Bernoulli distribution

A r.v. $X$ has the Bernoulli distribution with a probability $0<p<1$ if $P(X=1)=p$ and $P(X=0)=1-p$.

• Note that $p$ cannot be exactly $0$ or $1$, so both of these probabilities are positive.
• We say that $X$ has the Bernoulli distribution. The textbook writes it as $X\sim \text{Bern}(p)$.

Bernoulli trial

An experiment represented by an Indicator variable $I_A$ for some event $A$ that we’re testing.

• $I_A=1$ if $A$ occurs, and equals 0 if $A$ doesn’t occur.
• If $0<P(A)<1$, then $I_A\sim \text{Bern}(p)$, where $p=P(A)$.

PMF

Since $X$ can only take on a value of $0$ or $1$, the PMF of $X$ can be given, for $k\in \{0,1\}$, by

$$P(X=k)=\{\begin{array}{ll}p& \text{if }k=1\\ 1-p& \text{if }k=0\end{array}$$

variance

If $X$ is an indicator variable with $P(X=1)=p$, then its Variance $\text{Var}(X)=p(1-p)$.

We can prove this by showing $\mathbb{E}(X^2)=p$, because the support $\{0^2,1^2\}=\{0,1\}$ and so $\mathbb{E}(X)=\mathbb{E}(X^2)$. We have $\text{Var}(X)=\mathbb{E}(X^2)-\mathbb{E}(X{)}^2=p-p^2$.

The largest possible variance of an indicator r.v. is $\text{Var}(X)=0.25$, if $p=0.5$.

MGF

Use LOTUS to calculate the MGF. We have

$$M_X(t)=\mathbb{E}(e^{tX})=e^{0\cdot t}(1-p)+e^{1\cdot t}p=e^{t}p+(1-p)$$

for any $t\in \mathbb{R}$.