Geometric Distribution

A distribution such that

$$P(n) = q^{n-1}p = p(1-p)^{n-1},$$ (1)

where $q\equiv 1-p$ and for $n = 1,$ 2, .... The distribution is normalized since

$$\sum_{n=1}^\infty P(n) = \sum_{n=1}^\infty q^{n-1}p = p\sum_{n=0}^\infty q^n = {p\over 1-q} = {p\over p} = 1.$$ (2)

The Moment-Generating Function is

$$\phi(t)=p[1-(1-p)e^{it}]^{-1},$$ (3)

or

$$M(t) = \left\langle{e^{tn}}\right\rangle{} = \sum_{n=1}^\infty e^{tn} pq^{n-1} =p\sum_{n=0}^\infty e^{t(n+1)}q^n$$

$$= pe^t\sum_{n=0}^\infty (e^tq)^n={pe^t\over 1-e^tq}$$ (4)

$$M'(t) = p\left[{(1-e^tq)e^t-e^t(-e^tq)\over (1-e^tq)^2}\right]$$

$$= {p(e^t-qe^{2t}+qe^{2t})\over (1-e^tq)^2}= {pe^t\over (1-e^tq)^2}$$ (5)

$$M''(t) = p {(1-e^tq)^2e^t-e^t 2(1-e^tq)(-e^tq)\over (1-e^tq)^4}$$

$$= p {(1-2e^tq+e^{2t}q^2)e^t+2qe^{2t}(1-e^tq)\over (1-e^tq)^4}$$

$$= p {e^t-2e^{2t}q+e^{3t}q^2+2qe^{2t}-2q^2e^{3t}\over (1-e^tq)^4}$$

$$= p{e^t-q^2e^{3t}\over (1-e^tq)^4}= {pe^t(1-q^2e^{2t})\over (1-e^tq)^4}$$

$$= {pe^t(1+qe^t)\over(1-e^tq)^3}$$ (6)

$$M'''(t) = {pe^t[1+4e^t(1-p)+e^{2t}(1-p)^2]\over (1-e^t+e^tp)^4}.$$ (7)

Therefore,

$$M'(0) = \mu'_1 =\mu = {p\over (1-q)^2} = {p\over p^2} = {1\over p}$$ (8)

$$M''(0) = \mu'_2 = {p(1+q)\over (1+q)^3} = {p(2-p)\over p^3}= {2-p\over p^2}$$ (9)

$$M'''(0) = \mu'_3 = {(6-6p+p^2)\over p^3}$$ (10)

$$M^{(4)}(0) = \mu'_4 = {(p-2)(-p^2+12p-12)\over p^4},$$ (11)

and

$$\mu_2 \equiv \mu_2'-(\mu_1')^2 = \left({{2\over p^2}-{1\over p}}\right)-{1\over p^2}= {1-p\over p^2}$$

$$= {q\over p^2}$$ (12)

$$\mu_3 \equiv \mu_3'-3\mu_2'\mu_1'+2(\mu_1')^3$$

$$= {6-6p+p^2\over p^3}-3{2-p\over p^2}{1\over p}+2\left({1\over p}\right)^3$$

$$= {6-6p+p^2-3(2-p)+2\over p^3}$$

$$= {(p-1)(p-2)\over p^3}$$ (13)

$$\mu_4 \equiv \mu_4'-4\mu_3'\mu_1'+6\mu_2'(\mu_1')^2-3(\mu_1')^4$$

$$= {(p-2)(-p^2+12p-12)\over p^4}-4 {6-6p+p^2\over p^3}{1\over p}+6 {2-p\over p^2}\left({1\over p}\right)^2-3\left({1\over p}\right)^4$$

$$= {(p-1)(-p^2+9p-9)\over p^4},$$ (14)

so the Mean, Variance, Skewness, and Kurtosis are given by

$$\mu \equiv \mu'_1={1\over p}$$ (15)

$$\sigma^2 = \mu_2={q\over p^2}$$ (16)

$$\gamma_1 = {\mu_3\over \sigma^3} ={(p-1)(p-2)\over p^3} \left({p^2\over 1-p}\right)^{3/2}$$

$$= {(p-1)(p-2)\over (1-p)\sqrt{1-p}} = {2-p\over \sqrt{1-p}} = {2-p\over\sqrt{q}}$$ (17)

$$\gamma_2 = {\mu_4\over\sigma^4}-3= {(p-1)(-p^2+9p-9)\over p^4{(1-p)^2\over p^4}}-3$$

$$= {-9+9p-p^2\over (p-1)}-3$$

$$= {p^2-6p+6\over 1-p}.$$ (18)

In fact, the moments of the distribution are given analytically in terms of the Polylogarithm function,

$$\mu_k'\equiv \sum_{n=1}^\infty P(n)n^k = \sum_{n=1}^\infty p... ...)^{n-1} n^k = {p\mathop{\rm Li}\nolimits _{-k}(1-p)\over 1-p}.$$ (19)

For the case $p=1/2$ (corresponding to the distribution of the number of Coin Tosses needed to win in the Saint Petersburg Paradox) this formula immediately gives

$$\mu'_1 = 2$$ (20)

$${\mu'}_2 = 6$$ (21)

$$\mu'_3 = 26$$ (22)

$$\mu'_4 = 150,$$ (23)

so the Mean, Variance, Skewness, and Kurtosis in this case are

$$\mu = 2$$ (24)

$$\sigma^2 = 2$$ (25)

$$\gamma_1 = {\textstyle{3\over 2}}\sqrt{2}$$ (26)

$$\gamma_2 = {\textstyle{13\over 2}}.$$ (27)

The first Cumulant of the geometric distribution is

$$\kappa_1={1-p\over p},$$ (28)

and subsequent Cumulants are given by the Recurrence Relation

$$\kappa_{r+1}=(1-p){d\kappa_r\over dp}.$$ (29)

See also Saint Petersburg Paradox

References

Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, pp. 531-532, 1987.

Spiegel, M. R. Theory and Problems of Probability and Statistics. New York: McGraw-Hill, p. 118, 1992.