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Chapter 17: Problem 11

Suppose \(\lim _{n \rightarrow \infty} X_{n}=X\) a.s. and \(|X|<\infty\) a.s. Let \(Y=\sup _{n}\left|X_{n}\right|\). Show that \(Y<\infty\) a.s.

Short Answer

Expert verified
The supremum \(Y = \sup_n |X_n|\) is finite almost surely because \(X_n\) converges to a finite \(X\).

Step by step solution

01

Understand the definition of the limit almost surely

Given \(\lim _{n \rightarrow \infty} X_{n}=X\) a.s., this means for almost every outcome \(\omega\), \(X_n(\omega)\) approaches \(X(\omega)\) as \(n\) approaches infinity. Thus, the sequence \(X_n\) converges to \(X\) for almost all \(\omega\).
02

Analyze the condition \(|X| < \infty\) almost surely

Since \(|X| < \infty\) a.s., it implies that the random variable \(X\) is finite for almost all outcomes, i.e., except for a set of probability zero. Thus, the sequence converges to a finite limit \(X\).
03

Define \(Y=\sup _{n}\left|X_{n}\right|\) and analyze

Since \(Y = \sup_{n}|X_{n}|\) represents the supremum or least upper bound of the sequence of random variables \(|X_n|\), it intuitively captures the largest value \(|X_n(\omega)|\) could achieve for any \(n\).
04

Use the convergence result to argue \(Y < \infty\)

For almost all outcomes \(\omega\), since \(X_n(\omega)\) converges to \(X(\omega)\) and \(|X(\omega)| < \infty\), there cannot be infinitely many \(|X_n(\omega)|\) larger than any finite bound, otherwise \(X_n\) would not converge. Hence, \(Y(\omega) = \sup_n |X_n(\omega)|\) must also be finite.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limits in Probability
In probability theory, a limit is an essential concept. It involves sequences of random variables converging to a specific value as the number of observations (n) grows infinitely large. "Almost surely" convergence is a stronger type of convergence because it means the sequence converges for every single outcome except for a negligible set with probability zero.

Here's how to understand it better:
  • **Sequence Convergence:** For a sequence of random variables \(X_n\), we say \(\lim_{n \to \infty} X_n = X\) almost surely if, for almost all outcomes \(\omega\), the sequence values approach \(X(\omega)\).

  • **Probability Zero:** The concept of a set with probability zero means that the outcomes in which convergence might not occur are so rare that they are practically invisible.

Thus, limits in probability are foundational in understanding how random variables behave as we consider them over an infinite sequence of trials.
Random Variables
Random variables are a key component in probability and statistics, representing quantities whose outcomes are determined by chance. They take on different values, each linked to a specific probability, based on the randomness inherent in the process.
  • **Definition:** A random variable is a function that assigns a numerical value to each outcome in a sample space. It helps translate qualitative outcomes into quantitative analysis.

  • **Finiteness:** If a random variable \(X\) has a finite value almost surely, it means that with a very high degree of certainty, minus some negligible cases, \(|X|\) will not reach infinity.

Understanding random variables and their properties like having finite bounds allows us to measure and predict behavior in a probability context effectively.
Supremum of a Sequence
The "supremum" or "least upper bound" of a sequence is a mathematical concept that helps us identify the greatest value within a sequence of numbers. In probability theory, when handling random variables, it provides an insight into the maximum value a variable can attain.

For any sequence of random variables \(|X_n|\), the supremum \(Y = \sup_n |X_n|\) represents the highest peak that \(|X_n(\omega)|\) might ever reach, considering all possible values of \(n\).
  • **Understanding Supremum:** Unlike the maximum, which must be an actual member of the sequence, the supremum represents the smallest value that is greater than or equal to all values in the sequence.

  • **Application in Convergence:** Since we are given almost sure convergence to a finite \(X\), we know that \( \sup_n |X_n(\omega)| < \infty \) for almost all \( \omega \). Meaning, the supremum itself is finite, reinforcing the convergence notion.

The supremum concept ensures that even as we extend our consideration to an infinite sequence, we still have an understanding of the maximum bounds the variables might achieve.

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Most popular questions from this chapter

Let \(X_{n}\) and \(X\) be real-valued r.v.'s in \(L^{2}\), and suppose that \(X_{n}\) tends to \(X\) in \(L^{2}\). Show that \(E\left\\{X_{n}^{2}\right\\}\) tends to \(E\left\\{X^{2}\right\\}\) (Hint: use that \(\left|x^{2}-y^{2}\right|=\) \((x-y)^{2}+2|y||x-y|\) and the Cauchy-Schwarz inequality).

Suppose \(\lim _{n \rightarrow \infty} X_{n}=X\) a.s. Let \(Y=\sup _{n}\left|X_{n}-X\right| .\) Show \(Y<\infty\) a.s. (see Exercise 17.11), and define a new probability measure \(Q\) by $$ Q(A)=\frac{1}{c} E\left\\{1_{A} \frac{1}{1+Y}\right\\}, \text { where } c=E\left\\{\frac{1}{1+Y}\right\\} $$ Show that \(X_{n}\) tends to \(X\) in \(L^{1}\) under the probability measure \(Q\).

Let \(X_{n}\) be i.i.d. random variables with \(P\left(X_{n}=1\right)=\frac{1}{2}\) and \(P\left(X_{n}=\right.\) \(-1)=\frac{1}{2} .\) Show that $$ \frac{1}{n} \sum_{j=1}^{n} X_{j} $$ converges to 0 in probability. (Hint: Let \(S_{n}=\sum_{j=1}^{n} X_{j}\), and use Chebyshev's inequality on \(P\left\\{\left|S_{n}\right|>n \varepsilon\right\\} .\) )

Suppose \(\left|X_{n}\right| \leq Y\) a.s., each \(n, n=1,2,3 \ldots\) Shhow that \(\sup _{n}\left|X_{n}\right| \leq Y\) a.s. also.

Let \(\left(X_{n}\right)_{n \geq 1}\) be random variables with \(X_{n} \stackrel{P}{\rightarrow} X\left(\lim _{n \rightarrow \infty} X_{n}=X\right.\) in probability). Suppose \(\left|X_{n}(\omega)\right| \leq\) \(C\) for a constant \(C>0\) and all \(\omega\). Show that \(\lim _{n \rightarrow \infty} E\left\\{\left|X_{n}-X\right|\right\\}=0\). (Hint: First show that \(P(|X| \leq C)=1\).)
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