Question

The number of hours between successive train arrivals at thestation is uniformly distributed on (0, 1). Passengers arriveaccording to a Poisson process with rate 7 per hour. Suppose atrain has just left the station. Let X denote the number of peoplewho get on the next train. Find
(a) E(X)

(b) Var(X)

Answer 1

Answer:

a) $E[x]= E[E[X|T]]=E[7T]=7E[T]=\frac{7}{2}$

b) $Var[x] = 7E[T]+ 49 Var[T]=\frac{7}{2}+\frac{49}{12}=\frac{91}{12}$

Step-by-step explanation:

A uniform distribution, sometimes also known as a rectangular distribution, is a distribution that has constant probability and is defined on a interval [a,b].

The Poisson distribution is the discrete probability distribution in order to describe the number of events occurring in a given time period.   And is defined with a parameter calles usually $\lambda$.

Let T the random variable that represent the number of hours between successive train arrivals at the station, and we know that the distribution for T is given by:

$T \sim U(0,1)$

The expected value and the variance for an uniform random variable is given by:

$E[T] = \frac{b-a}{2}=\frac{1-0}{2}=\frac{1}{2}$

$Var(T) = \frac{(b-a)^2}{12}=\frac{1}{12}$

ANd let X the random variable that represent the  number of people who get on the next train, so the conditional distribution X|T is given by:

$X|T \sim Poisson(\lambda T)=Poisson(7T)$

Since we have a Poisson distribution we can find the expected value and the variance for the random variable X|T

$E[X|T]=7T$ $Var[X|T]=7T$

We can use this result in order to answer the questions like this:

Part a

Using the properties for expected value we have this:

$E[x]= E[E[X|T]]=E[7T]=7E[T]=\frac{7}{2}$

Part b

We are interested in Var (X) and in order to find this we can use the following property from conditional probability:

$Var[X]=E[Var[X|T]] + Var[E[X|T]]$

And since we already found Var[X|T] and E[X|T] we have this:

$Var[X]=E[7T] +Var[7T]$

$Var[x] = 7E[T]+ 49 Var[T]=\frac{7}{2}+\frac{49}{12}=\frac{91}{12}$