞P(AΔAn)=0\lim_{n \to \infty} P(A \Delta A_n) = 0, where Δ\Delta denotes the symmetric difference between sets
step 3
Given that AnA_n is an increasing sequence, we have A1⊆A2⊆…A_1 \subseteq A_2 \subseteq \ldots and A=⋃n=1∞AnA = \bigcup_{n=1}^{\infty} A_n
step 4
For any nn, AΔAn=A∖AnA \Delta A_n = A \setminus A_n because An⊆AA_n \subseteq A
step 5
Since An⊆An+1A_n \subseteq A_{n+1}, it follows that A∖An⊇A∖An+1A \setminus A_n \supseteq A \setminus A_{n+1}, which implies P(A∖An)≥P(A∖An+1)P(A \setminus A_n) \geq P(A \setminus A_{n+1})
step 6
The probability P(A∖An)P(A \setminus A_n) is non-increasing as nn increases and is bounded below by 0
step 7
By the continuity of probability measures, lim⁡n→∞P(A∖An)=P(lim⁡n→∞(A∖An))\lim_{n \to \infty} P(A \setminus A_n) = P(\lim_{n \to \infty} (A \setminus A_n))
step 8
Since A=⋃n=1∞AnA = \bigcup_{n=1}^{\infty} A_n, the limit lim⁡n→∞(A∖An)=A∖A=∅\lim_{n \to \infty} (A \setminus A_n) = A \setminus A = \emptyset
step 9
Therefore, lim⁡n→∞P(AΔAn)=P(∅)=0\lim_{n \to \infty} P(A \Delta A_n) = P(\emptyset) = 0
Answer
The sequence of events AnA_n converges in probability to the event A=⋃n=1∞AnA = \bigcup_{n=1}^{\infty} A_n.
Key Concept
Convergence in Probability for Events
Explanation
A sequence of events AnA_n converges in probability to an event AA if the probability of the symmetric difference between AnA_n and AA goes to zero as nn approaches infinity. In this case, the increasing nature of the sequence and the continuity of probability measures ensure the convergence in probability.
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