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Magazine Chapter 5: Problem 14
Let \(X\) be an exponential random variable with rate \(\lambda .\)
(a) Use the definition of conditional expectation to determine \(E[X \mid X
Short Answer
Expert verified
The conditional expectation \(E[X \mid X
Step by step solution
01
Define the Conditional PDF
Given that \(X\) follows exponential distribution with rate \(\lambda\), the probability density function of \(X\) is:
$$
f_X(x) = \lambda e^{-\lambda x}, x \ge 0
$$
Since we are given \(X < c\) as the condition, we'll have:
$$
f_{X\mid X
02
Calculate \(P(X
To find the denominator \(P(X
03
Find the Conditional Expectation
Now plug the conditional PDF and \(P(X
04
Find Expected Value \(E[X]\)
Given that \(X\) follows an exponential distribution with rate \(\lambda\), we have:
$$
E[X] = \frac{1}{\lambda}
$$
05
Use Identity to Find Conditional Expectation
By substituting the known values into the given identity:
$$
\frac{1}{\lambda} = E[X \mid Xc] P[X>c]
$$
We already found \(P(Xc) = e^{-\lambda c}\), we have:
$$
\frac{1}{\lambda} = \left(\frac{1}{1 - e^{-\lambda c}} \left[1 - (c+1)e^{-\lambda c}\right]\right)(1 - e^{-\lambda c}) + E[X \mid X>c] e^{-\lambda c}
$$
Solving for \(E[X \mid X>c]\) and substituting back into the given identity, we find that:
$$
E[X \mid X
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Probability Density Function
Understanding the probability density function (PDF) is crucial when dealing with continuous random variables such as the time until a next event in a process, often modelled by an exponential distribution. The PDF describes the likelihood of the random variable taking on a specific value. In mathematical terms, the PDF of an exponential random variable with a rate parameter \(\lambda\) is written as \(f_X(x) = \lambda e^{-\lambda x}\), where \(x \) must be nonnegative since exponential variables cannot take negative values.
The exponential PDF decreases exponentially as \(x\) increases, which intuitively makes sense because in many real-world phenomena, the likelihood of waiting a long time for an event (like a bus arrival) decreases as the waiting time increases. When calculating the conditional expectation, we start by focusing on the interval \( [0, c)\) and within this interval, the PDF remains unchanged but is reinterpreted as the conditional PDF given \(X < c\).
By integrating the PDF, you can derive various probabilities, such as \(P(X
The exponential PDF decreases exponentially as \(x\) increases, which intuitively makes sense because in many real-world phenomena, the likelihood of waiting a long time for an event (like a bus arrival) decreases as the waiting time increases. When calculating the conditional expectation, we start by focusing on the interval \( [0, c)\) and within this interval, the PDF remains unchanged but is reinterpreted as the conditional PDF given \(X < c\).
By integrating the PDF, you can derive various probabilities, such as \(P(X
Random Variable
A random variable, often symbolized as \(X\), \(Y\), or \(Z\), is a numerical value resulting from some random process. In this exercise, \(X\) represents the time until an event occurs and follows an exponential distribution. This type of variable is continuous, meaning it can assume an infinite number of values within a given range.
When we talk about \(E[X \mid Xexpected value or the mean of the random variable \(X\) given that it is less than a certain threshold \(c\). This is known as the conditional expectation, which provides us with the average outcome we would expect under these constraints. It integrates the concept of a weighted average into the continuous domain, where the 鈥榳eights鈥� are the probabilities expressed by the PDF.
Random variables allow us to quantify uncertainty and calculate the likelihood of various outcomes in processes that are inherently unpredictable, like radioactive decay, or waiting times in queues. Being able to calculate with these variables, especially their expected values, gives us powerful insight into the behavior and characteristics of such stochastic processes.
When we talk about \(E[X \mid X
Random variables allow us to quantify uncertainty and calculate the likelihood of various outcomes in processes that are inherently unpredictable, like radioactive decay, or waiting times in queues. Being able to calculate with these variables, especially their expected values, gives us powerful insight into the behavior and characteristics of such stochastic processes.
Integration by Parts
When you're tackling integrals, especially those involving exponential functions and polynomials such as in our exercise, integration by parts is an effective technique. Putting the method to practice involves recognizing parts of the integrand that can be separately differentiated and integrated. The rule comes from the product rule in differentiation and is formally stated as \(\int u dv = uv - \int v du\).
In the given solution, to calculate \(E[X \mid X
The strength of integration by parts lies in its ability to transform a hard-to-solve integral into smaller, more manageable pieces. However, choosing \(u\) and \(dv\) wisely is key. A good rule of thumb is to pick \(u\) to be the function that gets simpler when differentiated, as we see with \(x\) becoming \(1\) in our example. This tactic can significantly streamline solving integrals, particularly in probability and statistical contexts.
In the given solution, to calculate \(E[X \mid X
The strength of integration by parts lies in its ability to transform a hard-to-solve integral into smaller, more manageable pieces. However, choosing \(u\) and \(dv\) wisely is key. A good rule of thumb is to pick \(u\) to be the function that gets simpler when differentiated, as we see with \(x\) becoming \(1\) in our example. This tactic can significantly streamline solving integrals, particularly in probability and statistical contexts.
