The discrete random variable $\mathrm{X}$ can take all possible integer values from 1 to $\mathrm{k}$, each with a probability $\frac{1}{\mathrm{k}}$, then its variance is

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{E}(\mathrm{X})=\sum_{\mathrm{i}=1}^{\mathrm{k}} x_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=\frac{1}{\mathrm{k}}+\frac{2}{\mathrm{k}}+\ldots+\frac{\mathrm{k}}{\mathrm{k}}=\frac{\mathrm{k}+1}{2} \ & \mathrm{E}\left(\mathrm{X}^2ight)=\sum_{\mathrm{i}=1}^{\mathrm{k}} x_{\mathrm{i}}^2 \mathrm{p}_{\mathrm{i}}=\frac{1^2+2^2+3^2+\ldots+\mathrm{k}^2}{\mathrm{k}} \ & =\frac{(\mathrm{k}+1)(2 \mathrm{k}+1)}{6} \ & 	herefore \quad 	ext { Variance }=\mathrm{E}\left(\mathrm{X}^2ight)-[\mathrm{E}(\mathrm{X})]^2 \ & =\frac{2 \mathrm{k}^2+3 \mathrm{k}+1}{6}-\frac{\mathrm{k}^2+2 \mathrm{k}+1}{4} \ & \end{aligned}$ $\begin{aligned} & =\frac{4 \mathrm{k}^2+6 \mathrm{k}+2-3 \mathrm{k}^2-6 \mathrm{k}-3}{12} \ & =\frac{\mathrm{k}^2-1}{12}\end{aligned}$

The discrete random variable $\mathrm{X}$ can take all possible integer values from 1 to $\mathrm{k}$, each…

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