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Chapter 5: Problem 3

Let \(X\) be an exponential random variable. Without any computations, tell which one of the following is correct. Explain your answer. (a) \(E\left[X^{2} \mid X>1\right]=E\left[(X+1)^{2}\right]\) (b) \(E\left[X^{2} \mid X>1\right]=E\left[X^{2}\right]+1\) (c) \(E\left[X^{2} \mid X>1\right]=(1+E[X])^{2}\)

Short Answer

Expert verified
The correct answer is option (c): \(E\left[X^{2} \mid X>1\right]=(1+E[X])^{2}\). This is because both sides of the equation involve the expected value, \(E[X]\), and by comparison, this option is more plausible than the other two options.

Step by step solution

01

Consider Option (a) - Comparing the expressions

Option (a) states that: \(E\left[X^{2} \mid X>1\right]=E\left[(X+1)^{2}\right]\). We have a conditional expectation on the left-hand side, while it's an unconditional expectation on the right-hand side. These two expressions are not equal because the left-hand side is an expected value given a condition, whereas the right-hand side is an expected value without any conditions. This option can be eliminated.
02

Consider Option (b) - Comparing the expressions

Option (b) states that: \(E\left[X^{2} \mid X>1\right]=E\left[X^{2}\right]+1\). The left-hand side of the equation represents the conditional expected value of \(X^2\) given \(X>1\). The right-hand side is the expected value of \(X^2\) plus one. These two expressions are not equal. They could be equal if there were the same condition on the right-hand side, but there isn't, and therefore this option is wrong and can be eliminated.
03

Consider Option (c) - Comparing the expressions

Option (c) states that: \(E\left[X^{2} \mid X>1\right]=(1+E[X])^{2}\). If we analyze this equation, both sides involve the expected value \(E[X]\). The left-hand side has a conditional expectation of \(X^2\) given \(X > 1\). The right-hand side contains the square of the expected value of the random variable \(X\), offset by 1. By comparison, this option seems more plausible than the other two options. Thus, we can conclude that option (c) is the correct choice: \(E\left[X^{2} \mid X>1\right]=(1+E[X])^{2}\).

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Most popular questions from this chapter

There are two types of claims that are made to an insurance company. Let \(N_{i}(t)\) denote the number of type \(i\) claims made by time \(t\), and suppose that \(\left\\{N_{1}(t), t \geqslant 0\right\\}\) and \(\left\\{N_{2}(t), t \geqslant 0\right\\}\) are independent Poisson processes with rates \(\lambda_{1}=10\) and \(\lambda_{2}=1 .\) The amounts of successive type 1 claims are independent exponential random variables with mean \(\$ 1000\) whereas the amounts from type 2 claims are independent exponential random variables with mean \(\$ 5000 .\) A claim for \(\$ 4000\) has just been received; what is the probability it is a type 1 claim?

(a) Let \(\\{N(t), t \geqslant 0\\}\) be a nonhomogeneous Poisson process with mean value function \(m(t) .\) Given \(N(t)=n\), show that the unordered set of arrival times has the same distribution as \(n\) independent and identically distributed random variables having distribution function $$ F(x)=\left\\{\begin{array}{ll} \frac{m(x)}{m(t)}, & x \leqslant t \\ 1, & x \geqslant t \end{array}\right. $$ (b) Suppose that workmen incur accidents in accordance with a nonhomogeneous Poisson process with mean value function \(m(t) .\) Suppose further that each injured man is out of work for a random amount of time having distribution \(F\). Let \(X(t)\) be the number of workers who are out of work at time \(t .\) By using part (a), find \(E[X(t)]\).

There are three jobs and a single worker who works first on job 1 , then on job 2 , and finally on job 3 . The amounts of time that he spends on each job are independent exponential random variables with mean \(1 .\) Let \(C_{i}\) be the time at which job \(i\) is completed, \(i=1,2,3\), and let \(X=\sum_{i=1}^{3} C_{i}\) be the sum of these completion times. Find (a) \(E[X]\), (b) \(\operatorname{Var}(X)\).

Let \(X\) and \(Y\) be independent exponential random variables with respective rates \(\lambda\) and \(\mu\). (a) Argue that, conditional on \(X>Y\), the random variables \(\min (X, Y)\) and \(X-Y\) are independent. (b) Use part (a) to conclude that for any positive constant \(c\) $$ \begin{aligned} E[\min (X, Y) \mid X>Y+c] &=E[\min (X, Y) \mid X>Y] \\ &=E[\min (X, Y)]=\frac{1}{\lambda+\mu} \end{aligned} $$ (c) Give a verbal explanation of why \(\min (X, Y)\) and \(X-Y\) are (unconditionally) independent.

Consider a post office with two clerks. Three people, \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\), enter simultaneously. A and B go directly to the clerks, and \(C\) waits until either \(\mathrm{A}\) or \(\mathrm{B}\) leaves before he begins service. What is the probability that \(\mathrm{A}\) is still in the post office after the other two have left when (a) the service time for each clerk is exactly (nonrandom) ten minutes? (b) the service times are \(i\) with probability \(\frac{1}{3}, i=1,2,3 ?\) (c) the service times are exponential with mean \(1 / \mu ?\)
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