Problem 16 - Applied Probability

Question

A junction point in a network has two incoming lines and two outgoing lines. The number of incoming messages \(N_{1}\) on line one in one hour is Poisson (50)\(;\) on line 2 the number is \(N_{2} \sim\) Poisson (45). On incoming line 1 the messages have probability \(p_{1 a}=0.33\) of leaving on outgoing line a and \(1-p_{1 a}\) of leaving on line \(b\). The messages coming in on line 2 have probability \(p_{2 a}=0.47\) of leaving on line a. Under the usual independence assumptions, what is the distribution of outgoing messages on line a? What are the probabilities of at least 30,35,40 outgoing messages on line a?

Step-by-Step Solution

Answer

The distribution of outgoing messages on line a is Poisson(\(50 \cdot 0.33 + 45 \cdot 0.47\)). To find the probabilities for at least 30, 35, and 40 messages, calculate \(1 - P(X < k)\) using the CDF of the determined Poisson distribution.

Step 1: Define the Distribution of Messages on Line 1

Identify the distribution of the number of messages leaving on outgoing line a from incoming line 1. Since it is given that the number of incoming messages on line 1 is Poisson(50) and the probability of each message leaving on line a is 0.33, the number of messages leaving on line a from line 1 follows a binomial-like behavior. However, due to the thinning property of the Poisson distribution, the actual distribution is Poisson with parameter \(50 \cdot 0.33\).

Step 2: Define the Distribution of Messages on Line 2

Identify the distribution of the number of messages leaving on outgoing line a from incoming line 2. Like step 1, this is based on the thinning property of the Poisson distribution. Since the number of incoming messages is Poisson(45) and the probability of leaving on line a is 0.47, the distribution is Poisson with parameter \(45 \cdot 0.47\).

Step 3: Determine the Distribution for Outgoing Line a

Combine the distributions from step 1 and step 2. The number of outgoing messages on line a is the sum of two independent Poisson random variables, which is also a Poisson random variable. The parameter for outgoing line a is the sum of the individual parameters, so it is Poisson with parameter \(50 \cdot 0.33 + 45 \cdot 0.47\).

Step 4: Calculate the Parameter for Outgoing Line a

Calculate the parameter \(\lambda_a\) for the Poisson distribution of outgoing messages on line a. \(\lambda_a = 50 \cdot 0.33 + 45 \cdot 0.47\) gives the exact value of \(\lambda_a\).

Step 5: Determine the Probability of at Least k Messages

Use the cumulative distribution function (CDF) of the Poisson distribution to find the probability of at least 30, 35, 40 messages on line a. The CDF value for a Poisson random variable with parameter \(\lambda\) at \(k\) is \(P(X < k)\). To find the probability of at least \(k\) messages, compute \(1 - P(X < k)\), where \(X\) follows Poisson(\(\lambda_a\)). This will require the use of a Poisson table, calculator, or software.

Key Concepts

Probability Theory

Probability theory forms the foundation of understanding random events and stochastic processes. It allows us to quantify the likelihood of various outcomes in an uncertain scenario. In its most fundamental sense, probability can be seen as the ratio of the number of favorable outcomes to the number of all possible outcomes for a given event.

For example, consider flipping a fair coin; the probability of it landing on heads is 0.5 because there are two possible outcomes, heads or tails, making the ratio of favorable to possible outcomes equal to 1 out of 2. When we deal with problems like the one given in the exercise, where we have incoming messages following a Poisson distribution, we're interested in the probabilities of certain counts of events, such as messages, happening in a fixed period or space.

Stochastic Processes

Stochastic processes deal with collections of random variables usually indexed by time or space. They model systems that evolve over time in a way that involves randomness. The Poisson processes, such as the one described in the exercise, are a classic example. These processes count the number of events happening over continuous time and are defined by two properties: independence of increments and stationarity.

A Poisson process assumes that events occur independently over time. That is, the number of events happening in one time frame does not affect the number in another. In the context of the network junction problem, each message on the network is an event in the Poisson process, and each incoming message line is treated as an independent Poisson process.

Random Variables

A random variable is a variable whose values result from the outcomes of a random phenomenon. It's a quantitative description of the outcomes of a stochastic process. There are two types of random variables: discrete, which can take on a countable number of values, and continuous, which can take on any value in a range of values.

In the exercise, the number of messages, both incoming and outgoing, are discrete random variables because they can only take on non-negative integer values. The number of messages on line a from both incoming lines can be described specifically by Poisson random variables because the messages arrive at a constant average rate independently of the time since the last event.

Cumulative Distribution Function

The cumulative distribution function (CDF) of a random variable gives the probability that the variable will take a value less than or equal to a certain value. For discrete random variables like the number of outgoing messages on line a, the CDF is a step function that increases at each integer value where the probability mass function (PMF) is not zero.

In order to calculate probabilities for 'at least' scenarios, such as finding the probability of at least 30 messages, you would use the CDF and subtract its value from 1, this is because the CDF gives the probability of being less than or equal to a value, and by taking the complement, we find the probability of being greater than the given value. The exercise requires using the CDF to find out the probabilities for at least 30, 35, and 40 outgoing messages, which is done using the cumulative probabilities up to one less than these values.