29  Bernoulli random variable

Figure 29.1: Distribution (left) and density (right) functions of a Bernoulli random variable.

29.1 Distribution and density

Definition 29.1 To construct a continuous distribution for a Bernoulli random variable, one can use the linear combination of two heavy side functions (Equation 1.10), i.e.  $$\begin{array}{}\text{(29.1)}& F_B(x)=H(x)(1-p)+H(x-1)p\text{.}\end{array}$$ Given that the distribution function is specified in terms of Heavyside function, it is possible to take the derivative with respect to $x$ to obtain the density function, i.e.  $$\begin{array}{}\text{(29.2)}& f_B(x)=\delta (x)(1-p)+\delta (x-1)p\text{,}\end{array}$$ where $\delta (x)$ is the Dirac function (Equation 1.11).

29.2 Characteristic function

Proposition 29.1 The characteristic function of a Bernoulli random variable reads explicitly: $$\begin{array}{}\text{(29.3)}& {\varphi }_B(t)=p{e}^{it}+(1+p)\text{,}\end{array}$$ and similarly the moment generating function $$\begin{array}{}\text{(29.4)}& M_B(t)=p{e}^t+(1+p)\text{.}\end{array}$$

Proof: Proposition 29.1

Proof. By the definition of characteristic function, one have to compute the following expectation, i.e.  $$\begin{array}{rl}{\varphi }_B(t)& =\mathbb{E}\{e^{itX}\}=\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{ity}[p\delta (y-1)+(1-p)\delta (y)]dy=\\ & =p{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{ity}\delta (y-1)dy+{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{ity}\delta (y)dy-p{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{ity}\delta (y)dy=\\ & =p{e}^{it\cdot 1}+e^{it\cdot 0}-p{e}^{it\cdot 0}=\\ & =p{e}^{it}+(1-p)\end{array}$$ where, thanks to the fundamental property (Equation 1.12) of the Dirac delta the final expression simplifies.

29.3 Moments

Proposition 29.2 The $n$-th moment of a Bernoulli random variable with probability $p$ reads $$\mathbb{E}\{B^n\}=\mathbb{P}(B=1)=p\text{,}$$ for all $n\in \mathbb{N}$.

Proof: Proposition 29.2

Proof. Let’s considering the characteristic function of the Bernoulli (Equation 29.3) and take the derivative with respect to $t$, i.e.  $${\partial }_t{\varphi }_B(t)=ip{e}^{it}⟹\mathbb{E}\{B\}=\frac{{\partial }_t{\varphi }_B(t)}{i}=p{e}^{it}\text{.}$$ Hence, the above expression evaluated in $t=0$ gives $$\mathbb{E}\{B\}=\frac{{\partial }_t{\varphi }_B(t)}{i^n}{|}_{t=0}=p\text{.}$$ Thanks to the exponential structure the $n$-th derivative reads $${\partial }_t^n{\varphi }_B(t)=i^{n}p{e}^{it}⟹\mathbb{E}\{B^n\}=\frac{{\partial }_t^n{\varphi }_B(t)}{i^n}=p{e}^{it}\text{.}$$ Therefore, all the $n$ moments are exactly equal to $p$. The same result is obtained with the moment generating function (Equation 29.4).