Question

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

The expected number of passengers who will get on the next train is \(E[X] = \frac{7}{2} \ (3.5)\), and the variance of the number of passengers is \(\operatorname{Var}(X) = \frac{49}{6}\).

Step-by-step solution

Find the conditional PMF of X given t

Let's find the PMF of the number of passengers \(X\) who will board the next train, given the time \(t\) between train arrivals. Since passengers arrive according to a Poisson process with rate 7 per hour, and \(t\) is the time interval between trains, the distribution of \(X\) conditional on \(t\) will be a Poisson distribution with parameter \(\lambda = 7t\). The PMF of a Poisson distribution is given by the formula: \( P(X=k | t) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \) Substitute the parameter \(\lambda = 7t\) into the formula: \( P(X=k | t) = \frac{e^{-7t} \cdot (7t)^k}{k!} \) Now we have the conditional PMF of \(X\) given \(t\).

Find the marginal PMF of X

To find the expected value and variance of \(X\), we need the marginal PMF of \(X\). To obtain this, we'll integrate the conditional PMF of \(X\) given \(t\) over all possible values of \(t\), which is the interval \((0, 1)\) since the time between train arrivals is uniformly distributed on \((0, 1)\). So, the marginal PMF of \(X\) is given by: \( P(X=k) = \int_{0}^{1} P(X=k | t) \, dt = \int_{0}^{1} \frac{e^{-7t} \cdot (7t)^k}{k!} \, dt \)

Find E[X]

To find the expected value of \(X\), denoted by \(E[X]\), we'll use the formula: \( E[X] = \sum_{k=0}^{\infty} k \cdot P(X=k) \) We already have the marginal PMF of \(X\) from the previous step, so we can directly compute the expected value: \( E[X] = \sum_{k=0}^{\infty} k \cdot \int_{0}^{1} \frac{e^{-7t} \cdot (7t)^k}{k!} \, dt \) Now, we can find the expected value by solving the above equation. The expected value of \(X\) is: \( E[X] = 7 \cdot \int_{0}^{1} t \, dt = 7 \cdot \frac{1}{2} = \frac{7}{2} \) So, the expected number of passengers who will get on the next train is 3.5.

Find E[X^2]

To find the variance of \(X\), we need to find the expected value of the square of \(X\), denoted by \(E[X^2]\). The formula for \(E[X^2]\) is: \( E[X^2] = \sum_{k=0}^{\infty} k^2 \cdot P(X=k) \) Now, substitute the marginal PMF of \(X\) into the formula: \( E[X^2] = \sum_{k=0}^{\infty} k^2 \cdot \int_{0}^{1} \frac{e^{-7t} \cdot (7t)^k}{k!} \, dt \) By solving the above equation, we find that: \( E[X^2] = 49 \cdot \frac{2}{3} = \frac{98}{3} \)

Find Var(X)

Now, to find the variance of \(X\), denoted by \(\operatorname{Var}(X)\), we'll use the formula: \( \operatorname{Var}(X) = E[X^2] - (E[X])^2 \) Substitute the values of \(E[X]\) and \(E[X^2]\) found in the previous steps: \( \operatorname{Var}(X) = \frac{98}{3} - \left( \frac{7}{2} \right)^2 = \frac{49}{6} \) So, the variance of the number of passengers who will get on the next train is \(\frac{49}{6}\).

Key concepts

Understanding the Probability Mass Function (PMF)

In probabilistic terms, the probability mass function (PMF) is a fundamental concept used to describe the probability distribution of a discrete random variable. A discrete random variable is one that has a countable number of possible values, unlike a continuous random variable, which can take on any value within a given range.

The PMF gives us the probability that a discrete random variable will take on a specific value. For example, in a Poisson process like the one described in the exercise, the PMF is given by the formula:
\[ P(X=k | t) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \]
This represents the probability that exactly k events occur in a fixed period of time, given the average rate of \(\lambda\) occurrences in that period. A key takeaway here is that understanding PMFs can help us in determining the likelihood of different outcomes for a process or an experiment. For a Poisson process, the PMF is particularly helpful because it's defined by a single parameter \(\lambda\)—the average rate at which events occur.

Here's a quick tip: when tackling homework or test questions involving PMFs, always make sure to first identify the type of distribution you're dealing with—it can vary from Poisson to binomial, or others. Next, write down the PMF formula specific to that distribution, as it can guide you through calculating probabilities for particular outcomes.

Predicting Outcomes with Expected Value

The expected value, denoted as E[X], is a powerful concept used to predict the average outcome of a random variable over many repetitions of an experiment. It provides a measure of the center of the probability distribution and can be thought of as the long-term average if the process were repeated indefinitely.

In a Poisson process, we calculate the expected value using the formula:
\[ E[X] = \sum_{k=0}^{\infty} k \cdot P(X=k) \]
The calculation involves multiplying each possible value that the random variable can assume by the probability of that value occurring, and then summing all these products. In the exercise provided, E[X] represents the average number of people who get on the train, and this figure is expected to be consistent across numerous train arrivals.

For instance, the expected number of passengers, \(E[X]\), is 3.5. This means that over time, you can expect an average of 3.5 people to board the train between arrivals. Understanding expected value is crucial because it informs decisions in uncertain situations and can help in strategizing under risk. It illustrates what one might anticipate in the long run, which in practical terms could help in resource allocation, for instance, the number of tickets a train station should prepare to sell between train arrivals.

Variance: Measuring Dispersion of Data

The concept of variance is integral to understanding the spread of probability distributions for random variables. It provides a measure of how much the values of the variable deviate from the expected value; in other words, it's a numerical value indicating how much the data is spread out or clustered around the mean.

The variance is denoted by Var(X) and typically calculated using the formula:
\[ \operatorname{Var}(X) = E[X^2] - (E[X])^2 \]
Here, E[X^2] represents the expected value of the square of the random variable, and subtracting the square of the expected value of X gives us the variance.

In the context of the exercise, where passengers are arriving at a train station according to a Poisson process, knowing the variance helps in understanding the reliability of the expected number of passengers. A smaller variance suggests that the actual number of passengers you might see is likely to be close to the expected number, whereas a larger variance indicates greater unpredictability. In this specific scenario, the variance of the number of passengers boarding the next train was calculated to be \(\frac{49}{6}\), which gives us insight into the consistency of the number of passenger arrivals. A train station manager, for instance, might use this information for staffing needs or seating availability.

Keep in mind when interpreting variance that it is squared relative to the unit of measurement of the data. The square root of the variance, known as the standard deviation, can sometimes be more intuitive to work with as it is in the same units as the original data.