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Consider the following proposition. Part (i) i have no problem with. Its the proof of part (ii) that (because of my lack of knowledge of advanced measure theory) am having trouble in understanding.

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The proof goes as follows:

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Now the problematic issues are as follows:

  1. How does it follow that since seqquences of mean and variances are bounded we have that $ \sup_n \mathbb{E}[|X_n|^q] < \infty$ for all $q$?
  2. How does it follow that $\sup_n \mathbb{E}[|X_n|^q] < \infty \implies \sup_n \mathbb{E}[|X_n-X|^q] < \infty$ for all $q$?
  3. How do we know that the random variables $Y_n$ converge to 0 in probability?
  4. How does boundedness in $L^2$ imply unfirom integrability?

The last point (that 4 above implies convergence of $Y_n$ to 0 in $ L^1$) I understand as it is a standard result in measure theory.

The book which I am studying is Brownian Motion, Martingales and Stochastic Processes by Jean-François Le Gall.

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1 Answer 1

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  1. This can be proved in many ways. First way: Note that $$E(|X_n|^q) = E(|\sigma_nN + m_n|^q) \leq E((\sigma_n|N| + |m_n|)^q).$$ For $a, b \geq 0$, $(a + b)^q \leq (2\max(a, b))^q = 2^q\max(a, b)^q \leq 2^q(a^q + b^q)$. Hence $$E(|X_n|^q) \leq 2^q(\sigma_n^qE(|N|^q) + |m_n|^q) \to 2^q(\sigma^qE(|N|^q) + |m|^q) < \infty.$$ Second way: Use DCT to conclude that $E(|X_n|^q) = \int |x|^q \frac{1}{\sqrt{2\pi}\sigma_n}\exp(-\frac{(x - m_n)^2}{2\sigma_m^2})\,dx \to \int |x|^q \frac{1}{\sqrt{2\pi}\sigma}\exp(-\frac{(x - m)^2}{2\sigma^2})\,dx < \infty$.

  2. This is again due to $(a + b)^q \leq 2^q(a^q + b^q)$ for $a, b \geq 0$.

  3. You know that $X_n \to X$ in $L^2$, so $X_n \to X$ in probability. This means $|X_n - X| \to 0$ in probability. Then by continuous mapping theorem for convergence in probability, you get $|X_n - X|^p \to 0^p = 0$ in probability.

  4. On a probability space, uniform integrability of a collection $\{X_i : i \in I\}$ is equivalent to existence of a function $f : [0, \infty) \to [0, \infty)$ such that $f(x)/x \to \infty$ as $x \to \infty$ and $\sup_{i \in I}E(f(|X_i|)) < \infty$. See theorem 6.19 on page 154 of "Probability Theory" by Klenke. Now take $f(x) = x^2$.

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