Remove the restriction that two or more customers cannot arrive at the same time, (i.e., remove orderliness property)
be a Poisson Process with rate
, and let
be the i.i.d random variable, then
is a compound Poisson process
Buses arrive according to a Poisson process. Let
be the number of people on bus i, and let
be the total number of people arriving by time t.
Insurance claims arrive according to a Poisson process. Let
be the size of the claim (in dollars), and let
be the total amount due to all claims by time t.
var[X(t)] = var[E[X(t)|N(t)]] + E[var[X(t)|N(t)]]
and we have
1. Definition of Poisson Random Variable
Def: A Poisson random variable with mean A and probability mass function:
2. Binomial Approximate Poisson
be an indicator variable of either 0 or 1. And assume
for all i. Then
Based on the assumptions, we have a binomial distribution:
(A Poisson random variable)
is approximately Poisson with mean 400/365
4. Priliminary Definitions
Def : A stochastic process is a collection of random variable (RV) indexed by time .
- If T is continuous set, the process is a continuous time stochastic process (e.g., Poisson Process)
- If T is countable, then the process is a discrete time stochastic process (e.g., Markov chain)
Def. A counting process
is a stochastic process
Def: A counting process has stationary increments
if the distribution of the number of events in an interval depends on the length of the interval, but not on the starting point of the interval. That is,
does not depend on s. Intuitively, the interval can be “slide” around without changing its stochastic nature.
– Stochastic Process course notes
A random variable X has the memoryless property if
Def`2: A random variable X has the memoryless property if
The exponential distribution is the only distribution that has the memoryless property (Satisfy definition 2)
3. Useful Properties: First occurrence among events
Assume are exponential variable with rate .
Then what is the probability that .
4. Distribution of time of first event
This is the CDF of an exponential RV with rate
5. Distribution of time of last event (maximum)
Excess Distribution of Renewal Process
– Course notes of Stochastic Process, 2014 Fall
Excess of renewal process is defined as
(time until next event)
In the example of average time waiting bus, we drived
As we want to determine the fraction of time that
, and let
: Fraction of time that
= Fraction of “on” time for
be “on” time during cycle
Note: the ON time and OFF time for each cycle are dependent. A longer ON time implies a shorter OFF time.
Then we have
is an alternating renewal process, fraction of “on” time is
This is sometimes called equilibrium distribution,
3. Example of different distribution of
As expected (by memoryless property), excess distribution is an exponential distribution.
Example: Pareto Distribution
is also called the tail distribution
(mean only exist if
This is the CCDF of a uniform distribution on
Note: the above assume that
, then the result is 0.
Alternating Renewal Process
Consider a process
with “on” periods and “off” periods
- Let ON time of cycle j
- Let OFF time of cycle j
must satisfy the following property in order for
to be regenerative
must be i.i.d; in particular,
. That is
Then by the regenerative process, we have
the average “up” time =
Cars pass a point on highway according to Poisson process with rate
of cars are speeding (>10 mph over the pretend speed limit). Assume time to issue a ticket~UNIF[10,14] minutes(one officer)
Question: What fraction of speeding cars pass when the officer is busy.
Answer: The answer to the question is equivalent to the fraction of time the officer is busy.
- Let = time spent waiting to give a ticker
- Let = time spent giving a ticker
Speeders arrive according to Poisson process with rate
Fraction of time busy =
Motivation: Find the cost of buying options to prevent arbitrage opportunity.
be the price of a stock at time s, considered on a time horizon
. The following actions are available:
Approach: By the arbitrage theorem, this requires finding a probability measure so that each bet has zero expected pay off.
We try the following probability measure on
: suppose that
is geometric Brownian motion. That is,
2. Analyze Bet 0: Purchase 1 share of stock at time s
Now, we consider a discount factor
. Then, the expected present value of the stock at time t is
Choose such that
Thus the expected present value of the stock at t equals the present value of the stock when it is purchased at time s. That is, the discount rate exactly equals to the expected rate of the stock returns.
3. Analyze Bet i: Purchase 1 share of option i (at time s =0)
First, we drop the subscript i to simplify notation.
Expected present value of the return on this bet is
Setting this equal to zero implies that
has value 0 when , i.e., when
Thus the integral becomes
Now, apply a change of variables:
No arbitrage exists if we find costs and a probability distribution such that expected outcome of every bet is 0.
- If we suppose that the stock price follow geometric Brownian motion and we choose the option costs according to the above formula, then the expected outcome of every bet is 0.
- Note: the stock price does not actually need to follow geometric Brownian motion. We are saying that if stock price follow Brownian motion, then the expected outcome of very bet would be 0, so no arbitrage exists.
A few sanity checks
- When , cost of option is
- When , cost of option is
- As t increases, c increases
- As k increases, c decreases
- As increases, c increases
- As increase, c increase (assuming )
- As , c decreases