Expectation of Discrete Uniform Distribution

Theorem

Let $X$ be a discrete random variable with the discrete uniform distribution with parameter $n$.

Then the expectation of $X$ is given by:

$\expect X = \dfrac {n + 1} 2$

Proof

From the definition of expectation:

$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$

Thus:

\(\ds \expect X\)

\(=\)

\(\ds \sum_{k \mathop = 1}^n k \paren {\frac 1 n}\)

Definition of Discrete Uniform Distribution

\(\ds \)

\(=\)

\(\ds \frac 1 n \sum_{k \mathop = 1}^n k\)

\(\ds \)

\(=\)

\(\ds \frac 1 n \frac {n \paren {n + 1} } 2\)

from Closed Form for Triangular Numbers

\(\ds \)

\(=\)

\(\ds \frac {n + 1} 2\)

$\blacksquare$