FORMULA AND DERIVATION
Probability distribution:-
As stated above, a Bernoulli random variable takes the value 1 in case of a success, with probability p, and takes the value 0 in case of a failure, with probability \(q=1-p\ \). This can be written as follows:-
\(P\left(X=1\right)=p\ \) and \(\ P\left(X=0\right)=q\)
Therefore, the probability function can be written as follows:-
\[P\left(X=x\right)=p^xq^{\left(1-x\right)}\ for\ x=0,1\]
Mean:-
The mean or expected value for the Bernoulli distribution can be calculated from first principles as follows:-
\[E\left(X\right)=\mu =\sum{xP\left(X=x\right)}\]
\[\mu =\left(0\times q\right)+\left(1\times p\right)\]
\[\mu =p\]
The mean of the Bernoulli distribution is simply the probability of success, p.
Variance:-
Variance of the Bernoulli distribution can be derived from first principles using the formula:
\(Var\left(X\right)=E\left[{\left(x-\mu \right)}^2\right]=\sum{{\left(x-\mu \right)}^2P\left(X=x\right)}\)
or, using a simpler formula:
\(Var\left(X\right)=E\left(X^2\right)-E^2\left(X\right)\)
\(E\left(X^2\right)\)can be calculated as follows:-
\[E\left(X^2\right)=\sum{x^2}P\left(X=x\right)\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ =\left(0^2\times q\right)+\left(1^2\times p\right)\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ =p\]
Substituting this into the formula for variance, we get:-
\[Var\left(X\right)={\sigma }^2=E\left(X^2\right)-E^2\left(X\right)\]
\[{\sigma }^2=p-p^2\]
\[\ \ \ =p\left(1-p\right)\]
\[\ \ \ =pq\]
Probability Generating Function(PGF):-
The PGF for a Bernoulli distribution is
\[G_X\left(t\right)=pt+q\]
Moment Generating Function(MGF):-
The MGF for a Bernoulli distribution is
\[M_X\left(t\right)={pe}^t+q\]
Cumulant Generating Function(CGF):-
The CGF for a uniform distribution is
\[C_X\left(t\right)={\mathrm{ln} \left({pe}^t+q\right)\ }\]
EXAMPLES
Example 1
There are four cards marked A, B, C and D. A gambler makes a bet that on drawing one of the four cards, he would get the card marked D.
In this case, getting the card marked D would be considered a success. The probability of success is nothing but the probability of getting card D, which is ¼.
Therefore,
\[p={1}/{4}=0.25\]
\[q=1-{1}/{4}={3}/{4}=0.75\]
The probability distribution can be written as follows:-
\(x\) |
\(P(X=x)\) |
\(0\) |
\(\frac{3}{4}=0.75\) |
\(1\) |
\(\frac{1}{4}=0.25\) |
Mean:-
\[E\left(X\right)=\mu =p=0.25\]
Variance:-
\[{\sigma }^2=pq=0.25\times 0.75=0.1875\]