Question

Cars get parked in a lot (with infinite capacity) according to a λ-rate Poisson process, and indepen- dently stay parked for a random duration. The parking time duration of a car follows a common distribution X, with cdf F(x) = P(X ≤ x). Let N(t) be the number of cars parked at time t.

1a. What is the distribution of N(t)?

1b. Assuming a car arrival rate of 1 per minute, and X (in min) ∼ Gamma(3, 1) (as defined in class), what is the expected number of cars parked after 1 hour?

2. In the long run (t → ∞), what is the expected number of cars, as a function of λ and the moments of X?

Answer 1

Here the cars arrive at a Parking Lot at a λ-rate Poisson Process independently. So, the Parking Time of a car Follows Exponential Distribution.

i.e. X~ exp(λ)

(The above is implied from the Famous result: The “interarrival” times are typically exponentially distributed for a Poisson Process i.e. the Experimental Units arrive at the system at a Poisson Rate.)

So, if we consider N(t) be the number of cars parked at time t

Here, N(0)=0

N(t) has independent increments

The counting process {N(t),t∈[0,∞)} is called a Poisson process with rates λ.

1. The PMF of N(t) converges to a Poisson distribution with rate λt. More generally, we can argue that the number of arrivals in any interval of length τ follows a Poisson(λτ)

(The above can be proved from convergence of Distribution)

So, N(t) ~ Poi(λt)

2. Now the car arrival rate is provided to be 1 per minute.

i.e. λ = 1

[We know, Z ~ Gamma(α, λ) and if we consider α=1 then Z ~ exp(λ)]

Here, X (in min) ~ Gamma(3, 1)

We have to find E[N(t)].

t = 1 hour or 60 minutes

=> E[N(t)] = λt = 180

So, the Expected Number of Cars to be parked within 1 hr. is 180.

2. As t → ∞ i.e. in the long run then E[N(t)] = λt → ∞

So, in the long run as there is infinite Parking space the Expected Number of cars will tend to infinity. (even if the expected parking time is 1/λ)

The moments will cease to Exist.