- #1

O_o

- 32

- 3

## Homework Statement

-A bus is moving along an infinite route and has stops at n = 0, 1, 2, 3, ...

-The bus can hold up to B people

-The number of people waiting at stop n, Yn, is distributed Poisson(10) and is independent from the number waiting at all the other stops.

-At any given stop each passenger has probability p of getting off and the passengers are mutually independent.

-Assume after the passengers get off at stop n, everyone waiting at that stop gets on as long as there are spots remaining

[tex] \{X_n; n \geq 0\} [/tex] is a Markov Chain representing the number of passengers on the bus after it leaves stop n.1) What are the transition probabilities, pij?

2) What is [tex] E[X_n] [/tex] for n = 0, 1, 2, ..., 20 if p = 0.4 and B = 20

## Homework Equations

## The Attempt at a Solution

(1)

[tex]X_n = \min\{X_{n-1} - L_n + Y_n, B\}[/tex]

Where L_n ~ Binomial(X_{n-1}, p) is the number of people who got off at stop n

The transition probabilities need to be broken up into cases. For i > 0, j < i [tex] p_{i,j} = \sum_{k=i-j}^i {i \choose k} p^k(1-p)^{i-k} \left( \frac{10^{j-i+k} e^{-10}}{(j-i+k)!}\right)[/tex]

for i>0, B > j >= i [tex]

p_{i,j} = \sum_{k=0}^i {i \choose k} p^k(1-p)^{i-k} \left( \frac{10^{j-i+k} e^{-10}}{(j-i+k)!}\right)[/tex]

for i>=0 j = B[tex]

p_{i,j} = \sum_{k=0}^i {i \choose k} p^k(1-p)^{i-k} \left(\sum_{h=j-i+k}^{\infty}\frac{10^{h} e^{-10}}{(h)!}\right)[/tex]

(2)

I feel reasonably okay about question (1) although maybe it's completely wrong. My main concern was part 2. If I want to use [tex]X_n = \min\{X_{n-1} - L_n + Y_n, B\} [/tex] to find the expected value, I get this [tex] E[X_n] =\min\{E[X_{n-1}] - E[L_n] + E[Y_n], B\}[/tex]

which seems like it should be doable Except L_n ~ binomial (X_{n-1}, p) so the parameter is a random variable. Am I allowed to say[tex]E[L_n] = E[X_{n-1}]p[/tex]

We haven't learned anything about having random variables in sum boundaries so I'm a bit clueless. Thank you.

Last edited: