Poisson

1
Poisson

• Poisson is similar to binomial
• Some important differences
• Poisson counts the number of events in a time
  period
• Not Success/Failure
• Not trials
• Time period has decimal places
• Turns out to be very important

2
Poisson

• Assumptions
• 1. For small time interval, Prob(gt1 event) is
  negligible
• 2. For small time, Prob(1 event) ? t
• 3. Non overlapping time intervals are independent

3
Poisson

• Prob(no events in time t)
• For (not small) time t, divide 0,t into N
  subintervals that are small
• Each subinterval has either 0 or 1 event
• Prob(0 events in a subinterval) 1-? t/N
• Prob(0 events overall) (1-? t/N)N
• Take limit as N-gt inf
• e-? t

4
Poisson

• Prob(1 event in time t)
• Divide as before
• Prob(1 event) N? (t/N)(1-? t/N)N-1
• ? t (1-? t/N)N-1
• Limit ? t e-? t

5
Poisson

• General formula for PMF
• Prob(k events) e-? t (? t)k/k!
• Note Book uses only ? instead of ?t
• He assumes t1

6
Poisson

• Recursive formula for PMF
• PMF(k) PMF(k-1) ?t/k
• PMF(0) e-? t
• Shows that PMF increases when ?t/kgt1
• I.e., klt?t
• Decreases for larger k
• If k?t, then 2 PMF values are the same

7
Poisson

• Mean ?t
• SD?Mean

8
Poisson

• Matlab
• pprob(t,lambda,lo,hi)
• For Prob(at least X events), use large value for
  HI

9
Poisson

• Exercises
• 1. Example 27, p. 175
• 2. p. 176, 3.4.4
• 3. Suppose rate1.5 and T3.5, find prob of being
  more than 2 SD above the mean
• 4. Repeat for T6

10
Erlang

• An Erlang RV is the time until the R-th event
  occurs
• R1 is called an exponential RV
• Prob(Erlang lt T) Prob(at least R events in time
  T)
• pprob(t,lambda,R,999)
• erlangprob(r,rate,0,T)

11
Erlang

• Exercises
• 1. Suppose I get calls at a rate of 2.5/hr. Find
  prob that time to next call is gt0.75 hrs
• 2. Find prob that time to the 3rd call is lt1.1
  hrs
• 3. 90 of time, the time to the 3rd call is AT
  LEAST how much?
• 4. 75 of time, the time to the 2nd call is AT
  MOST how much?

12
Erlang

• What if we change T very slightly?
• The probability will change slightly
• T is a continuous parameter
• Cannot make slight changes in N for the binomial

13
Erlang

• Conditional prob for waiting time
• Rate2.5, R6
• Prob(timegt4) erlangprob(6,2.5,4,99)
• 0.2414
• Mean6/2.5 2.4
• So reasonable that Timegt4 is not so likely

14
Erlang

• Prob(timegt6 timegt4)

15
Erlang
16
Erlang
17
Erlang
18
Erlang
19
Erlang
20
Erlang
21
Erlang

• Prob(timegt6 timegt4) 0.0148/ 0.2414
• 0.0613
• gt 0.0148

22
Erlang

• Exercises
• 1. R1, find Prob(Tgt4Tgt1)
• 2. Compare to Prob(Tgt3), since 4 is 3 more than 1

23
Inference

• Inference is as before
• We observe 3 events in time 2.3
• Find 90 upper confidence bound for the rate
• For large rate, it would be unusual to have so
  few events
• Find rate so Prob(lt3 events) 0.10
• gtgt bisect(_at_(r)(pprob(2.3,r,0,3)),0.1,0,6,.0001)
• 2.9047

24
Inference

• Find 90 lower confidence bound for the rate
• For small rate, it would be unusual to have so
  many events
• Find rate so Prob(gt3 events) 0.10
• gtgt bisect(_at_(r)(pprob(2.3,r,3,99)), 0.1,0,3,.001)
• 0.4792

25
Inference

• Suppose we know that the time for 3 events is 2.3
• 90 upper confidence
• Unusual for time to be long
• Prob(timegt2.3) 0.10
• gtgt bisect(_at_(r)(erlangprob(3,r,2.3,99))
  ,0.1,0,3,.0001)
• 4.0941

26
Inference

• 90 lower confidence
• Unusual for time to be short
• Prob(timelt2.3) 0.10
• gtgt bisect(_at_(r)(erlangprob(3,r,0,2.3)),
  0.1,0,3,.0001)
• 0.4792

27
Inference
28
Inference

• Lower bounds are the same
• Upper bounds are different
• Two observations are not exactly the same
• When we have fixed time, it is not assumed that
  an event occurs at the very end of the interval
• When we are timing events, it is assumed that we
  stop timing at the last event

29
Inference

• Exercises
• 1. Suppose we have 5 events in 2.2 hrs. Find 95
  upper and lower bounds for the rate.
• 2. Repeat 1 if the time for 5 events is 2.2 hrs.
• 3. Repeat if the time for 10 events is 4.4 hrs.

30
Inference

• P-values are as before
• PV Prob(obs or more extreme when H0 true)

31
Inference

• H0 rate2.5/hr vs Ha rategt2.5
• Observe 3 events in ½ hr
• PV pprob(2.5, .5, 3, 99)
• 0.1315
• If the time to the 3rd event is ½ hr
• PVerlangprob(3,2.5, 0,.5)
• 0.1315

32
Inference 2 sided

• Suppose we wish to test H0 ? 3 vs Ha not 3
• In 2 hrs, we have 11 events
• How to define or something more extreme?
• Obviously, gt11 is more extreme

33
Inference 2 sided

• But there should be some small numbers of events
  that are more extreme than 11
• Should we base it on how far they are from the
  mean?
• Better to use values that are less likely

34
Inference 2 sided

• We have been doing that all along
• Events gt11 are less likely than 11 events
• To determine if we should include 0 or 1 or 2 as
  more extreme we need to know how likely they
  are, compared to 11 events

35
Inference 2 sided

• gtgt pprob(2,3,11,11)
• 0.0225
• gtgt pprob(2,3,2,2)
• 0.0446 ------ NOT less likely than 11
• gtgt pprob(2,3,1,1)
• 0.0149 ------ IS less likely than 11
• gtgt pprob(2,3,0,0)
• 0.0025 ------ IS less likely than 11
• So, in addition to 11 or more, we should also
  include 0 and 1

36
Inference 2 sided

• gtgt pprob(2,3,11,99)pprob(2,3,0,1)
• 0.0600

37
Inference 2 sided

• For symmetric distns, this wont be a problem
• Helps to understand what  more extreme  means