Vsub Segment 41

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Calculating

Poisson and Exponential

1. Show that the waiting times (times between events) in a Poisson process are Exponentially distributed. (I think we've done this before.)

This follows from the definition of a Poisson process. A Poisson process (is a stochastic process) that counts the number of events and the time that these events occur in a given interval. The time between each pair of consecutive events has an exponential distribution. Mathematically:

The number of events (of a constant rate Failed to parse (unknown error): \lambda ) that occur in a given time interval is a Poisson random variable given by

   Failed to parse (unknown error):  P [(N(t+ \tau) - N(t)) = k] = \frac{e^{-\lambda \tau} (\lambda \tau)^k}{k!}  \qquad k= 0,1,\ldots,

N(t) denotes the number of events that have occured in time t.

The waiting time between two events: we want the distribution of Failed to parse (unknown error): \tau (when k=1 in the above definition). Let the arrival times of events in the Poisson process be given by Failed to parse (unknown error): T_1, T_2, \ldots . Then, we can say that the probability of the Failed to parse (unknown error): n^{th} event occurring after time t is equal to the probability of there having occurred fewer than Failed to parse (unknown error): n events in the time upto t. Mathematically,

 Failed to parse (unknown error):  P(T_n > t) = P (N(t) < n) 

Now, consider the first event (n=1):

 Failed to parse (unknown error):  P(T_1 > t) = P (N(t) = 0 ) = P[(N(t) - N(0)) = 0] = \frac{e^{-\lambda t} (\lambda t)^{0}}{0!} = e^{-\lambda t} 

This shows that the waiting time (t) until the first arrival has an exponential distribution. Similarly,

 Failed to parse (unknown error):  P((T_n - T_{n-1}) > t) = P[(N(t_n) - N(t_{n-1})) = 0] = \frac{e^{-\lambda t} (\lambda t)^{0}}{0!} = e^{-\lambda t} 

Hence, the waiting times between events in a Poisson process are exponentially distributed (and the expected wait time is Failed to parse (unknown error): 1/\lambda )

Wait times

2. Plot the pdf's of the waiting times between (a) every other Poisson event, and (b) every Poisson event at half the rate.


wait times poisson process. wait times poisson process. wait times poisson process.

Poisson and Gamma

3. Show, using characteristic functions, that the waiting times between every Nth event in a Poisson process is Gamma distributed. (I think we've also done one before, but it is newly relevant in this segment.)