## 3.5 Properties of pmfs and cdfs

Let’s define the **support** of a random variable \(X\) — this is the set of all possible values that \(X\) can possibly take. We’ll use the notation \(\mathcal{X}\) to denote the support of \(X\).

**Example 3.5 **Suppose \(X\) is the outcome from a roll of a die. Then, the support of \(X\) is given by \(\mathcal{X} = \{1,2,3,4,5,6\}\). In other words, the only possible values for \(X\) are from \(1,\ldots,6\).

**Example 3.6 **Suppose \(X\) is the number of years of education that a person has. The support of \(X\) is given by \(\mathcal{X} = \{0, 1, 2, \ldots, 20\}\). Perhaps I should have chosen a larger number than 20 to be the maximum possible value that \(X\) could take, but you will get the idea — a person’s years of education can be 0 or 1 or 2 or up to some maximum value.

**Properties of pmfs**

For any \(x\), \(0 \leq f_X(x) \leq 1\)

In words: the probability of \(X\) taking some particular value can’t be less than 0 or greater than 1 (neither of those would make any sense)

\(\sum_{x \in \mathcal{X}} f_X(x) = 1\)

In words: if you add up \(\mathrm{P}(X=x)\) across all possible values that \(X\) could take, they sum to 1.

**Properties of cdfs for discrete random variables**

For any \(x\), \(0 \leq F_X(x) \leq 1\)

In words: the probability that \(X\) is less than or equal to some particular value \(x\) has to be between 0 and 1.

If \(x_1 < x_2\), then \(F_X(x_1) \leq F_X(x_2)\)

In words: the cdf is increasing in \(x\) (e.g., it will always be the case that \(\mathrm{P}(X \leq 3) \leq \mathrm{P}(X \leq 4)\)).

\(F_X(-\infty)=0\) and \(F_X(\infty)=1\)

In words: if you choose small enough values of \(x\), the probability that \(X\) will be less than that is 0; similar (but opposite) logic applies for big values of \(x\).

**Connection between pmfs and cdfs**

\(F_X(x) = \displaystyle \sum_{z \in \mathcal{X} \\ z \leq x} f_X(z)\)

In words: you can “recover” the cdf from the pmf by adding up the pmf across all possible values that the random variable could take that are less than or equal to \(x\). This will be clearer with an example:

**Example 3.7 **Suppose that \(X\) is the outcome of a roll of a die. Earlier we showed that \(F_X(3) = 1/2\). We can calculate this by

\[ \begin{aligned} F_X(3) &= \sum_{\substack{z \in \mathcal{X} \\ z \leq 3}} f_X(z) \\ &= \sum_{z=1}^3 f_X(z) \\ &= f_X(1) + f_X(2) + f_X(3) \\ &= \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \\ &= \frac{1}{2} \end{aligned} \]