Bernoulli Distribution

probability-theory

Definition

Bernoulli Distribution

The Bernoulli distribution models any setting with two possible outcomes, commonly denoted as success vs. failure, parameterised with success probability $p\in [0,1]$.

$$X\sim \mathrm{Bernoulli}(p)$$

Probability Mass Function

The probability mass function $f:\mathfrak{S}\to [0,1]$ of a success event is given by:

$$f(x)=p^x\cdot (1-p{)}^{1-x}$$

for $x\in \{0,1\}$ where $x=1$ denotes a success and $x=0$ a failure.

Expected Value

The expectation of $X\sim \mathrm{Bernoulli}(p)$ is given by:

$$\mathbb{E}[X]=\sum _{x=0}^{1}x\cdot f(x)=0\cdot f(0)+1\cdot f(1)=0+p^1(1-p{)}^0=p$$

Variance

The variance of $X\sim \mathrm{Bernoulli}(p)$ is given by:

$$\begin{array}{rl}\mathbb{V}ar(X)& =\mathbb{E}[(X-\mathbb{E}[X]{)}^2]\\ & =\sum _{x=0}^1(x-\mathbb{E}[X]{)}^{2}f(x)\\ & =p^{2}f(0)+(1-p{)}^{2}f(1)\\ & =p^2(1-p)+(1-p{)}^{2}p\\ & =p(1-p)\end{array}$$

Cumulative Distribution Function

The cdf of $X\sim \mathrm{Bernoulli}(p)$ is given by:

$$F_X(x)=P(X\le x)=\{\begin{array}{ll}0& x<0\\ f(0)=1-p& 0\le x<1\\ f(0)+f(1)=1& 1\le x\end{array}$$

Example: Let $y=g(x)=2x-1\in \{-1,+1\}$ for $x\in \{0,1\}$. What’s the cdf $F_Y$ of $Y=g(X)$?

$$\begin{array}{rl}{F}_Y(y)& =P(Y\le y)\\ & =P(2X-1\le y)\\ & =P\left(X\le \frac{y+1}{2}\right)\\ & =F_X\left(\frac{y+1}{2}\right)\\ & =\{\begin{array}{ll}0& y<-1\\ 1-p& -1\le y<1\\ 1& 1\le y\end{array}\end{array}$$

Example: Let $y\in \{-1,+1\}$. We get:

$$\begin{array}{rl}{F}_Y(y)& =P(Y=y)\\ & =P(2X-1=y)\\ & =P\left(X=\frac{y+1}{2}\right)\\ & =f_X\left(\frac{y+1}{2}\right)\\ & =p^{(y+1)/2}(1-p{)}^{1-(y+1)/2}\end{array}$$