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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 \(t\) was launched before time \(s\), where \(s<t\), is given by the following expression: \[P(N_0(t-s) = N(t)) = \sum_{k=0}^\infty P(N_0(t-s) = N(t) | N(t) = k) P(N(t) = k).\] It is calculated by finding the probability of all satellites launched after time \(s\) being in space at time \(t\) and then conditioning on the number of satellites launched up to time \(t\).

Step by step solution

01

Define the relevant random variables

Let \(N(t)\) represent the number of satellites launched up to time \(t\). Since the satellite launches follow a Poisson process with rate \(\lambda\), we have \(N(t) \sim \text{Poisson}(\lambda t)\). Additionally, let \(X_i\) be the random variable representing the time in space for the \(i\)-th satellite. The distribution of \(X_i\) is given by \(G\). Now, we need to find the probability \(P(N_0(t-s) = N(t))\), where \(N_0(t-s)\) is the number of satellites launched after time \(s\) and still in the air at time \(t\).
02

Find the probability of a satellite being in space at time \(t\) given it was launched at time \(x\)

If a satellite is launched at time \(x\) and is still in space at time \(t\), then it must be the case that it spent at least \(t-x\) time units in space. We can write this probability using the distribution \(G\): \[P(t-X_i < x) = P(X_i > t-x) = 1 - G(t-x).\]
03

Find the probability of all satellites launched after time \(s\) being in space at time \(t\)

We want to find the probability that all satellites launched after time \(s\) and up to time \(t\) are still in space at time \(t\). If the satellite is launched at time \(x\) with \(s \leq x \leq t\), then it must be in space at time \(t\). We can use the result of Step 2 to compute this probability: \[P(X_i > t-x | x \in [s, t]) = 1 - G(t-x).\] Since the satellite launches are a Poisson process, they are independent of each other, and the time spent in space for each satellite is also independent. Therefore, the probability that all satellites launched after time \(s\) are still in space at time \(t\) is given by the product of their individual probabilities: \[P(N_0(t-s) = N(t)) = \prod_{i=1}^{N(t)} P(X_i > t-x_i | x_i \in [s, t]).\]
04

Compute the final probability

Finally, we find the probability \(P(N_0(t-s) = N(t))\) by conditioning on the number of satellites launched up to time \(t\): \[P(N_0(t-s) = N(t)) = \sum_{k=0}^\infty P(N_0(t-s) = N(t) | N(t) = k) P(N(t) = k).\] Using the results from Step 3 and the fact that \(N(t) \sim \text{Poisson}(\lambda t)\), we can calculate this probability.

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Key Concepts

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

Random Variables
In the context of this satellite launch exercise, a random variable is a numerical representation of a random phenomenon. Here, we focus on two main random variables:
  • \(N(t)\): Represents the number of satellites launched up to a specific time \(t\). This is a classic example of counting the number of independent events that happen until a given time.
  • \(X_i\): Indicates the random time spent in space by the \(i\)-th satellite. Each \(X_i\) follows the same distribution \(G\), which could be any valid probability distribution describing how long a satellite remains in space.
Understanding these random variables is crucial because they form the foundation of evaluating probabilities in the given Poisson process framework. Each \(N(t)\) and \(X_i\) is instrumental in determining the specifics of how many satellites could still be airborne after time \(t\).
When working with random variables, it's vital to identify which are dependent and which are independent, as these relationships guide the calculations in probability tasks.
Probability Distribution
Probability distributions describe how the probabilities are spread over the values of a random variable. In our satellite launch scenario, the key distributions are:
  • Poisson Distribution: The number of satellites launched by time \(t\), \(N(t)\), follows a Poisson distribution with parameter \(\lambda t\). This describes the probabilities of different numbers of launches occurring within the time interval, based on a constant average rate \(\lambda\).
  • Distribution \(G\): Each satellite spends a time in space characterized by this distribution. The actual form of \(G\) is not specified and could be an exponential, normal, or another type of distribution. What matters is how \(G\) is used in calculations, as it informs us about satellite longevity in orbit.
In a typical Poisson process, the Poisson distribution plays a crucial role in determining the frequency of independent events. Thus, understanding these distributions helps predict the satellite behavior based on probabilistic models.
Conditional Probability
Conditional probability reflects the likelihood of an event occurring given that another event has already occurred. For this satellite launch problem:
It is used to compute the probability of a satellite being still in space at time \(t\), provided that it was launched after time \(s\). This yields:
  • \(P(X_i > t-x | x \in [s, t])\): This expresses the conditional probability that a satellite launched after time \(s\) remains in space at time \(t\).
Using conditional probability, we focus only on those satellites launched in the specific time frame \([s, t]\). By narrowing the focus, we apply conditional probability to accurately assess situations given complex criteria.
This clarifies how certain events' likelihoods change upon receiving new information. In particular, knowing the satellite's launch time directly influences predictions about its presence in space.
Independent Events
Independence between events implies that the occurrence of one does not affect the likelihood of the other. In this problem setup:
  • The launch of each satellite is independent, following the Poisson process.
  • Each \(X_i\), representing time in space, is independent from other satellites’ \(X_i\).
Independence is critical as it allows us to multiply probabilities of individual events to find joint probabilities, as seen in Step 3. Here, the probability that all satellites remain in space is the product of individual satellite stay probabilities. This simplifies computations significantly.
Recognizing independent events helps understand cumulative behaviors of systems composed of many components, like these satellites, without complex interdependencies that would otherwise complicate calculations.

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