/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 Let \(h(x)=P\left(\sum_{i=1}^{T}... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 57

Let \(h(x)=P\left(\sum_{i=1}^{T} X_{i}>x\right\\}\) where \(X_{1}, X_{2}, \ldots\) are independent random variables having distribution function \(F_{e}\) and \(T\) is independent of the \(X_{i}\) and has probability mass function \(P[T=n\\}=\rho^{n}(1-\rho), n \geqslant 0 .\) Show that \(h(x)\) satisfies Equation (7.53). Hint: Start by conditioning on whether \(T=0\) or \(T>0\).

Short Answer

Expert verified
In summary, we have shown that the given function \(h(x)\) can be expressed as: \[h(x) = \left(P(0>x)\cdot(1-\rho)\right) + \left(P\left(\sum_{i=1}^{T} X_{i} > x | T > 0\right)\cdot1\right).\] This is done by conditioning on \(T=0\) or \(T>0\) and using the probability mass function and distribution function provided. Although we couldn't prove the relation to the unspecified Equation (7.53), we demonstrated a step-by-step approach to breaking down the exercise based on the given hint.

Step by step solution

01

Calculate the Probabilities for \(T=0\) and \(T>0\)

To calculate \(P(T=0)\) and \(P(T>0)\), use the given probability mass function \(P[T=n]=\rho^n(1-\rho), n \geqslant 0\). We have: 1. For \(n=0\): \[P(T=0) = \rho^0(1-\rho) = 1-\rho.\] 2. For \(n>0\): \[P(T>0) = \sum_{n=1}^{\infty} \rho^n(1-\rho) = 1-\rho + \rho^2(1-\rho) + \rho^3(1-\rho) + \cdots .\] This is an infinite geometric series with the first term \(a = (1-\rho)\) and common ratio \(r=\rho\). The sum formula for an infinite geometric series is: \[\sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r}.\] So with this: \[P(T>0) = \frac{1-\rho}{1-\rho} =1.\]
02

Calculate Independent Probabilities

Now, let's calculate \(P\left(\sum_{i=1}^{T} X_{i} > x | T = 0\right)\) and \(P\left(\sum_{i=1}^{T} X_{i} > x | T > 0\right)\): 1. If \(T=0\), there is no random variable \(X_i\), so the sum is equal to 0: \[P\left(\sum_{i=1}^{T} X_{i} > x | T = 0\right) = P(0>x).\] 2. If \(T>0\), the result can be obtained via the given distribution function \(F_{e}\): \[P\left(\sum_{i=1}^{T} X_{i} > x | T > 0\right).\]
03

Calculate the Function \(h(x)\)

Finally, substitute the calculated probabilities in the formula of \(h(x)\): \[h(x) = \left(P(0>x)\cdot(1-\rho)\right) + \left(P\left(\sum_{i=1}^{T} X_{i} > x | T > 0\right)\cdot1\right).\] Since we don't have the exact equation (7.53), we could not prove the direct relation. However, we have demonstrated how to break down the exercise by following the given hint and considering the conditions on \(T=0\) and \(T>0\).

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Key Concepts

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

Infinite Geometric Series
Understanding an infinite geometric series is key to solving many problems in probability and other areas of mathematics. An infinite geometric series is a sum of the form \(\sum_{n=1}^{\infty} ar^{n-1}\), where \(a\) is the first term and \(r\) is the common ratio. For the series to converge, the common ratio \(r\) must satisfy \(|r| < 1\). The sum of an infinite geometric series can be found using the formula \(S = \frac{a}{1-r}\) if \(r\) is strictly between -1 and 1.

In our exercise, we have a probability mass function that references such a series. When calculating the probability that \(T > 0\), we've identified an infinite geometric series with the first term \(a = (1-\rho)\) and common ratio \(r = \rho\). Using the sum formula for an infinite geometric series, we can easily find that \(P(T>0)\) equals 1. This is a critical step in probability calculations, especially when dealing with random processes over an indefinite number of trials or time periods.
Random Variables
A random variable is a valuable concept in statistics and probability theory. It's essentially a variable whose value is subject to variations due to chance. Random variables can be classified into two categories: discrete and continuous. Discrete random variables take on a countable number of distinct values. Continuous random variables, on the other hand, can take on any value within a given range.

In the context of our exercise, \(X_1, X_2, ..., X_T\) are independent random variables with distribution function \(F_e\), and these variables represent distinct outcomes of a probabilistic event. Independence means that the result of one variable does not influence the others. The sum \(\sum_{i=1}^{T} X_{i}\) is then a new random variable, and its behavior is essential when calculating the function \(h(x)\), which represents a specific probability regarding the sum of these variables exceeding a certain value \(x\).
Distribution Function
The distribution function, also known as the cumulative distribution function (CDF), is a fundamental concept in probability. It describes the probability that a real-valued random variable \(X\) will have a value less than or equal to \(x\). Mathematically, for a given value of \(x\), the distribution function \(F(x)\) is calculated as \(F(x) = P(X \leq x)\).

In our exercise, \(F_e\) is the distribution function of the random variables \(X_i\). This function helps to find probabilities related to individual random variables out of a sequence of independent random variables. By using this distribution function, we can compute probabilities like \(P\left(\sum_{i=1}^{T} X_{i} > x | T > 0\right)\), which is a necessary element in deriving our desired function \(h(x)\). The understanding of CDF is crucial in determining probabilities associated with sums of random variables, as it allows for assessing the accumulated likelihoods up to a certain point.

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Most popular questions from this chapter

Random digits, each of which is equally likely to be any of the digits 0 through 9 , are observed in sequence. (a) Find the expected time until a run of 10 distinct values occurs. (b) Find the expected time until a run of 5 distinct values occurs.

Consider a system that can be in either state 1 or 2 or \(3 .\) Each time the system enters state \(i\) it remains there for a random amount of time having mean \(\mu_{i}\) and then makes a transition into state \(j\) with probability \(P_{i j} .\) Suppose $$ P_{12}=1, \quad P_{21}=P_{23}=\frac{1}{2}, \quad P_{31}=1 $$ (a) What proportion of transitions takes the system into state \(1 ?\) (b) If \(\mu_{1}=1, \mu_{2}=2, \mu_{3}=3\), then what proportion of time does the system spend in each state?

Let \(U_{1}, U_{2}, \ldots\) be independent uniform \((0,1)\) random variables, and define \(N\) by $$ N=\min \left(n: U_{1}+U_{2}+\cdots+U_{n}>1\right\\} $$ What is \(E[N] ?\)

Let \(X_{1}, X_{2}, \ldots\) be a sequence of independent random variables. The nonnegative integer valued random variable \(N\) is said to be a stopping time for the sequence if the event \(\\{N=n\\}\) is independent of \(X_{n+1}, X_{n+2}, \ldots .\) The idea being that the \(X_{i}\) are observed one at a time-first \(X_{1}\), then \(X_{2}\), and so on-and \(N\) represents the number observed when we stop. Hence, the event \(\\{N=n\\}\) corresponds to stopping after having observed \(X_{1}, \ldots, X_{n}\) and thus must be independent of the values of random variables yet to come, namely, \(X_{n+1}, X_{n+2}, \ldots\) (a) Let \(X_{1}, X_{2}, \ldots\) be independent with $$ P\left[X_{i}=1\right\\}=p=1-P\left(X_{i}=0\right\\}, \quad i \geqslant 1 $$ Define $$ \begin{aligned} &N_{1}=\min \left[n: X_{1}+\cdots+X_{n}=5\right\\} \\ &N_{2}=\left\\{\begin{array}{ll} 3, & \text { if } X_{1}=0 \\ 5, & \text { if } X_{1}=1 \end{array}\right. \\ &N_{3}=\left\\{\begin{array}{ll} 3, & \text { if } X_{4}=0 \\ 2, & \text { if } X_{4}=1 \end{array}\right. \end{aligned} $$ Which of the \(N_{i}\) are stopping times for the sequence \(X_{1}, \ldots ?\) An important result, known as Wald's equation states that if \(X_{1}, X_{2}, \ldots\) are independent and identically distributed and have a finite mean \(E(X)\), and if \(N\) is a stopping time for this sequence having a finite mean, then $$ E\left[\sum_{i=1}^{N} X_{i}\right]=E[N] E[X] $$ To prove Wald's equation, let us define the indicator variables \(I_{i}, i \geqslant 1\) by $$ I_{i}=\left\\{\begin{array}{ll} 1, & \text { if } i \leqslant N \\ 0, & \text { if } i>N \end{array}\right. $$ (b) Show that $$ \sum_{i=1}^{N} X_{i}=\sum_{i=1}^{\infty} X_{i} I_{i} $$ From part (b) we see that $$ \begin{aligned} E\left[\sum_{i=1}^{N} X_{i}\right] &=E\left[\sum_{i=1}^{\infty} X_{i} I_{i}\right] \\ &=\sum_{i=1}^{\infty} E\left[X_{i} I_{i}\right] \end{aligned} $$ where the last equality assumes that the expectation can be brought inside the summation (as indeed can be rigorously proven in this case). (c) Argue that \(X_{i}\) and \(I_{i}\) are independent. Hint: \(I_{i}\) equals 0 or 1 depending on whether or not we have yet stopped after observing which random variables? (d) From part (c) we have $$ E\left[\sum_{i=1}^{N} X_{i}\right]=\sum_{i=1}^{\infty} E[X] E\left[I_{i}\right] $$ Complete the proof of Wald's equation. (e) What does Wald's equation tell us about the stopping times in part (a)?

There are three machines, all of which are needed for a system to work. Machine \(i\) functions for an exponential time with rate \(\lambda_{i}\) before it fails, \(i=1,2,3 .\) When a machine fails, the system is shut down and repair begins on the failed machine. The time to fix machine 1 is exponential with rate \(5 ;\) the time to fix machine 2 is uniform on \((0,4) ;\) and the time to fix machine 3 is a gamma random variable with parameters \(n=3\) and \(\lambda=2 .\) Once a failed machine is repaired, it is as good as new and all machines are restarted. (a) What proportion of time is the system working? (b) What proportion of time is machine 1 being repaired? (c) What proportion of time is machine 2 in a state of suspended animation (that is, neither working nor being repaired)?
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