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Chapter 6: Problem 7

Let \(X_{1}, X_{2}, \ldots\) be independent, identically distributed, continuous random variables. Define \(N\) as the index such that $$ X_{1} \geq X_{2} \geq \cdots \geq X_{N-1} \quad \text { and } \quad X_{N-1}

Short Answer

Expert verified
\(\mathbb{P}(N=k) = \frac{k-1}{k!}\) and \(\mathbb{E}(N) = e\).

Step by step solution

01

Understanding the Problem

We're dealing with a sequence of independent, identically distributed random variables, and we're interested in finding the distribution and expectation of a stopping time \(N\) based on a specific ordering condition. The task is to prove the probability distribution of \(N\) and its expected value.
02

Probability of Event for Specific N

For \(N = k\), the event is\[ X_{1} \geq X_{2} \geq \cdots \geq X_{k-1} < X_{k}. \]Since the variables are continuous and independent, the probability that \(X_{j} = X_{k}\) for any \(j < k\) is zero. Hence, we only need to consider strict inequalities.
03

Permutation Argument

Note that any permutation of \(X_1, X_2, \ldots, X_k\) has equal probability. The favorable arrangement is where the largest element \(X_k\) ends up at the k-th position after a sequence of non-increasing elements. There are \((k-1)!\) such arrangements because the first \(k-1\) elements can be any permutation of \(k-1\) elements.
04

Total Arrangements and Probability

Out of all possible \(k!\) arrangements of \(X_1, X_2, \ldots, X_k\), only \((k-1)!\) satisfy our required condition. Therefore, the probability is:\[\mathbb{P}(N = k) = \frac{(k-1)!}{k!} = \frac{1}{k}.\]
05

Verification by Summation

To verify this probability, if \(\mathbb{P}(N = k) = (k-1)/k!\), we check:\[\sum_{k=1}^{\infty} \frac{k-1}{k!} = 1.\]Rewriting:\[\sum_{k=1}^{\infty} \frac{1}{k} - \frac{1}{(k-1)!} = \sum_{k=2}^{\infty} \frac{1}{k-1} - \frac{1}{(k-1)!} = 1,\]which converges correctly, confirming the probabilities sum to 1.
06

Calculating the Expected Value

To find the expected value, compute:\[\mathbb{E}(N) = \sum_{k=1}^{\infty} k \cdot \frac{k-1}{k!} = \sum_{k=1}^{\infty} \frac{k-1}{(k-1)!} = \sum_{k=1}^{\infty} \frac{1}{(k-2)!} = e.\]Here, the final step uses the series expansion for \(e\).

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Key Concepts

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

Independent Identically Distributed Variables
Independent identically distributed variables, often abbreviated as i.i.d., are a fundamental concept in probability and statistics. These variables are "independent" which means that the occurrence of one does not influence the occurrence of another. Each variable in the sequence has the same probability distribution, hence "identically distributed".
  • **Independence** ensures that knowing the outcome of one variable, say \(X_1\), does not provide any information about another variable, such as \(X_2\).
  • **Identically Distributed** means all variables \(X_1, X_2, X_3, \ldots\) follow the same probability distribution. This often simplifies analysis since statistical properties apply uniformly across the sequence.
The significance of i.i.d. variables comes into play in the law of large numbers and central limit theorem, both of which rely on sequences of such variables. In the context of the exercise, i.i.d. variables allow us to apply symmetry and make simplifications in calculating probabilities and expectations.
Continuous Random Variables
Continuous random variables can take any value within a given range, a feature distinguishing them from discrete random variables, which are limited to countable values. When dealing with continuous variables, probabilities are usually determined over intervals rather than distinct points.
  • The **probability density function (PDF)** is used instead of a simple probability for individual values, as the probability of a specific outcome is effectively zero in continuous distributions.
  • The **cumulative distribution function (CDF)** gives the probability that the variable takes a value less than or equal to a particular number, enabling us to calculate probabilities for ranges of outcomes.
In the problem scenario, the continuous nature of the random variables means any two will almost surely not be equal, simplifying calculations when establishing ordering conditions like \(X_{1} \geq X_{2} \geq \cdots \geq X_{k-1} < X_{k}\).
Expected Value
The expected value, or mean, of a random variable encapsulates the center or "average" behavior of the random variable when repeated often. For both theoretical analysis and practical applications, it provides insight into the "typical" outcome.
  • The expected value \(\mathbb{E}(X)\) is calculated by integrating the product of variable value and its probability over the distribution for continuous variables.
  • For sums of independent variables, the expected value is simply the sum of individual expected values, simplifying complexity in analysis.
In this exercise, when we consider the expected stopping time \(\mathbb{E}(N)\) as \(e\), it implies that the average position where the ordering changes (from non-increasing to increasing) happens around the value of \(e\). Calculating this involves infinite series summation, a skill often encountered in probability theory.
Probability Distribution
A probability distribution explains how the probabilities of a random variable are spread across its possible values. It offers a complete description of the variable's range and how each outcome occurs with associated probability.
  • **Distributions for Discrete Variables**: Assigned probabilities to specific values (e.g., tossing a die). For continuous variables, we focus on densities and ranges rather than specific outcomes.
  • **Forms of Distribution**: There are many types, notably uniform, normal, exponential, and others, each with unique properties dictating the behavior of the variables they model.
In the context of this exercise, the distribution \(\mathbb{P}(N = k)\) is unusual but finds its foundation in factorial mathematics, specifically \(\frac{(k-1)!}{k!} = \frac{1}{k}\). This highlights the role of permutations, as the way variables are ordered (or permuted) drastically influences probabilities, especially with ordered stopping times.

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Most popular questions from this chapter

Is the function \(G\), defined by $$ G(x, y)= \begin{cases}1 & \text { if } x+y \geq 0 \\ 0 & \text { otherwise }\end{cases} $$ the joint distribution function of some pair of random variables? Justify your answer.

(a) Suppose that the continuous random variables \(X\) and \(Y\) are independent with probability density functions \(f\) and \(g\), both of which are symmetric about zero. (i) Find the joint probability density function of \((U, V)\), where \(U=X\) and \(V=Y / X\). (ii) Show that the marginal density function of \(V\) is $$ f_{V}(v)=2 \int_{0}^{\infty} x f(x) g(x v) d x $$ (iii) Let \(X\) and \(Y\) be independent normal random variables, each with mean 0 , and with non-zero variances \(a^{2}\) and \(b^{2}\), respectively. Show that \(V=Y / X\) has probability density function $$ f_{V}(v)=\frac{c}{\pi\left(c^{2}+v^{2}\right)} \quad \text { for }-\infty

Show that there exists a constant \(c\) such that the function $$ f(x, y)=\frac{c}{\left(1+x^{2}+y^{2}\right)^{3 / 2}} \quad \text { for } x, y \in \mathbb{R} $$ is a joint density function. Show that both marginal density functions of \(f\) are the density function of the Cauchy distribution.

An aeroplane drops medical supplies to two duellists. With respect to Cartesian coordinates whose origin is at the target point, both the \(x\) and \(y\) coordinates of the landing point of the supplies have normal distributions which are independent. These two distributions have the same mean 0 and variance \(\sigma^{2}\). Show that the expectation of the distance between the landing point and the target is \(\sigma \sqrt{\pi / 2}\). What is the variance of this distance? (Oxford \(1976 \mathrm{M}\) )

\(X\) and \(Y\) are independent random variables normally distributed with mean zero and variance \(\sigma^{2}\). Find the expectation of \(\sqrt{X^{2}+Y^{2}}\). Find the probabilities of the following events, where \(a, b, c\), and \(\alpha\) are positive constants such that \(b0\). (Consider various cases depending on the relative sizes of \(a, b\), and \(c\).) (Oxford 1981M)
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