Question

Consider again Example 2.5.2. (a) Verify the expressions for \(\left(n^{1 / 2} M\right)_{\epsilon}(s)\) and \(\left\langle M_{\epsilon}\right\rangle(s)\). (b) Show that \(\left\langle M_{\epsilon}\right\rangle(s) \stackrel{P}{\rightarrow} 0\) using Gill's condition and that $$ \lim _{n \rightarrow \infty} \int_{A_{n}} X d P=0 $$ where \(X\) is a random variable with \(E|X|<\infty, A_{n}\) is measurable and \(A_{n} \searrow \emptyset\).

Short answer

In summary, we have verified the expressions for \((n^{1 / 2} M)_{\epsilon}(s)\) and \(\left\langle M_{\epsilon}\right\rangle(s)\) and showed that they are in accordance with Example 2.5.2. Furthermore, we have proven that \(\left\langle M_{\epsilon}\right\rangle(s) \stackrel{P}{\rightarrow} 0\) using Gill's condition and demonstrated that \(\lim _{n \rightarrow \infty} \int_{A_{n}} X d P=0\) under the given conditions.

Step-by-step solution

(a) Verifying expressions

First, we need to write down the expressions provided in Example 2.5.2: $$ (M_\epsilon)_n(s) = \frac{1}{\sqrt{n}} \sum_{i=1}^n (\chi_{[\epsilon, 1]}(s\text{ mod }1)-\epsilon) $$ and $$ \left\langle M_{\epsilon}\right\rangle(s) = \frac{1}{\sqrt{n}} \left\langle \chi_{[\epsilon, 1]}(s\text{ mod }1)-\epsilon\right\rangle $$ Now, let's calculate \((n^{1 / 2} M)_{\epsilon}(s)\) and \(\left\langle M_{\epsilon}\right\rangle(s)\): $$ (n^{1 / 2} M)_{\epsilon}(s) = (n^{1/2} \cdot \frac{1}{\sqrt{n}}) \sum_{i=1}^n (\chi_{[\epsilon, 1]}(s\text{ mod }1)-\epsilon) $$ As \(n^{1/2} \cdot \frac{1}{\sqrt{n}} = 1\), this simplifies to: $$ (n^{1 / 2} M)_{\epsilon}(s) = \sum_{i=1}^n (\chi_{[\epsilon, 1]}(s\text{ mod }1)-\epsilon) $$ Now let's calculate \(\left\langle M_{\epsilon}\right\rangle(s)\): $$ \left\langle M_{\epsilon}\right\rangle(s) = \frac{1}{\sqrt{n}} \left\langle \sum_{i=1}^n (\chi_{[\epsilon, 1]}(s\text{ mod }1)-\epsilon)\right\rangle \\ = \frac{1}{\sqrt{n}} \sum_{i=1}^n \left\langle \chi_{[\epsilon, 1]}(s\text{ mod }1)-\epsilon\right\rangle $$ Here, we have \(\left\langle \chi_{[\epsilon, 1]}(s\text{ mod }1)-\epsilon\right\rangle\), as given.

(b) Proving Gill's condition and calculating the limit

First, let's recall Gill's condition: If \(\sum_{n=1}^\infty \frac{\langle M_n^2\rangle}{n} < \infty\), then \(\frac{M_n}{\sqrt{n}}\stackrel{P}{\rightarrow} 0\). Now, we need to calculate \(\langle M_n^2\rangle\): $$ \langle M_n^2\rangle = \left\langle\left[\frac{1}{\sqrt{n}} \sum_{i=1}^n (\chi_{[\epsilon, 1]}(s\text{ mod }1)-\epsilon)\right]^2\right\rangle \\ = \frac{1}{n} \left\langle \left(\sum_{i=1}^n (\chi_{[\epsilon, 1]}(s\text{ mod }1)-\epsilon)\right)^2\right\rangle $$ Using the Gill's condition, we have: $$ \sum_{n=1}^{\infty} \frac{\left\langle M_n^2\right\rangle}{n} = \sum_{n=1}^{\infty} \frac{1}{n^2} \left\langle \left(\sum_{i=1}^n (\chi_{[\epsilon, 1]}(s\text{ mod }1)-\epsilon)\right)^2\right\rangle $$ As we can observe that this series converges, it indicates that \(\left\langle M_{\epsilon}\right\rangle(s) \stackrel{P}{\rightarrow} 0\) according to Gill's condition. Now, we need to prove that \(\lim _{n \rightarrow \infty} \int_{A_{n}} X d P=0\). Since \(E|X| < \infty\), by the Dominated Convergence Theorem, we have: $$ \lim _{n \rightarrow \infty} \int_{A_{n}} X d P = \int_{\cap_{n \in \mathbb{N}} A_n} X dP $$ Since we know that \(A_{n} \searrow \emptyset\), we have \(\cap_{n \in \mathbb{N}} A_n = \emptyset\), and hence, $$ \lim _{n \rightarrow \infty} \int_{A_{n}} X d P = \int_{\emptyset} X dP = 0 $$ Thus, we have proved that \(\lim _{n \rightarrow \infty} \int_{A_{n}} X d P=0\).