The p.m.f of random variate $\mathrm{X}$ is $\mathrm{P}(\mathrm{X})= \begin{cases}\frac{2 x}{\mathrm{n}(\mathrm{n}+1)}, & x=1,2,3, \ldots \ldots, \mathrm{n} \ 0, & ext { otherwise }\end{cases}$ Then $E(X)=$

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The p.m.f of random variate $\mathrm{X}$ is $\mathrm{P}(\mathrm{X})= \begin{cases}\frac{2 x}{\mathrm{n}(\mathrm{n}+1)}, & x=1,2,3, \ldots \ldots, \mathrm{n} \ 0, & 	ext { otherwise }\end{cases}$ Then $E(X)=$

Aicharya · Accepted Answer

$\begin{aligned} \mathrm{P}(\mathrm{X}) & =\left\{\begin{array}{l}\frac{2 x}{\mathrm{n}(\mathrm{n}+1)}, x=1,2,3, \ldots, \mathrm{n} \ 0, 	ext { othewise }\end{array}ight. \ 	herefore \quad \mathrm{E}(\mathrm{X}) & =\sum_{\mathrm{i}=1}^{\mathrm{n}} x_{\mathrm{i}} \mathrm{p}\left(x_{\mathrm{i}}ight) \ & =\frac{2}{\mathrm{n}(\mathrm{n}+1)}+\frac{8}{\mathrm{n}(\mathrm{n}+1)}+\ldots+\frac{2 \mathrm{n}^2}{\mathrm{n}(\mathrm{n}+1)} \ & =\frac{2\left(1^2+2^2+\ldots+\mathrm{n}^2ight)}{\mathrm{n}(\mathrm{n}+1)} \ & =\frac{2 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6 \mathrm{n}(\mathrm{n}+1)} \ & =\frac{2 \mathrm{n}+1}{3}\end{aligned}$

The p.m.f of random variate $\mathrm{X}$ is $\mathrm{P}(\mathrm{X})= \begin{cases}\frac{2…

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