Bernoulli Distribution

A Distribution given by

$$P(n) = \left\{\begin{array}{ll} q\equiv 1-p & \mbox{for $n = 0$}\\ p & \mbox{for $n = 1$}\end{array}\right.$$ (1)

$$= p^n(1-p)^{1-n}\qquad \hbox{for }n = 0, 1.$$ (2)

The distribution of heads and tails in Coin Tossing is a Bernoulli distribution with $p=q=1/2$. The Generating Function of the Bernoulli distribution is

$$M(t)=\left\langle{e^{tn}}\right\rangle{}=\sum_{n=0}^1 e^{tn} p^n(1-p)^{1-n} = e^0(1-p)+e^t p,$$ (3)

so

$$M(t) = (1-p)+pe^t$$ (4)

$$M'(t) = pe^t$$ (5)

$$M''(t) = pe^t$$ (6)

$$M^{(n)}(t) = pe^t,$$ (7)

and the Moments about 0 are

$$\mu_1' = \mu = M'(0)=p$$ (8)

$$\mu_2' = M''(0) = p$$ (9)

$$\mu_n' = M^{(n)}(0) = p.$$ (10)

The Moments about the Mean are

$$\mu_2 = \mu_2'-(\mu_1')^2 = p-p^2=p(1-p)$$ (11)

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

$$= p(1-p)(1-2p)$$ (12)

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

$$= p-4p^2+6p^3-3p^4$$

$$= p(1-p)(3p^2-3p+1).$$ (13)

The Mean, Variance, Skewness, and Kurtosis are then

$$\mu = \mu_1' = p$$ (14)

$$\sigma^2 = \mu_2 = p(1-p)$$ (15)

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

$$= {1-2p\over\sqrt{p(1-p)}}$$ (16)

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

$$= {6p^2-6p+1\over p(1-p)}.$$ (17)

To find an estimator for a population mean,

$$\left\langle{p}\right\rangle{} = \sum_{Np=0}^N p{N\choose Np}\theta^{Np}(1-\theta)^{Nq}$$

$$= \theta \sum_{Np=1}^N {N-1\choose Np-1}\theta^{Np-1}(1-\theta)^{Nq}$$

$$= \theta[\theta+(1-\theta)]^{N-1} = \theta,$$ (18)

so $\left\langle{p}\right\rangle{}$ is an Unbiased Estimator for $\theta$. The probability of $Np$ successes in $N$ trials is then

$${N\choose Np}\theta^{Np}(1-\theta)^{Nq},$$ (19)

where

$$p={\hbox{[number of successes]}\over N} \equiv {n\over N}.$$ (20)

See also Binomial Distribution