/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Let \(X_{1}, \ldots, X_{n}\) be ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(X_{1}, \ldots, X_{n}\) be integrable i.i.d. random variables. Let \(S_{n}=X_{1}+\) \(\ldots+X_{n}\). Show that $$ \mathbf{E}\left[X_{i} \mid S_{n}\right]=\frac{1}{n} S_{n} \quad \text { for every } i=1, \ldots, n $$

Short Answer

Expert verified
\(\mathbf{E}[X_i | S_n] = \frac{1}{n} S_n\) for each \(i = 1, \ldots, n\).

Step by step solution

01

Understanding Expectations and Conditional Expectations

The problem involves the expectation and conditional expectation of i.i.d. random variables. Here, \(X_1, X_2, \ldots, X_n\) are identically distributed, so their expectations and distributions are the same. Conditional expectation \( \mathbf{E}[X_i \mid S_n] \) relates to the expected value of a random variable \(X_i\), given the sum \(S_n\).
02

Definition of Conditional Expectation

The conditional expectation \( \mathbf{E}[X_i | S_n] \) is the expected value of \(X_i\) given that we know the sum \(S_n = X_1 + X_2 + \cdots + X_n\). Here, each \(X_i\) contributes equally to \(S_n\), since they are identically distributed.
03

Using Symmetry of i.i.d. Random Variables

Since all \(X_i\) are i.i.d., each one contributes the same amount to the sum \(S_n\). Thus, the contribution of each \(X_i\) can be intuitively represented as \(\frac{S_n}{n}\), given the symmetric role they play in the sum.
04

Writing the Conditional Expectation in Terms of Symmetric Functions

Consider the symmetric nature of the sum and the i.i.d. variables: since each \(X_i\) is identically distributed, \(\mathbf{E}[X_i | S_n] = \mathbf{E}[X_j | S_n]\) for any \(i, j\). Thus, averaging over \(n\) terms, each should contribute equally to \(S_n\) in expectation: \(\mathbf{E}[X_i | S_n] = \frac{1}{n}S_n\).
05

Finalizing the Expectation Result

By applying symmetry and understanding the contribution of each \(X_i\) to the sum \(S_n\), we conclude that \(\mathbf{E}[X_i | S_n] = \frac{1}{n} S_n\) for each \(i = 1, \ldots, n\). This satisfies the condition provided in the problem statement.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

i.i.d. Random Variables
i.i.d. stands for independent and identically distributed. These terms are crucial for understanding how certain random variables behave.
Independent random variables don't affect each other. If you know the outcome of one, it doesn't tell you anything about another. This independence simplifies calculations in probability and statistics.
Identically distributed means all random variables follow the same probability distribution. They have the same average (expected value), variance, and higher moments. In simpler terms, they behave the same way statistically.
  • Example: Imagine tossing a fair coin multiple times. Each toss is independent of the others and has the same probability (1/2 for heads and 1/2 for tails).
  • Application: In our problem, if we have multiple instances of an i.i.d. variable, they are all going to contribute equally to the sum because they share the same distribution.
Symmetric Functions
Symmetric functions refer to scenarios where the roles of input variables are interchangeable without affecting the outcome. This is important for i.i.d. variables because their identical nature means treating them differently doesn't change the result.
In the context of our problem, the sum of i.i.d. variables, represented as \(S_n = X_1 + X_2 + \dots + X_n\), is symmetric because swapping any two \(X_i\) values doesn't change the value of the sum.
This symmetry implies that each \(X_i\) contributes equally to \(S_n\), leading to the intuitive and neat result of the expected value calculation \(\mathbf{E}[X_i | S_n] = \frac{1}{n} S_n\). By this result, each variable's impact on the total is equally averaged across all \(n\) variables.
  • Insights: Understanding symmetric functions can help us realize situations in mathematical problems where every component shares equal impact or importance in the final result.
Expectation
Expectation, or expected value, is a fundamental concept in probability, representing the average outcome if you repeated an experiment many times. It's calculated using the probability-weighted sum of all possible outcomes.
Mathematically expressed, the expectation of a random variable \(X\) is \(\mathbf{E}[X] = \sum x P(X = x)\). This means you take each possible value \(x\), multiply it by how likely \(x\) is to occur, and sum up these products.
The concept of expectation applies to conditional scenarios too, as seen in our problem. With conditional expectation, the focus is on finding the expected value given some condition or previous knowledge, like knowing the sum \(S_n\):
\(\mathbf{E}[X_i | S_n]\) tells us what average value to expect for \(X_i\) if we already know the value of \(S_n\). Here, symmetry in i.i.d. variables results in a straightforward \(\frac{1}{n} S_n\) expectation for each \(X_i\).
  • Useful Tip: Expectations provide a way to simplify complex random processes into understandable averages, which are easier to work with and interpret.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \((E, \mathcal{E})\) be a Borel space and let \(\mu\) be an atom-free measure (that is, \(\mu(\\{x\\})=0\) for any \(x \in E\) ). Show that for any \(A \in \mathcal{E}\) and any \(n \in \mathbb{N}\), there exist pairwise disjoint sets \(A_{1}, \ldots, A_{n} \in \mathcal{E}\) with \(\biguplus_{k=1}^{n} A_{k}=A\) and \(\mu\left(A_{k}\right)=\mu(A) / n\) for any \(k=1, \ldots, n\).

Let \(X\) and \(Y\) be real random variables with joint density \(f\) and let \(h: \mathbb{R} \rightarrow \mathbb{R}\) be measurable with \(\mathbf{E}[|h(X)|]<\infty\). Denote by \(\lambda\) the Lebesgue measure on \(\mathbb{R}\). (i) Show that almost surely $$ \mathbf{E}[h(X) \mid Y]=\frac{\int h(x) f(x, Y) \lambda(d x)}{\int f(x, Y) \lambda(d x)} $$ (ii) Let \(X\) and \(Y\) be independent and \(\exp _{\theta}\)-distributed for some \(\theta>0\). Compute \(\mathbf{E}[X \mid X+Y]\) and \(\mathbf{P}[X \leq x \mid X+Y]\) for \(x \geq 0\)

Assume the random variable \((X, Y)\) is uniformly distributed on the disc \(B:=\left\\{(x, y) \in \mathbb{R}^{2}: x^{2}+y^{2} \leq 1\right\\}\) and on \([-1,1]^{2}\), respectively. (i) In both cases, determine the conditional distribution of \(Y\) given \(X=x\). (ii) Let \(R:=\sqrt{X^{2}+Y^{2}}\) and \(\Theta=\arctan (Y / X) .\) In both cases, determine the conditional distribution of \(\Theta\) given \(R=r\).

Let \(X_{1}\) and \(X_{2}\) be independent and exponentially distributed with parameter \(\theta>0\). Compute \(\mathbf{E}\left[X_{1} \wedge X_{2} \mid X_{1}\right]\)

(Rejection sampling for generating random variables) Let \(E\) be a countable set and let \(P\) and \(Q\) be probability measures on \(E\). Assume there is a \(c>0\) with $$ f(e):=\frac{Q(\\{e\\})}{P(\\{e\\})} \leq c \quad \text { for all } e \in E \text { with } P(\\{e\\})>0 $$ Let \(X_{1}, X_{2}, \ldots\) be independent random variables with distribution \(P\). Let \(U_{1}, U_{2}, \ldots\) be i.i.d. random variables that are independent of \(X_{1}, X_{2}, \ldots\) and that are uniformly distributed on \([0,1]\). Let \(N\) be the smallest (random) nonnegative integer \(n\) such that \(U_{n} \leq f\left(X_{n}\right) / c\) and define \(Y:=X_{N}\) Show that \(Y\) has distribution \(Q\). Remark. This method for generating random variables with a given distribution \(Q\) is called rejection sampling, as it can also be described as follows. The random variable \(X_{1}\) is a proposal for the value of \(Y\). This proposal is accepted with probability \(f\left(X_{1}\right) / c\) and is rejected otherwise. If the first proposal is rejected, the game starts afresh with proposal \(X_{2}\) and so on.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.