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Chapter 3: Problem 18

Telegrafenprozess. Sei \(\left(N_{t}\right)_{t \geq 0}\) ein Poisson-Prozess zur Intensit盲t \(\alpha>0\) sowie \(Z_{t}=\) \((-1)^{N_{t}}\). Zeigen Sie: \(P\left(Z_{s}=Z_{t}\right)=\left(1+e^{-2 \alpha(t-s)}\right) / 2\) f眉r \(0 \leq s

Short Answer

Expert verified
The solution is that \(P(Z_s = Z_t) = \frac{1}{2}(1 + e^{-2\alpha(t-s)})\) for \(0 \leq s < t\).

Step by step solution

01

Define the Event

First, understand what the event \(Z_s = Z_t\) means. Given that \(Z_t = (-1)^{N_t}\), the event \(Z_s = Z_t\) happens when \(((-1)^{N_s})= ((-1)^{N_t})\). For this to happen, the number of jumps, \(N_t - N_s\), must be even.
02

Calculate the Probability of an Even Number of Jumps

The number of additional events from \(s\) to \(t\) in a Poisson process with rate \(\alpha\) is a Poisson random variable with parameter \(\alpha (t-s)\). The probability that this is an even number is given by \(P(N_{t-s} \text{ is even}) = \sum_{k=0}^{\infty}\frac{e^{-\alpha (t-s)} (\alpha (t-s))^{2k}}{(2k)!}\), which we rewrite using an exponential function.
03

Simplify the Probability to Exponential Form

The formula for the probability of having an even number of jumps becomes \(\sum_{k=0}^{\infty}\frac{(\alpha (t-s))^{2k}}{(2k)!} = \frac{1}{2}(1 + e^{-2\alpha(t-s)})\). This can be derived using the properties of cosine hyperbolic function: \(\cosh(x) = \frac{e^x + e^{-x}}{2}\).
04

Combine Results

Finally, combine the above steps to conclude that \(P(Z_s = Z_t) = \frac{1}{2}(1 + e^{-2\alpha(t-s)})\), as this reflects the probability of the number of jumps being even.

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Key Concepts

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

Stochastic Processes
Stochastic processes are mathematical objects used to describe systems evolving over time under uncertainty. They are processes that involve randomness, and often model the evolution of random variables. The Poisson process, given as an example in the Original Exercise, is a prime example of a stochastic process.
In a stochastic process like the Poisson process, events occur continuously and independently. These types of models are particularly useful in fields such as physics, finance, and telecommunications.
  • The notation \(ig(N_t\big)_{t \geq 0}\) represents a Poisson process, detailing the number of events occurring up to time \(t\).
  • It is characterized by its rate \(\alpha\), indicating the average number of events in a given interval.
  • It's important to remember that events in a Poisson process are not influenced by when prior events occurred.

Thus, stochastic processes are invaluable for scenarios requiring the modeling of random occurrences over time.
Probability Theory
At the heart of understanding stochastic processes is probability theory, a branch of mathematics concerned with analyzing random phenomena. It allows us to quantify the likelihood of potential outcomes and is critical for finding solutions to various problems, such as the one presented in the Original Exercise.
Probability theory equips us with essential tools:
  • Random variables that are integral to forming probability distributions.
  • The Poisson distribution, key for modeling the number of events occurring in a fixed period in the context of our Poisson process problem.
  • The concept of independence which ensures event occurrences are unrelated to each other.

In the given problem, we use probability theory to determine the likelihood of having an even number of events, using techniques such as the sum of probabilities and properties like those of the exponential function.
Random Variables
Random variables are fundamental entities in probability theory. They are used to model unknown or random outcomes. In the problem discussed, \(N_t\) is a random variable depicting the number of events by time \(t\) in a Poisson process.
Key characteristics of random variables include:
  • They can typically be described by probability distributions.
  • Discrete random variables have countable outcomes, like the number of events in a Poisson process.
  • Continuous random variables have outcomes over a continuum.

In our example, \(N_t\) is discrete because it counts the number of events. To solve the problem, we evaluate when this count results in an even number by leveraging its distribution properties and ensure appropriate calculations to determine probabilities tied to the Poisson process.

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Most popular questions from this chapter

In einem Laden ist eine Alarmanlage eingebaut, die im Falle eines Einbruchs mit Wahrscheinlichkeit \(0.99\) die Polizei alarmiert. In einer Nacht ohne Einbruch wird mit Wahrscheinlichkeit \(0.002\) Fehlalarm ausgel枚st (z. B. durch eine Maus). Die Einbruchswahrscheinlichkeit f眉r eine Nacht betr盲gt \(0.0005 .\) Die Anlage hat gerade Alarm gegeben. Mit welcher Wahrscheinlichkeit ist ein Einbruch im Gange?

W眉rfelparadox. Zwei W眉rfel \(W_{1}\) und \(W_{2}\) seien wie folgt beschriftet: $$ W_{1}: 633333, \quad W_{2}: 555222 $$ Anton und Brigitte w眉rfeln mit \(W_{1}\) bzw. \(W_{2}\). Wer die h枚here Augenzahl erzielt, hat gewonnen. (a) Zeigen Sie, dass Anton die besseren Gewinnchancen hat; wir schreiben daf眉r \(W_{1} \succ W_{2}\). (b) Brigitte bemerkt dies und schl盲gt Anton vor: 鈥濱ch beschrifte jetzt einen dritten W眉rfel. Du darfst dir dann einen beliebigen W眉rfel aussuchen, ich w盲hle mir einen der beiden anderen." Kann Brigitte den dritten W眉rfel so beschriften, dass sie in jedem Fall die besseren Gewinnchancen hat (d. h. so dass \(W_{1} \succ W_{2} \succ W_{3} \succ W_{1}\), also die Relation \(\succ\) nicht transitiv ist)?

Beim zweimaligen Wurf mit einem fairen Tetraeder-W眉rfel, dessen Fl盲chen mit \(1,2,3\), 4 beschriftet seien, bezeichne \(X\) die Summe und \(Y\) das Maximum der jeweils unten liegenden Augenzahl. (a) Bestimmen Sie die gemeinsame Verteilung \(P \circ(X, Y)^{-1}\) von \(X\) und \(Y\). (b) Konstruieren Sie zwei Zufallsvariablen \(X^{\prime}\) und \(Y^{\prime}\) 眉ber einem geeigneten Wahrscheinlichkeitsraum \(\left(\Omega^{\prime}, \mathscr{F}^{\prime}, P^{\prime}\right)\) mit denselben Verteilungen wie \(X\) und \(Y\) (d. h. \(P \circ X^{-1}=\) \(\left.P^{\prime} \circ X^{\prime-1}, P \circ Y^{-1}=P^{\prime} \circ Y^{\prime-1}\right)\), f眉r die jedoch \(X^{\prime}+Y^{\prime}\) eine andere Verteilung besitzt als \(X+Y\).

Ausd眉nnung eines Poisson-Prozesses. Sei \(\alpha>0,\left(L_{i}\right)_{i \geq 1}\) eine Folge von unabh盲ngigen, zum Parameter \(\alpha\) exponentialverteilten Zufallsvariablen, sowie \(T_{k}=\sum_{i=1}^{k} L_{i}, k \geq 1\) Sei ferner \(\left(X_{k}\right)_{k \geq 1}\) eine (von den \(L_{i}\) unabh盲ngige) Bernoulli- Folge zum Parameter \(\left.p \in\right] 0,1[.\) Zeigen Sie: Die Zufallsvariablen $$ N_{t}^{X}:=\sum_{k \geq 1} X_{k} 1_{] 0, t]}\left(T_{k}\right), \quad t \geq 0 $$ bilden einen Poisson-Prozess zum Parameter \(p \alpha .\) Insbesondere ist \(T_{1}^{X}:=\inf \left\\{t>0: N_{t}^{X} \geq 1\right\\}\) exponentialverteilt zum Parameter \(p \alpha\).

Seien \(X, Y\) unabh盲ngige, zu einem Parameter \(\alpha>0\) exponentialverteilte Zufallsvariablen. Bestimmen Sie die Verteilungsdichte von \(X /(X+Y)\).
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