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Satellites are launched into space at times distributed according to a Poisson process with rate \(\lambda .\) Each satellite independently spends a random time (having distribution \(G\) ) in space before falling to the ground. Find the probability that none of the satellites in the air at time \(t\) was launched before time s, where \(s

Short Answer

Expert verified
The probability that none of the satellites in the air at time t were launched before time s can be found by summing the product of the probabilities from Steps 2 and 3 over all possible values of n (from 0 to infinity): \(P(\text{No satellites launched before s}) = \sum_{n=0}^{\infty} e^{-\lambda (t-s)} \frac{(\lambda (t-s))^n}{n!} \cdot (1 - G(t-s))^n\)

Step by step solution

01

Calculate the probability a satellite is still in the air at time t

If a satellite is launched at time s, it spends a random time with distribution G in the air. Let X be the random variable representing the time spent by a satellite in space. We want to find the probability that X > t-s, that is, the satellite is still in the air at time t. This probability can be calculated as \(P(X > t-s) = 1 - P(X \leq t-s)\), where \(P(X \leq t-s)\) is the cumulative distribution function of G, denoted by \(G(t-s)\).
02

Calculate the number of satellites launched between time s and t

The number of satellites launched between times s and t follows a Poisson distribution with rate λ. Let Y be the random variable representing the number of satellites launched between time s and t. The probability that Y = n is given by the Poisson distribution: \(P(Y = n) = e^{-\lambda (t-s)} \frac{(\lambda (t-s))^n}{n!}\)
03

Calculate the probability that all satellites launched between time s and t are still in the air at time t

Using the probability calculated in Step 1, we can find the probability that all n satellites launched between time s and t are still in the air at time t: \(P(\text{All satellites in the air}) = (1 - G(t-s))^n\)
04

Multiply probabilities of Steps 2 and 3 and sum over all possible values of n

Now, we multiply the probabilities from Steps 2 and 3 and sum these products over all possible values of n (from 0 to infinity) to find the probability that none of the satellites in the air at time t were launched before time s: \(P(\text{No satellites launched before s}) = \sum_{n=0}^{\infty} P(Y = n) \cdot P(\text{All satellites in the air})\) \(= \sum_{n=0}^{\infty} e^{-\lambda (t-s)} \frac{(\lambda (t-s))^n}{n!} \cdot (1 - G(t-s))^n\) This is the final expression for the probability that none of the satellites in the air at time t were launched before time s.

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Key Concepts

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

Poisson Distribution
Understanding the Poisson distribution is crucial when dealing with events that occur independently and at a constant average rate. It's commonly used to model the number of times an event occurs within a fixed interval of time or space.

The key parameter in a Poisson distribution is the rate \(\lambda\), which represents the average number of events in the given interval. This distribution helps predict the probability of a specific number of events happening within a defined period. For example, it can estimate the number of satellites launched within a certain timeframe when launches are occurring continuously and independently.

To calculate the probability of exactly \(n\) events (in our case, satellite launches) happening in a given interval, we use the formula:\[P(Y = n) = e^{-\lambda(t-s)} \frac{(\lambda(t-s))^n}{n!}\]Here, \(t-s\) represents the interval of interest, \(e\) is the base of natural logarithms, and \(n!\) is the factorial of \(n\). Since Poisson distribution is discrete, we use factorial and powers of \(\lambda\) to represent the distinct number of occurrences. It's also worth noting that the sum of probabilities for all possible values of \(n\) in a Poisson distribution equates to one, which is a property of all probability distributions.
Cumulative Distribution Function
The cumulative distribution function (CDF), denoted as \(G\) in our example, is a fundamental concept in probability that describes the probability that a real-valued random variable \(X\), such as the time a satellite spends in space, takes on a value less than or equal to a specific point \(t-s\).

The function is expressed as:\[G(t-s) = P(X \leq t-s)\]The CDF is powerful because it gives a complete description of the distribution of a random variable. For continuous variables, the CDF is the integral of the probability density function (PDF), while for discrete variables, it is the accumulation of probabilities up to the specified value.

Understanding the CDF helps us deduce the likelihood of various outcomes. In our satellite case, we utilized the CDF to determine the probability of a satellite being in the air for a more extended period than the time since its launch. By subtracting this value from one, we calculate the complementary probability needed for further analysis:\[P(X > t-s) = 1 - G(t-s)\]This step is essential for subsequently deriving the overall probability of interest in the case of our satellite problem.
Random Variable
In probability and statistics, a random variable is a variable whose possible values are numerical outcomes of a random phenomenon. Essentially, it's a quantitative variable that assigns a numerical value to each possible outcome of an experiment.

Random variables can be discrete or continuous. Discrete random variables, such as the number of satellites launched in a given time frame, have countable values. On the other hand, continuous random variables, like the amount of time a satellite remains operational before re-entry, can take any value within an interval and are associated with measurements.

In our Poisson process example, we have two random variables: \(X\), representing the random time a satellite stays in space, and \(Y\), the count of satellites launched within a specific time interval. These random variables are the backbone of the probability calculations. They allow us to employ the functions of the Poisson distribution and the CDF to analyze and predict different scenarios based on the underlying random processes at play.

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Most popular questions from this chapter

Customers arrive at a two-server service station according to a Poisson process with rate \(\lambda .\) Whenever a new customer arrives, any customer that is in the system immediately departs. A new arrival enters service first with server 1 and then with server 2\. If the service times at the servers are independent exponentials with respective rates \(\mu_{1}\) and \(\mu_{2}\), what proportion of entering customers completes their service with server 2?

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be independent and identically distributed exponential random variables. Show that the probability that the largest of them is greater than the sum of the others is \(n / 2^{n-1}\). That is, if $$ M=\max _{j} X_{j} $$ then show $$ P\left\\{M>\sum_{i=1}^{n} X_{i}-M\right\\}=\frac{n}{2^{n-1}} $$ Hint: What is \(P\left[X_{1}>\sum_{i=2}^{n} X_{i}\right\\} ?\)

Let \(\\{N(t), t \geqslant 0\\}\) be a Poisson process with rate \(\lambda\). For \(sN(s))\) (b) \(P(N(s)=0, N(t)=3)\); (c) \(E[N(t) \mid N(s)=4] ;\) (d) \(E[N(s) \mid N(t)=4]\).

A set of \(n\) cities is to be connected via communication links. The cost to construct a link between cities \(i\) and \(j\) is \(C_{i j}, i \neq j .\) Enough links should be constructed so that for each pair of cities there is a path of links that connects them. As a result, only \(n-1\) links need be constructed. A minimal cost algorithm for solving this problem (known as the minimal spanning tree problem) first constructs the cheapest of all the (in) links. Then, at each additional stage it chooses the cheapest link that connects a city without any links to one with links. That is, if the first link is between cities 1 and 2, then the second link will either be between 1 and one of the links \(3, \ldots, n\) or between 2 and one of the links \(3, \ldots, n .\) Suppose that all of the \(\left(\begin{array}{c}n \\\ 2\end{array}\right)\) costs \(C_{i j}\) are independent exponential random variables with mean \(1 .\) Find the expected cost of the preceding algorithm if (a) \(n=3\), (b) \(n=4\).

Consider the coupon collecting problem where there are \(m\) distinct types of coupons, and each new coupon collected is type \(j\) with probability \(p_{j}, \sum_{j=1}^{m} p_{j}=1\). Suppose you stop collecting when you have a complete set of at least one of each type. Show that $$ P\\{i \text { is the last type collected }\\}=E\left[\prod_{j \neq i}\left(1-U^{\lambda_{i} / \lambda_{i}}\right)\right] $$ where \(U\) is a uniform random variable on \((0,1)\).

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