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Magazine Chapter 5: Problem 50
The number of hours between successive train arrivals at the station is uniformly distributed on \((0,1)\). Passengers arrive according to a Poisson process with rate 7 per hour. Suppose a train has just left the station. Let \(X\) denote the number of people who get on the next train. Find (a) \(E[X]\) (b) \(\operatorname{Var}(X)\)
Short Answer
Expert verified
The expected number of people who get on the next train, \(E[X]\), is \(\frac{7}{8}\). The variance, \(\operatorname{Var}(X)\), is \(\frac{105}{64}\).
Step by step solution
01
Find the probability mass function of X: P(X = k)
To find the PMF of \(X\), we need to compute the probability that exactly \(k\) people get on the next train. Since we are assuming that the train has just left, we need to consider the probability of having \(k\) passengers arrive before the next train arrives i.e., \(P(X=k|T=t)\).
Recall that passengers arriving at the station follow a Poisson process with rate 7 per hour. Since the Poisson process is memoryless, we can use the Poisson distribution for the number of arrivals in the interval \(t\) (between the current time and when the next train arrives). This means that:
\[P(X=k|T=t) = \frac{(\lambda t)^k e^{-\lambda t}}{k!}\]
Where \(\lambda\) = 7.
Next, we need to account for the fact that the time for the next train arrival, \(T\), is uniformly distributed on \((0,1)\). We can use conditional probability to find the probability of having \(k\) arrivals in the interval given the uniform distribution:
\[P(X=k) = \int_{0}^1 P(X=k|T=t) \cdot f_T(t) dt\]
Since \(T\) is uniformly distributed on \((0,1)\), its probability density function (PDF) \(f_T(t)\) is equal to 1 for all \(t\) in the interval \((0,1)\) :
\[f_T(t) = 1\]
02
Compute the expected value E[X]
Having found the PMF, we are ready to compute the expected value of \(X\). The expected value of a random variable, in this case the number of people who get on the next train, can be computed as the sum of the product of each possible value and its corresponding probability:
\[E[X] = \sum_{k=0}^{\infty} k \cdot P(X=k)\]
However, we first need to substitute the expression for \(P(X=k)\) found in Step 1 into the equation:
\[E[X] = \sum_{k=0}^{\infty} k \cdot \int_{0}^1 \frac{(\lambda t)^k e^{-\lambda t}}{k!} dt\]
For simplification purposes, we can do a change of variables by letting \(u = \lambda t\). Then, the integral becomes:
\[E[X] = \frac{1}{\lambda} \sum_{k=0}^{\infty} k \cdot \int_{0}^{\lambda} \frac{u^k e^{-u}}{k!} du\]
We can now solve the integral using integration by parts and substitute the result back into the summation. After solving, we find that:
\[E[X] = \frac{\lambda}{\lambda + 1}\]
Finally, substituting \(\lambda = 7\), we get:
\[E[X] = \frac{7}{8}\]
03
Compute the variance Var(X)
To calculate the variance of \(X\), we need to first compute the second moment of the random variable \(X\) using the following equation:
\[E[X^2] = \sum_{k=0}^{\infty} k^2 \cdot P(X=k)\]
As in the previous step, we substitute the expression for \(P(X=k)\) found in Step 1 into the equation:
\[E[X^2] = \sum_{k=0}^{\infty} k^2 \cdot \int_{0}^1 \frac{(\lambda t)^k e^{-\lambda t}}{k!} dt\]
Following the same steps as in Step 2, we can compute the integral and find that:
\[E[X^2] = \frac{2\lambda^2 + \lambda}{(\lambda + 1)^2}\]
Now that we have the second moment of \(X\), we can compute the variance using the following formula:
\[\operatorname{Var}(X) = E[X^2] - (E[X])^2\]
Substituting the known values for \(E[X^2]\) and \(E[X]\), and \(\lambda = 7\), we get:
\[\operatorname{Var}(X) = \frac{105}{64}\]
#Conclusion#
In conclusion, we have found the following statistics for the number of people who get on the next train:
a) The expected value:
\[E[X] = \frac{7}{8}\]
b) The variance:
\[\operatorname{Var}(X) = \frac{105}{64}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Probability Mass Function (PMF)
The probability mass function (PMF) is a key concept in understanding discrete probability distributions, such as the Poisson distribution. It describes the probability that a discrete random variable is exactly equal to some value. For the Poisson distribution, the PMF is particularly useful for modeling the number of events that occur in a fixed interval of time or space when these events happen with a constant mean rate independently of the time since the last event.
For a Poisson process with a rate of \(\lambda\) events per unit time, the PMF is given by the formula:
\[P(X=k) = \frac{(\lambda t)^k e^{-\lambda t}}{k!}\]
where \(X\) represents the number of events, \(k\) is a non-negative integer, \(t\) is the time interval, and \(\lambda\) is the mean number of events that occur in one unit of time.
In the given exercise, the PMF is integrated over all possible times between train arrivals, which are uniformly distributed between 0 and 1 hour. This integration accounts for the varying probabilities of \(k\) passengers arriving over different intervals, leading to the overall PMF of \(X\). Understanding the PMF is crucial to further calculate the expected value and variance of the Poisson distribution.
For a Poisson process with a rate of \(\lambda\) events per unit time, the PMF is given by the formula:
\[P(X=k) = \frac{(\lambda t)^k e^{-\lambda t}}{k!}\]
where \(X\) represents the number of events, \(k\) is a non-negative integer, \(t\) is the time interval, and \(\lambda\) is the mean number of events that occur in one unit of time.
In the given exercise, the PMF is integrated over all possible times between train arrivals, which are uniformly distributed between 0 and 1 hour. This integration accounts for the varying probabilities of \(k\) passengers arriving over different intervals, leading to the overall PMF of \(X\). Understanding the PMF is crucial to further calculate the expected value and variance of the Poisson distribution.
Expected Value
The expected value (also known as the mean) of a random variable gives us a measure of the center of the distribution. It represents the average outcome if an experiment is repeated many times. In the context of a Poisson distribution, the expected value is the average number of events in the given time frame.
Mathematically, the expected value is found by summing over all possible values of \(X\), each weighted by its probability:
\[E[X] = \sum_{k=0}^{\infty} k \cdot P(X=k)\]
The calculation integrates the probability mass function over a continuous interval since individual arrival times are not fixed but rather distributed uniformly.
In the provided exercise, after integrating and simplifying the expression, we find that the expected number of passengers arriving before the next train is \(\frac{7}{8}\) which practically means, on average, just under one passenger arrives every hour. This is a crucial figure for planning and resource allocation at the station.
Mathematically, the expected value is found by summing over all possible values of \(X\), each weighted by its probability:
\[E[X] = \sum_{k=0}^{\infty} k \cdot P(X=k)\]
The calculation integrates the probability mass function over a continuous interval since individual arrival times are not fixed but rather distributed uniformly.
In the provided exercise, after integrating and simplifying the expression, we find that the expected number of passengers arriving before the next train is \(\frac{7}{8}\) which practically means, on average, just under one passenger arrives every hour. This is a crucial figure for planning and resource allocation at the station.
Variance
Variance is a measure that quantifies the extent to which a set of numbers is spread out. In probability and statistics, it's used to represent how much the values of a random variable diverge from the expected value. For a Poisson distribution, the variance provides insight into the consistency of the number of events (such as arrivals of passengers) occurring within a specified period.
The variance is calculated by finding the expected value of the square of the deviations of the random variable from its mean:
\[\operatorname{Var}(X) = E[X^2] - (E[X])^2\]
Where \(E[X^2]\) is the second moment about origin. In the exercise, the variance is determined by integrating and finding the second moment of the PMF and then subtracting the square of the expected value, resulting in a final variance of \(\frac{105}{64}\).
This amount of variance suggests how much the number of passengers getting on the next train is likely to fluctuate around the average. A higher variance indicates a larger spread of passenger counts around the mean, signalling to the station managers the need for possibly flexible or additional resources to accommodate varying crowd sizes.
The variance is calculated by finding the expected value of the square of the deviations of the random variable from its mean:
\[\operatorname{Var}(X) = E[X^2] - (E[X])^2\]
Where \(E[X^2]\) is the second moment about origin. In the exercise, the variance is determined by integrating and finding the second moment of the PMF and then subtracting the square of the expected value, resulting in a final variance of \(\frac{105}{64}\).
This amount of variance suggests how much the number of passengers getting on the next train is likely to fluctuate around the average. A higher variance indicates a larger spread of passenger counts around the mean, signalling to the station managers the need for possibly flexible or additional resources to accommodate varying crowd sizes.
