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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 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

01

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 \(t\).
02

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: \( P(X=k) = \int_{0}^{1} P(X=k | t) \, dt = \int_{0}^{1} \frac{e^{-7t} \cdot (7t)^k}{k!} \, dt \)
03

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.
04

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: \( E[X^2] = 49 \cdot \frac{2}{3} = \frac{98}{3} \)
05

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 \(\frac{49}{6}\).

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Most popular questions from this chapter

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