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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 Xc] P\\{X>c\\} $$

Short Answer

Expert verified
In summary, for an exponential random variable \(X\) with rate \(\lambda\), we find the conditional expectation \(E[X \mid X<c]\) using two methods: (a) Direct calculation using truncated distribution yields: \(E[X \mid X<c] = \frac{1}{\lambda}\). (b) Applying the Law of Total Expectation also gives: \(E[X \mid X<c] = \frac{1}{\lambda}\).

Step by step solution

01

Understanding the Exponential Distribution

The exponential distribution is a statistical distribution used to model the time elapsed between events. It is defined by its rate parameter \(\lambda\), which is the average number of events per unit time. The probability density function (PDF) of an exponential random variable is given by \(f(x)=\lambda e^{-\lambda x}\) for \(x \geq 0\), and E[X] = \(\frac{1}{\lambda}\).
02

Defining Conditional Expectation

Conditional expectation, or the expected value of a random variable given some condition, is described by integrating over the region of interest, the product of the variable and its conditional distribution. It is given as \(E[X|B] = \int_{B} xf(x)dx \) for any event B that has positive probability.
03

Solve for \( E[X \mid X

To calculate \(E[X \mid X<c]\) we will use the truncated distribution which describes the random variable \(X\) given \(X<c\). In mathematical terms, this is \(\int_{0}^{c} x \cdot f(x) dx\), where \(f(x) = \lambda e^{-\lambda x }\). Integrate by parts (let \(u=x\) and \(dv=e^{-\lambda x}\)dx to get \(du=dx\), \(v=-\frac{1}{\lambda}e^{-\lambda x }\)) to get \(-\frac{x}{\lambda} e^{-\lambda x} \Biggr|_0^{c} + \int_{0}^{c} \frac{1}{\lambda} e^{-\lambda x } dx\). Evaluate this to find that \(E[X \mid X<c] = \frac{1}{\lambda}\).
04

Use the Law of Total Expectation to solve for \( E[X \mid X

The Law of Total Expectation allows us to compute \(E[X \mid Xc] P\{X>c\}\). We already know that \(E[X] = \frac{1}{\lambda}\) and \(P\{Xc\} = e^{-\lambda c}\). For \(E[X \mid X>c]\), it can be shown that for any exponential distribution with rate \(\lambda\) if \(X>c\), then \(X-c\) is also exponentially distributed with rate \(\lambda\). Hence, \(E[X \mid X>c] = c + \frac{1}{\lambda}\). By substituting all these values into the initial equation, and solving for \(E[X \mid X<c]\), we also get that \(E[X \mid X<c] = \frac{1}{\lambda}\).

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