Geometric and Negative Binomial Distributions

Pascal’s Triangle Practice Problems Geometric Distribution Negative Binomial Distribution

Pascal’s Triangle

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Pascal’s Triangle Practice Problems Geometric Distribution Negative Binomial Distribution

Pascal’s Triangle

(nk

)=(

n− 1k − 1

)+(

n− 1k

)(k = 1, . . . , n− 1)

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Pascal’s Triangle Practice Problems Geometric Distribution Negative Binomial Distribution

Chebyshev’s inequality example

Example (#4.33 p. 132)The number of equipment breakdowns in a manufacturing plant is closelymonitored by the supervisor of operations, because it is critical to the productionprocess. The number averages 5 per week, witha a standard deviation of 0.8 perweek.

1 Find an interval that includes at least 90% of the weekly figures for thenumber of breakdowns.

2 The supervisor promises that the number of breakdowns will rarely exceed8 in a one-week period. Is the director safe in making this claim? Why?

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Pascal’s Triangle Practice Problems Geometric Distribution Negative Binomial Distribution

Binomial Example

Example (#4.3 p. 144)Every hospital has backup generators for critical systems should the electricitygo out. Independent but identical backup generators are installed so that theprobability that at least one system will operate correctly when called upon is noless than .99. Let n denote the number of backup generators in a hospital. Howlarge must n be to achieve the specified probability of at least one generatoroperating, if

1 p = .95?2 p = .8?

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Pascal’s Triangle Practice Problems Geometric Distribution Negative Binomial Distribution

Binomial Example

Example (#8 from an old test )A hospital has seven backup generators, and three are required to sustain theoperations of the hospital upon blackout. If the probability a given generatorworks is .8 and each of the backup generators are independent, what is theprobability that at least three backup generators will be working if a blackoutoccurs?

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Pascal’s Triangle Practice Problems Geometric Distribution Negative Binomial Distribution

Geometric Distribution – Number of Failures to FirstSuccess

When flipping a coin, we count the number of tails before the first headsappears.

When setting off fireworks, we count the number of successfully firedfireworks before the first dud appears.

When rolling two dice, we count the number of rolls before the dice sum toseven.

Number of Failures to First Success(or equivalently, number of success to first failure)

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Pascal’s Triangle Practice Problems Geometric Distribution Negative Binomial Distribution

Geometric Distribution built from Bernoulli rv’s

Let Y1, Y2, . . . be iid Bernoulli(p) random variables, i.e.

Yi =

{1, w.p. p0, w.p. 1− p.

Let X be the random variable that counts the number of failures of the Bernoullitrials before the first success.Therefore

if Y1 = 1 then X = 0if Y1 = 0 and Y2 = 1 then X = 1if Y1 = 0, Y2 = 0, and Y3 = 1 then X = 2...

...

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Pascal’s Triangle Practice Problems Geometric Distribution Negative Binomial Distribution

Probability Mass Function of the Geometric Distribution

P(X = k) = p(k)= P(Y1 = 0, Y2 = 0, . . . , Yk = 0, Yk+1 = 1)= P(Y1 = 0)P(Y2 = 0) · · ·P(Yk = 0)P(Yk+1 = 1)= (1− p)(1− p) · · · (1− p)p

= (1− p)kp

= qkp

The geometric decay of the pmf gives rise to the name geometric distribution.

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Pascal’s Triangle Practice Problems Geometric Distribution Negative Binomial Distribution

Summing the pmf

Let’s check that the pmf sums to one.∞∑

k=0

p(k) =∞∑

k=0

(1− p)kp

= p∞∑

k=0

(1− p)k

=p

1− (1− p)= 1

And the cdf is computed as

F(x) = P(X ≤ x) =x∑

k=0

qkp = px∑

k=0

qk

= p(

1− qx+1

1− q

)= p

(1− qx+1

p

)= 1− qx+1

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Pascal’s Triangle Practice Problems Geometric Distribution Negative Binomial Distribution

Mean and Variance of the Geometric Distribution

E(X) =∞∑

k=0

kpqk = p∞∑

k=0

kqk

= p(0 + q + 2q2 + 3q3 + · · ·

)= pq(1 + 2q + 3q2 + · · · )

From the usual geometric series,1

1− q= 1 + q + q2 + q3 + · · ·

take the derivative of both sides with respect to q to get1

(1− q)2 = 1 + 2q + 3q2 + · · ·

Therefore E(X) is computed to be

E(X) = pq(

1(1− q)2

)=

pqp2 =

1− pp

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Pascal’s Triangle Practice Problems Geometric Distribution Negative Binomial Distribution

Memoryless Property of the Geometric Distribution

The geometric distribution is the only discrete distribution that has thememoryless property.(The exponential distribution is the only continuous distribution that has thememoryless property.)If we have observed j failures, then the probability of observing at lease k morefailures is independent of the event that j failures were just witnessed.

P(X ≥ j + k|X ≥ j) = P(X ≥ k)

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Pascal’s Triangle Practice Problems Geometric Distribution Negative Binomial Distribution

Other Parameterizations of the Geometric Distribution

Instead of number of failures until the first success, we may be interested in thetime to the first success. In other words,

if Y1 = 1 then X̃ = 1if Y1 = 0 and Y2 = 1 then X̃ = 2if Y1 = 0, Y2 = 0, and Y3 = 1 then X̃ = 3...

...

In this parametrization, pX̃(k) = pX(k − 1) = pqk−1 for k = 1, 2, . . ..The expected value of X̃ is

E(X̃) =∞∑

k=1

kpqk−1 =1q

∞∑k=1

kpqk =1q

E(X) =1p

And there are other parameterizations like in section 4.5.3 (p.158).

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Pascal’s Triangle Practice Problems Geometric Distribution Negative Binomial Distribution

Coupon Collector’s ProblemSuppose there are n distinct action figures in a set, and suppose a random actionfigure from this set is placed in every Wheaties cereal box. How many boxes ofWheaties are you expected to buy to have a full set of action figures? Let T bethe number of boxes opened when finally one has collected all of the actionfigures. Let Ti denote the number of boxes opened to collect the ith distinctaction figure after already having found i− 1 distinct action figures. Thus

T = T1 + T2 + · · ·+ Tn

and soE(T) = E(T1) + E(T2) + · · ·+ E(Tn)

Notice that Ti follows the geometric distribution (as parameterized in theprevious slide) with p = n−(i−1)

n . ThereforeE(T) = E(T1) + E(T2) + · · ·+ E(Tn)

=n∑

i=1

nn− i + 1

=nn

+n

n− 1+ · · ·+ n

1= n

(11

+12

+ · · ·+ 1n

)︸ ︷︷ ︸

Harmonic number

e.g. when n = 50, E(T) = 225Arthur Berg Geometric and Negative Binomial Distributions 14/ 17

Pascal’s Triangle Practice Problems Geometric Distribution Negative Binomial Distribution

Negative Binomial Distribution

The negative binomial distribution generalizes the geometric distribution to givethe number of failures before the rth success occurs. The pmf is given as

P(X = k) = P( the first (k+r-1) trials contain (r − 1) successes)

× P((k + r)th trial is a success)

=(

k + r − 1r − 1

)pr−1(1− p)k × p

=(

k + r − 1r − 1

)pr(1− p)k

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Pascal’s Triangle Practice Problems Geometric Distribution Negative Binomial Distribution

William Gosset, “Student” (1876-1937)

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Pascal’s Triangle Practice Problems Geometric Distribution Negative Binomial Distribution

Happy Valentine’s Day!

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