Transcript Lecture 19 Poisson Random Variables

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Consider the number of typos on each page
of your textbook…
Or the number of car accidents between exit
168 and exit 178 on IN-65.
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For binomial random variables, we know that
they describe outcomes of binomial
experiments.
Similarly, Poisson random variables describe
outcomes of Poisson experiment
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Properties of Poisson Experiment:
◦ It measures the number of occurrence of an event
over an interval, area or space.
◦ The probability of an occurrence is the same for any
two intervals of equal length/area.
◦ The occurrence or non-occurrence in any
interval/area (area A) is independent of the
occurrence or non-occurrence in any other
interval/area (area B), if there is no overlap between
A and B (or when A and B are disjoint)
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If a discrete random variable describes the outcome
of Poisson experiment, we call this random variable a
Poisson random variable.
X~POI(λ)
P(X=k)=λ^k*e^(-λ) / k!
– This is the probability of having k occurrences for a Poisson
random variable within a given interval/area.
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If X is a Poisson random variable,
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There is only ONE parameter for Poisson distribution,
λ.
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– X must be non-negative integer
– X has no upper limit
λ must be positive but does not have to be integer
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If X~POI(λ), λ is both the mean AND variance
of X. (Easier than binomial?)
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I get 1.5 visit during my office hour on
average. How likely will it be for me to have 3
visits during an office hour? (Suppose the
number of visits follows a Poisson
distribution)
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What is the chance that I will have more than
5 visits in my next office hour?
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On your textbook, there are 22 chapters and
a total of 330 typos. You are reading chapter
one and detected 20 typos, assuming each
chapter of the textbook has the same number
of pages, is that normal? (Suppose the
number of typos follows a Poisson
distribution).
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You continued to read chapter two and also
detected 20 typos. What is the probability
that you get 20 typos on both chapter one
and chapter two?
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A survey was conducted to research the
number of car accidents on inter-state
highway. It has found that there was an
average of 2.5 accidents over each 100 miles
of inter-state highway each year.
a. It is about 100 miles from Lafayette to
Edinburgh on IN-65, what is the probability
that you see no accidents during one trip?
(recall on your own experience.)
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b. The Indiana department of transportation
reported that during the year 2008, 4
accidents happened on IN-65 between
Lafayette and Edinburgh, what is the
probability for that?
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Compare part a and b in example III, what can
we say about using Poisson random variable
to analyze real life events?
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1. Suppose the number of typos in today’s
Exponent follows a Poisson distribution with
mean 18 and the number of typos in
yesterday’s Exponent follows another Poisson
distribution with mean 25, then the total
number of typos in these two days’ Exponent
still follows a Poisson distribution with mean
43 (=18+25)
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Let Xf be the number of times you miss a
class in fall semester and Xf~POI(35), let Xsp
be the number of times you miss a class in
spring semester and Xsp~POI(40), Xsu be the
number of times you miss a class in the
summer Xsu~POI(10),
Let Y be the total number of classes you miss
for a school year, find the distribution of Y
and the corresponding parameter.
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Suppose we have more than one independent
Poisson random variables, say n of them, X1,
X2, …, Xn, each with parameter (λ1, λ2, …, λn
etc). Here, independent means the
corresponding region/interval for each of
those variables are disjoint.
Then the sum of those variables X=X1+
X2+…+Xn, still follows a Poisson distribution
with parameter (λ1+λ2+…+λn).
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Sometimes, Poisson random variable
(POI(λ))is used to approximate Binomial
random variables (BIN(n,p)) when n is large
and p is small.
In this case, we simply set λ=np.
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Two empirical rules exist for using Poisson to
approximate binomial:
1. n ≥ 20 and p ≤ 0.05
Or
2. n ≥ 100 and np ≤ 10
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Suppose there is a weekly lottery which
everyone has a 1/1000 chance of winning.
What is the probability that you win this
lottery five times a year.
A. Using binomial.
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B. Using Poisson approximation.
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Take a fair coin and toss it as many times as
needed until you observe a head.
Let X= number of tosses that is needed.
Sample points={H, TH, TTH, TTTH, …}
Distribution of X
X
P(X)
1
(H)
1/2
2(TH 3(TTH) 4(TTT
)
H)
1/4 1/8
1/16
K(TTT…TH)
?
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X
Think about another example, if we keep
tossing a biased coin with 70% of getting a
tail and 30% of seeing a head, let X=number
of tosses needed to get the first head.
Then the distribution of X is:
1
(H)
P(X 0.3
)
2(TH)
3(TTH)
0.7*0. 0.7*0.7*0.
3
3
4(TTTH)
K(TTT…TH)
0.7*0.7*0.7*
0.3
0.7*0.7*…*0.7*
0.3
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If we repeat an experiment with two
outcomes, success/failure, with probability p
for success and q=1-p for failure, the
number of trials needed to get the first
success follows a Geometric distribution, say,
X~Geo(p).
P(X=k)=(1-p)^(k-1)*p
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Similarities and differences between Binomial
and Geometric random variable.
– Similarities: independent trials of the identical
experiment, probabilities of success/failure
consistent.
– Differences:
• For Binomial, we know how many trials we have in
total.
• For Geometric, we don’t know it, actually that number
is not of interest.
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Someone is trying to take the road test to get
a driver’s license. If the probability of passing
the test is 40%, what is the probability that
this person will pass the test at second shot?
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What is the probability that someone will pass
the road test in 5 trials?
Given that someone has taken the test 4
times and still has not got the license, what is
that person’s chance of passing it the next
time?
 If
X~Geo(p), then
◦ E(X)=1/p
◦ Var(X)=(1-p)/(p^2)
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On average, how many times does one have
to take the test to get the driver’s license?