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Prove that (a) \(\max \left(X_{1}, X_{2}\right)=X_{1}+X_{2}-\min \left(X_{1}, X_{2}\right)\) and, in general, $$ \text { (b) } \begin{aligned} \max \left(X_{1}, \ldots, X_{n}\right)=& \sum_{1}^{n} X_{i}-\sum_{i

Short Answer

Expert verified
To summarize, we have proven parts (a) and (b) using case analysis and induction, derived the probability formula for unions of events in part (c) by defining appropriate random variables and taking expectations, and finally, for part (d), we have derived an expression for the expected time until an event has occurred in all n independent Poisson processes using the cumulative distribution function and calculating the expectation.

Step by step solution

01

Proof of (a)

Our goal is to prove \(\max \left(X_{1}, X_{2}\right)=X_{1}+X_{2}-\min \left(X_{1},X_{2}\right)\). First, let's observe the possible cases: - \(X_1 \ge X_2\): In this case, \(\max(X_1, X_2) = X_1\) and \(\min(X_1, X_2) = X_2\). So the equation becomes \(X_1 = X_1 + X_2 - X_2\), which holds true. - \(X_2 > X_1\): In this case, \(\max(X_1, X_2) = X_2\) and \(\min(X_1, X_2) = X_1\). So the equation becomes \(X_2 = X_1 + X_2 - X_1\), which also holds true. Since all cases have been addressed, the equation in (a) has been proven.
02

Proof of (b)

The equation for part (b) involves more terms, but the reasoning is similar to part (a). Induction can be used to generalize the proof for n terms. Let \(P(n)\) denote the expression for n terms: \( \max \left(X_{1}, \ldots, X_{n}\right) = \sum_{1}^{n} X_{i} -\sum_{i<j} \min \left(X_{i}, X_{j}\right) + \sum_{i<j<k} \min \left(X_{i}, X_{j}, X_{k}\right) +\cdots + (-1)^{n-1} \min \left(X_{1}, \ldots, X_{n}\right) \) The base step has already been proven in part (a): \(P(2)\) holds true. Now consider \(P(n + 1)\): Using induction, we assume that the \(P(n)\) expression holds true. To prove \(P(n + 1)\), we have to show that the expression also holds for n+1 terms. \( \max \left(X_{1}, \ldots, X_{n+1}\right) = \max(\max\left(X_{1},\ldots,X_n\right), X_{n+1}) \) Using \(P(n)\) expression for \(\max(X_{1},\ldots,X_n)\), \( \max \left(X_{1}, \ldots, X_{n+1}\right) = \max(X_{1} + {\cdots} + X_n - \min \left(X_{1}, X_{n}\right)+\text{alt} \text{ terms}, X_{n+1}) \) Using part (a) proof, the relationship between max and min terms is established: \begin{align*} \max \left(X_{1}, \ldots, X_{n+1}\right) = & X_{1}+{\cdots}+X_n+X_{n+1}-\min(\min \left(X_{1}, X_{n}\right)+{\cdots},X_{n+1})\\ &+\text{alt terms}(-1)^n \min(\min \left(X_{1}, X_{n}\right), X_{n+1}) \end{align*} Which fits the expression with n+1 terms, and thus \(P(n + 1)\) holds true by induction.
03

Proof of (c)

To prove part (c), we will define random variables \(X_i, i=1,…,n\), where \(X_i\) is an indicator random variable for event \(A_i\) (\(X_i = 1\) if event \(A_i\) occurs, and \(X_i = 0\) otherwise). We use the expectation to derive the probability formula for unions: \(P(\bigcup_{1}^{n} A_{i}) = E(\max(X_1,\ldots,X_n))\) Applying part (b) equation for \(\max(X_1,\ldots,X_n)\), and taking the expectations inside, we get: \begin{align*} P\left(\bigcup_{1}^{n} A_{i}\right) &= E\left(\sum_{1}^{n} X_{i} -\sum_{i<j} \min \left(X_{i}, X_{j}\right) + \sum_{i<j<k} \min \left(X_{i}, X_{j}, X_{k}\right) +\cdots+(-1)^{n-1} \min \left(X_{1}, \ldots, X_{n}\right)\right) \\ &= \sum_{1}^{n} E(X_i) -\sum_{i<j} E\left(\min \left(X_{i}, X_{j}\right)\right) + \sum_{i<j<k} E\left(\min \left(X_{i}, X_{j}, X_{k}\right)\right) +\cdots+(-1)^{n-1} E\left(\min \left(X_{1}, \ldots, X_{n}\right)\right) \end{align*} Further simplification gives us the formula for the probability of the union of events: \( P\left(\bigcup_{1}^{n} A_{i}\right) = \sum_{i} P\left(A_{i}\right) -\sum_{i<j}\sum P\left(A_{i} A_{j}\right)+\cdots+(-1)^{n-1} P\left(A_{1} \cdots A_{n}\right) \)
04

Deriving Expression for (d)

To derive an expression for the expected time until an event has occurred in all n independent Poisson processes, let's consider \(Z_i\) as the time until the first event occurs in the ith Poisson process, where each \(Z_i\) follows an exponential distribution with rate \(\lambda_i\). Let \(W = \max(Z_1, Z_2, \ldots, Z_n)\), which is the time until an event has occurred in all processes. The cumulative distribution function of \(W\) is: \(P(W \leq w) = P(\max(Z_1,\ldots,Z_n)\leq w) = P(\bigcup_{1}^{n} (Z_i \leq w))\) Using the proof of part (c), we can derive the probability of the union: \(P(W \leq w) = \sum_{i} P(Z_i \leq w) -\sum_{i<j} P(Z_i\leq w, Z_j\leq w)+\cdots+(-1)^{n-1} P(Z_1 \leq w,\ldots, Z_n\leq w)\) For independent events, we have: \begin{align*} P(W \leq w) &= \sum_{i} (1 - e^{-\lambda_i w}) -\sum_{i<j} (1 - e^{-(\lambda_i + \lambda_j)w})+\cdots+(-1)^{n-1}(1 - e^{-\sum_{i=1}^{n} \lambda_i w}) \\ \end{align*} Now, we can calculate the expected time until an event has occurred in all processes by taking the derivative of the cumulative distribution function, i.e., probability density function \(f_W(w)\), and calculating the expectation: \begin{align*} f_W(w) &= \frac{d}{dw} P(W \leq w) \\ E(W) &= \int_{0}^{\infty} w f_W(w) dw \\ \end{align*}

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Key Concepts

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

Cumulative Distribution Function
The cumulative distribution function (CDF) is a fundamental concept in probability and statistics that describes the probability that a real-valued random variable X will be less than or equal to a certain value. More formally, the CDF, denoted as F(x), is defined by:
\( F(x) = P(X \leq x) \),
where P represents probability. Understanding the CDF is critical for interpreting probabilities and has practical applications in data analysis, risk assessment, and statistical inference.

Characteristics of CDFs

  • Non-decreasing: As x increases, F(x) does not decrease.
  • Right-continuous: There are no jumps or discontinuities when moving from left to right along the x-axis.
  • Limits: The CDF approaches 0 as x approaches negative infinity and 1 as x approaches positive infinity.
  • Useful for defining other statistics: The median, quartiles, and percentiles can be derived from the CDF.
The CDF can be relevant both for discrete and continuous random variables. For discrete variables, it shows the accumulated probabilities at each point, often taking the form of a step function. For continuous variables, it is a smoothly increasing function. In the step-by-step solution provided, we explore the CDF concept further in the context of finding the expected time until an event has occurred in all n independent Poisson processes, a scenario that involves understanding the intersection of probabilities across multiple processes.
Poisson Processes
Poisson processes are vital when dealing with the timing of events in various scenarios such as queues, traffic flow, and decay of radioactive material. The core idea is modeling discrete events happening continuously and independently at a constant average rate.

Key Points of Poisson Processes

  • Events occur independently of each other.
  • The average rate of event occurrence is constant (denoted by λ).
  • The number of events that occur in any interval follows a Poisson distribution.
The rate parameter λ is crucial; it defines the intensity or frequency of occurrences and uniquely determines the process's behavior. In the context of the exercise solution, each independent Poisson process has a rate λ_i, and we are interested in the expected time until the first event occurs in each process. Independent Poisson processes can be combined to determine collective behavior, an aspect that was used to derive an expression for the expected time until an event has occurred in all processes by utilizing their cumulative distribution functions.
Indicator Random Variables
Indicator random variables are simple yet powerful tools used in probability theory to capture the occurrence or non-occurrence of an event. An indicator random variable, often denoted by I or X with subscripts, is defined as follows:
\( X_i = \begin{cases} 1, & \text{if event } A_i \text{ occurs} \ 0, & \text{otherwise} \end{cases} \)

Uses of Indicator Random Variables

  • Modeling the presence or absence of characteristics.
  • Simplifying complex probability problems by splitting into individual outcomes.
  • Facilitating calculations involving unions and intersections of events.
In our provided proof, indicator random variables were employed to convert part (c) of the problem—dealing with the union of events—into an expectation problem. By defining such variables for the occurrence of each event Ai and applying expectation, we observed a formula parallel to the inclusion-exclusion principle. This illustrates how the seemingly abstract notion of indicator random variables can translate into concrete and intuitive results in probability.

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Most popular questions from this chapter

An insurance company pays out claims on its life insurance policies in accordance with a Poisson process having rate \(\lambda=5\) per week. If the amount of money paid on each policy is exponentially distributed with mean \(\$ 2000\), what is the mean and variance of the amount of money paid by the insurance company in a four-week span?

Let \(X_{1}\) and \(X_{2}\) be independent exponential random variables, each having rate \(\mu .\) Let $$ X_{(1)}=\operatorname{minimum}\left(X_{1}, X_{2}\right) \text { and } X_{(2)}=\operatorname{maximum}\left(X_{1}, X_{2}\right) $$ Find (a) \(E\left[X_{(1)}\right]\) (b) \(\operatorname{Var}\left[X_{(1)}\right]\) (c) \(E\left[X_{(2)}\right]\) (d) \(\operatorname{Var}\left[X_{(2)}\right]\)

Let \(\\{N(t), t \geqslant 0\\}\) be a Poisson process with rate \(\lambda\). For \(sN(s))\) (b) \(P(N(s)=0, N(t)=3)\); (c) \(E[N(t) \mid N(s)=4] ;\) (d) \(E[N(s) \mid N(t)=4]\).

One hundred items are simultaneously put on a life test. Suppose the lifetimes of the individual items are independent exponential random variables with mean 200 hours. The test will end when there have been a total of 5 failures. If \(T\) is the time at which the test ends, find \(E[T]\) and \(\operatorname{Var}(T)\).

Consider a nonhomogeneous Poisson process whose intensity function \(\lambda(t)\) is bounded and continuous. Show that such a process is equivalent to a process of counted events from a (homogeneous) Poisson process having rate \(\lambda\), where an event at time \(t\) is counted (independent of the past) with probability \(\lambda(t) / \lambda ;\) and where \(\lambda\) is chosen so that \(\lambda(s)<\lambda\) for all \(s\).

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