Variance of a uniform distribution
Task number: 3550
Let \(X\) be a random variable that takes on the values \(1{,}2,…,n\) with a uniform probability distribution. Determine the variance of this random variable.
Hint
Use the equation \(var(X)=E[(X-EX)^2]=E[X^2]-(EX^2)\).
Also note that \(\sum\limits_{i=1}^n i^2 =\frac{n(n+1)(2n+1)}6\).
Solution
\( EX=\sum\limits_{i=1}^n i\cdot \frac{1}{n}= \frac{n+1}2\).
\( EX^2=\sum\limits_{i=1}^n i^2\cdot \frac{1}{n}= \frac{(n+1)(2n+1)}6\).
And so \( var(X)=\left(\frac{n+1}2\right)^2-\frac{(n+1)(2n+1)}6=\frac{(n+1)(n-1)}{12}.\)
Answer
The variance of the random variable \(X\) is \(\frac{(n+1)(n-1)}{12}\).
