Problem 231
Question
Lack of Memory Property. Show that for an exponential random variable \(X,
P\left(X
Step-by-Step Solution
Verified Answer
By simplifying using the lack of memory property, we find that the two probabilities are equal: \(P(Xt_1) = P(X
1Step 1: 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
2Step 2: Write the Conditional Probability
Start with writing the required conditional probability: \[ P(Xt_1) = \frac{P(Xt_1)}{P(X>t_1)}. \]
3Step 3: Simplify the Intersection
The intersection of events \(Xt_1\) implies \(X\) is between \(t_1\) and \(t_1+t_2\). Therefore, \[ P(Xt_1) = P(t_1 < X < t_1 + t_2). \]
4Step 4: Use Memoryless Property
Apply the lack of memory property of the exponential distribution, which states: \[ P(X > t_1 + t_2) = P(X > t_2)P(X > t_1). \]
5Step 5: 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 \[ P(Xt_1) = P(X>t_1) - (1 - P(X>t_2)P(X>t_1)). \]
6Step 6: 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 \[ P(Xt_1) = P(X < t_2).\]
7Step 7: 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.
Key Concepts
Memoryless PropertyConditional ProbabilityJoint Probability Distributions
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:
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:
- \( P(X > t_1 + t_2 | X > t_1) = P(X > t_2) \)
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:
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.
The formula for conditional probability is given by:
- \( P(A | B) = \frac{P(A \cap B)}{P(B)} \)
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.
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.
Other exercises in this chapter
Problem 228
A bearing assembly contains 10 bearings. The bearing diameters are assumed to be independent and normally distributed with a mean of 1.5 millimeters and a stand
View solution Problem 229
Let the random variable \(X\) denote a measurement from a manufactured product. Suppose that the target value for the measurement is \(m .\) For example, \(X\)
View solution Problem 232
Determine the mean and variance of a beta random variable. Use the result that the probability density function integrates to \(1 .\) That is, \(\frac{\Gamma(\a
View solution Problem 234
A process is said to be of six-sigma quality if the process mean is at least six standard deviations from the nearest specification. Assume a normally distribut
View solution