Question : Find Mean and Variance of uniform probability distribution
\[f(x)\ =\frac{1}{k} \] for k= 1,2,3,….,k
Solution :
Given uniform probability distribution \[f\left(x_i\right)\ =\frac{1}{k} \] for i= 1,2,3,….,k This is a discrete random variable. Hence Mean of a discrete random variable are calculated by the formula \[\mu=\sum_{i=1}^{k}{x_if\left(x\right)}\] \[=1\ast\frac{1}{k}+2\ast\frac{1}{k}+3\ast\frac{1}{k}+\ldots.+k\ast\frac{1}{k}\] \[=\frac{1}{k}+\frac{2}{k}+\frac{3}{k}+\ldots.+1\] \[\mu=\frac{1}{k}\left(1+2+3+\ldots.+k\right)\] \[\mu=\frac{1}{k}\frac{k\left(k+1\right)}{2}\] \[\mu=\frac{k+1}{2}\] and Variance of a discrete random variable are calculated by the formula \[V\left(X\right)=E\left[X^2\right]-E\left[X\right]^2\] \[=\sum_{i=1}^{k}{x_i^2f\left(x_i\right)}-\mu^2\] \[=\left(1\right)^2\frac{1}{k}+\left(2\right)^2\frac{1}{k}+\left(3\right)^2\frac{1}{k}+\ldots.+\left(k\right)^2\frac{1}{k}-\left(\frac{k+1}{2}\right)^2\] \[=\frac{1}{k}\left(\left(1\right)^2+\left(2\right)^2+\left(3\right)^2+\ldots.+\left(k\right)^2\right)-\left(\frac{k+1}{2}\right)^2 \] \[=\frac{1}{6k}k\left(k+1\right)\left(2k+1\right)-\left(\frac{k+1}{2}\right)^2\] \[=\left(k+1\right)\left[\frac{2k+1}{6}-\frac{k+1}{4}\right]\] \[=\frac{\left(k-1\right)\left(k+1\right)}{12}\] \[V\left(X\right)=\frac{k^2-1}{12}\] Therefore Mean and Variance of uniform probability mass function are (k+1)/2 , (k^2-1)/12