제 6 절 음이항분포

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제 6 절 음이항분포 31

제 6 절 음이항분포

• The negative binomial distribution

Based on an experiment satisfying the following conditions:

1. The experiment consists of a sequence of independent trials.

2. Each trial can result in either a success (S) or a failure (F).

3. The probability of success is constant from trial to trial.

P (S on trial i) = p for i = 1, 2, 3, · · · 4. The experiment continues until a total of r successes.

X = the number of failures that precede the rth success negative binomial rv

p(x) =

µx + r − 1 r − 1

¶

p^r(1 − p)^x x = 0, 1, 2, · · ·

E(X) = r(1 − p)

p V (X) = r(1 − p) p²

Yutae Lee —————————– http://hyomin.deu.ac.kr/∼ylee/

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32 제 3 장 이산확률변수와 확률분포

제 7 절 포아송분포

Def X ∼ Poisson(λ) with parameter λ (λ > 0) if the pmf of X is p(x) = e^−λλ^x

x! x = 0, 1, 2 · · ·

• History

http://www.umass.edu/wsp/statistics/lessons/poisson/index.html The Poisson Distribution is named for its discoverer, who first applied it to the deliberations of juries; in that form it did not attract wide attention. More suggestive was Poisson’s application to the science of artillery. The distribution was later and independently discovered by von Bortkiewicz, Rutherford, and Gosset. It was von Bortkiewicz who called it The Law of Small Numbers, but as noted above, though it has a special usefulness at the small end of the range, a Poisson Distribution may also be computed for larger r. The fundamental trait of the Poisson is its asymmetry, and this trait it preserves at any value of λ.

• The Classic Example (Prussian horses)

http://www.umass.edu/wsp/statistics/lessons/poisson/index.html The classic Poisson example is the data set of von Bortkiewicz (1898), for the chance of a Prussian cavalryman being killed by the kick of a horse. Ten army corps were observed over 20 years, giving a total of 200 observations of one corps for a one year period. The period or module of observation is thus one year. The total deaths from horse kicks were 122, and the average number of deaths per year per corps was thus 122/200 = 0.61. This is a rate of less than 1. It is also obvious that it is meaningless to ask how many times per year a cavalryman was not killed by the kick of a horse. In any given year, we expect to Yutae Lee —————————– http://hyomin.deu.ac.kr/∼ylee/

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제 7 절 포아송분포 33 observe, well, not exactly 0.61 deaths in one corps (that is not possible;

deaths occur in modules of 1), but sometimes none, sometimes one, occasionally two, perhaps once in a while three, and (we might intu- itively expect) very rarely any more. Here, then, is the classic Poisson situation: a rare event, whose average rate is small, with observations made over many small intervals of time.

Bortkiewicj (1898)

# of fatalities kicked by horses

for 10 corps (large units of army over 10,000 solders) of Prussian cavalry over 20 years

λ = 0 × 109 + 1 × 65 + 2 × 22 + 3 × 3 + 4 × 1 200

= 0.61 This is how to get average.

X = # of deaths per year per corps ∼ Poisson(0.61)

P (X = 0) = e^−λλ⁰

0! = e^−0.61 ≈ 0.54335

# of deaths

per year Poisson prob. 200×Poisson prob. observed per corps

0 0.54335 108.67 109

1 0.33145 66.29 65

2 0.10110 20.22 22

3 0.02055 4.11 3

4 0.00315 0.63 1

5 0.00040 0.08 0

6 0.00005 0.01 0

200

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34 제 3 장 이산확률변수와 확률분포

• A (very useful) relation between binomial & Poisson distributions Thm X ∼ binomial(n, p)

large n, small p such that np ≈ λ Then the distr. of X =∼ Poisson(λ)

• Justification for Poisson distr.

Example (α-particles)

X ∼ binomial(n, p)

n = 480 large; p small; np ≈ λ = 3.87 X ∼ Poisson(λ) approximately by the Thm

Example (Prussian horses)

X ∼ binomial(n, p)

n large; p small; np ≈ λ = 0.61

X ∼ Poisson(λ) approximately by the Thm

Example (Phone calls) 8p.m. – 12p.m.

X = # of calls ∼ binomial(n, p)

n = 4 × 60 = 240 large; p small; np ≈ λ

∴ X ∼ Poisson(λ)

• Mean & Variance

X ∼ Poisson(λ) E(X) = λ And the good thing is V (X) = λ