# What is the domain and range of the PMF of the Bernoulli distribution with p = 0.1?

In probability theory and statistics, the Bernoulli distribution

is the discrete probability distribution of a random variable which takes the value 1 with probability p and the value 0 with probability $$q=1-p}$$.

The probability mass function f of this distribution, over possible outcomes k, is

$$f(k;p)={\begin{cases}p&{\text{if }}k=1,\\q=1-p&{\text{if }}k=0.\end{cases}}}$$

let p = 0.1, and then

$$f(k;0.1)={\begin{cases}0.1&{\text{if }}k=1,\\0.9&{\text{if }}k=0.\end{cases}}$$

So, the domain of this particular function f(k;0.1), its domain is the set {0,1} with 2 elements, and its range is the set {0.9,0.1} with 2 elements.

In this specific case, is my understanding right?

Domain is the set of values that you can literally evaluate the function. Given the definition, you can't evaluate the function with $$k$$ outside the set $$\{0,1\}$$, which means it is indeed the domain of it. Note that wikipedia doesn't explicitly comment on the domain, but the support. However, if a discrete function is written in piecewise form like this and you don't see an else statement, it typically means you can't evaluate values other than listed in the definition.
Range is the set of values you can obtain when you evaluate the function within its domain. And, in this case, it is $$\{p,1-p\}$$ in general. For $$p=0.5$$, the range has only one element. Otherwise, it has two as you noted for $$p=0.1$$.
• So, "can't evaluate values other than listed in the definition" only affects the generalization of this function, but the pmf $f(k;0.1)={\begin{cases}0.1&{\text{if }}k=1,\\0.9&{\text{if }}k=0.\end{cases}}$ here is legitimate, is it? – yaojp Aug 7 at 5:44