Expectation of Discrete Uniform Distribution

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

Then the expectation of $X$ is given by:

$E \left({X}\right) = \dfrac {n+1} 2$


From the definition of expectation:

$\displaystyle E \left({X}\right) = \sum_{x \in \Omega_X} x \Pr \left({X = x}\right)$


\(\displaystyle E \left({X}\right)\) \(=\) \(\displaystyle \sum_{k=1}^n k \left({\frac 1 n}\right)\) $\quad$ Definition of discrete uniform distribution $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 n \sum_{k=1}^n k\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 n \frac {n \left({n+1}\right)} 2\) $\quad$ from Closed Form for Triangular Numbers $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {n+1} 2\) $\quad$ $\quad$