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Lack of Memory Property. Show that for an exponential random variable \(X, P\left(Xt_{1}\right)=P\left(X

Short Answer

Expert verified
By simplifying using the lack of memory property, we find that the two probabilities are equal: \(P(Xt_1) = P(X<t_2)\).

Step by step solution

01

Identify Given Information

We have an exponential random variable \(X\) with a lack of memory property. We need to show that \(P\left(Xt_{1}\right)=P\left(X<t_{2}\right)\).
02

Write the Conditional Probability

Start with writing the required conditional probability: \[ P(Xt_1) = \frac{P(Xt_1)}{P(X>t_1)}. \]
03

Simplify the Intersection

The intersection of events \(Xt_1\) implies \(X\) is between \(t_1\) and \(t_1+t_2\). Therefore, \[ P(Xt_1) = P(t_1 < X < t_1 + t_2). \]
04

Use Memoryless Property

Apply the lack of memory property of the exponential distribution, which states: \[ P(X > t_1 + t_2) = P(X > t_2)P(X > t_1). \]
05

Evaluate the Probability

Using the exponential property, calculate the intersection: \[ P(t_1 < X < t_1 + t_2) = P(X > t_1) - P(X > t_1 + t_2). \] Then \[ P(Xt_1) = P(X>t_1) - (1 - P(X>t_2)P(X>t_1)). \]
06

Simplify to Reach Final Expression

Simplifying gives \[ P(X>t_1)P(X>t_2) = P(X>t_2)P(X>t_1), \] leading to \[ P(Xt_1) = P(X < t_2).\]
07

Conclude the Result

Thus, we have confirmed that the exponential random variable properties give us \[ P(Xt_1) = P(X < t_2), \] confirming the lack of memory property.

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Key Concepts

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

Memoryless Property
The memoryless property is a fascinating characteristic of the exponential distribution. It essentially means that the probability of an event occurring within a certain time does not depend on how much time has already elapsed. This feature is unique to exponential distributions among continuous probability distributions.
For example, if you are waiting for a bus and have already waited 10 minutes, the distribution of the remaining waiting time will be the same as it was at the beginning — thanks to this memoryless property. The past does not affect the future probability in such scenarios.
Formally, for an exponential random variable \( X \), the memoryless property states:
  • \( P(X > t_1 + t_2 | X > t_1) = P(X > t_2) \)
This implies that the conditional probability of \( X \) being greater than two times \( (t_1 + t_2) \) given that \( X \) has exceeded \( t_1 \) is the same as \( X \) being greater than \( t_2 \). This lack of memory can simplify many probabilistic calculations, making it a powerful tool in statistical analyses of life events, decay, or waiting times.
Conditional Probability
Conditional probability is a crucial concept in probability theory that allows us to calculate the probability of an event occurring given that another event has already happened. This is influential in scenarios where previous knowledge about an event can affect the likelihood of a subsequent event.
The formula for conditional probability is given by:
  • \( P(A | B) = \frac{P(A \cap B)}{P(B)} \)
Here \( P(A | B) \) represents the probability of event \( A \) happening given that \( B \) has occurred. The intersection \( P(A \cap B) \) models where both events \( A \) and \( B \) happen simultaneously.
In the context of the exponential distribution, when dealing with events such as waiting times or lifetimes, conditional probability helps in deriving further relations. Considering our exercise, it's essential in moving from the intersection of events to solving for a desired outcome using probability laws along with the exponential's properties.
Joint Probability Distributions
Joint probability distributions provide information about the likelihood of two or more random variables occurring simultaneously. This is crucial when analyzing outcomes that depend on multiple factors. A joint distribution gives us a broader perspective on how variables are interrelated.
For continuous random variables like those following an exponential distribution, a joint probability density function (pdf) is used. While in discrete cases we talk of pmf (probability mass function), in continuous cases, the pdf does the job. This function helps calculate the probability of an event falling within a specific range for both variables.
  • If \( X \) and \( Y \) are random variables, then their joint probability is dependent on how they interact with each other.
  • Their joint behavior can be explored further by integrating the joint pdf over a given space.
When understanding this in terms of exponential distributions, the way events overlap and the joint probability assists in understanding scenarios like system failures, where multiple factors influence the outcome. It deepens the knowledge of probabilistic relationships and enriches the analysis through identifying dependency or independence between events.

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Most popular questions from this chapter

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