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ST2334: SOME NOTES ON THE GEOMETRIC AND NEGATIVE

BINOMIAL DISTRIBUTIONS AND MOMENT GENERATING

FUNCTIONS

Geometric Distribution

Consider a sequence of independent and identical Bernoulli trials with successprobabilityp (0, 1). Define the random variableXas the number of trials until

we see a success, and we include the successful trial (for example, if I flip a coinwhich shows heads (a success) with probability p, then Xis the number of flips toobtain a head, including the successful flip). We know that:

X X ={1, 2, . . . , }

that is,Xis a positive integer. Now, what is the probability that Xtakes the valuex? Well, supposeX= 1, then we must have:

P(X= 1) =p.

This is because, we have only one Bernoulli trial, and it is a success. Suppose, nowX= 2; then:

P(X= 2) = (1 p)p.

This is because we have two Bernoulli trials, and the first is a failure and the seconda success. Similarly

P(X= 3) = (1 p)2p.

Thus, it follows that:

P(X= x) = f(x) = (1 p)x1p x X ={1, 2, . . . , }.

Any random variable with the above PMF is said to have a geometric distributionand we write X Ge(p).

The distribution function, for x X:

F(x) =x

y=1

(1 p)y1p= px

y=1

(1 p)y1 =p1 (1 p)x

p = 1 (1 p)x.

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Calculating the expectation is somewhat tedious:

E

[X] =

x=1 x(1 p)

x1

p

= px=1

d

dp

(1 p)x

= pd

dp

x=1

(1 p)x

= pd

dp

(1 p)

p

= 1

p.

Perhaps an easier way (and this is the case for E[Xq], q 1) is via the MGF:

M(t) = E[eXt ]

=x=1

ext(1 p)x1p

= p

1 p

x=1

[(1 p)et]x

= p

1 p(1 p)et

1

1 (1 p)et

= pet

1 (1 p)et

where we have assumed that (1 p)et

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ST2334 3

solution (that is the Geometric distribution). Let us suppose that r = 2. Then wehave, forx {2, 3, . . . }

P(X=x) = (x 1)(1 p)x2

p2

.The logic is as follows: we have to have two successes and hence x 2 failures,which accounts for the (1 p)x2p2 part, then we know that the last successfultrial is atx, so the first successful trial must lie in one of the first x 1 trials; thisis why we multiply by x 1 (remember the trials are identical). Now suppose thatr= 3, Then we have, for x {3, 4, . . . }

P(X= x) =

x 1

2

(1 p)x3p3.

The logic is as follows: we have to have three successes and hencex 3 failures,which accounts for the (1 p)x3p3 part, then we know that the last successfultrial is atx, so the first and second successful trials must lie in one of the first x 1trials; this is why we multiply by x12 which is the number of ways of picking twoout ofx 1 when the order does not matter. Then, following this reasoning, wehave for any r 1

P(X=x) = f(x) =

x 1

r 1

(1 p)xrpr x X ={r, r+ 1, . . . }.

A random variable with the above PMF is said to have a negative binomial distri-bution with parameters r, p, denoted X Ne(r, p).

The distribution function cannot typically be written down in terms of an an-alytic expression (i.e. without a summation) and computing the expectation fromthe definition is a very tedious and tricky exercise. We focus on calculating themoment generating function:

M(t) = E[eXt ]

=x=r

x 1r 1

(1 p)xrprext

=x=r

x 1

r 1

((1 p)et)xr(pet)r

= pet

1 (1 p)et

r x=r

x 1

r 1

((1 p)et)xr(1 (1 p)et)r

= pet

1 (1 p)et

r

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Moment Generating Functions

We note:

M(t) = ddt

M(t) = ddtE[eXt ] = E[d

dteXt ] = E[XeXt ].

Thus M(0) = E[X]. We are assuming that it is legitimate to swap the order ofsummation and differentiation, which holds for all cases in this course. Using asimilar approach, one can show thatM(2)(0) = E[X2].