# Vsub Segment 41

## 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.

### 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.)