Question

The number of hours between successive train arrivals at the station is uniformlydistributed on [0;1]. Passengers arrive according to a Poisson process with rate 7 perhour. Suppose a train has just left the station. LetXdenote the number of peoplewho get on the next train. Denote byTthe arrival time of the next train.(a)GivenT= 0:4, what is the conditional expectationE(XjT= 0:4).(b)FindE(X). Hint: useE(X) =E[E(XjT)].

Answer 1

E ( X ) = 3.5

Var ( X ) = 91 / 12

Explanation:

Solution:-

- Let X = M (T), where M (T) is the Random variable that denotes the number of arrivals in time T for the Poisson process. The parameter λ = rate of 7 per hour

- To find E ( X ), condition X on the random arrival time T of the next train.

- Note that if Y ~ Poisson ( μ ), then the T is defined by uniform distribution over the interval [ 0 , 1 ] :

E ( Y ) = Var ( Y ) = μ

E ( T ) = 1 / 2

Var ( T ) = 1 / 12

- We have, N ( T ) ~ Poisson ( λt ), where t ≥ 0 and λ = 7. Thus,

E ( N ( T ) / T ) = λ*T

Var ( N ( T ) / T ) = λ*T.

Therefore,

E ( X ) = E ( N ( T ) ) = E ( E ( N ( T ) / T ) )

= E ( λ*T ) = λ* E ( T ) = 7/ 2 = 3.5

- For two Random Variables U and V,

Var ( U ) = E ( Var ( U / V ) ) + Var ( E ( U / V ) )

Therefore,

Var ( X ) = E ( Var ( X / T ) ) + Var ( E ( N ( T ) / T ) )

= E ( λ*T) + Var ( λ*T )

= λ* E ( T ) + λ^2* Var ( T )

= 3.5 + 7^2 / 12

= 91 / 12