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# Two probability distributions of the discrete random variable X and Y are given below.

X 0 1 2 3 P(X) $\frac{1}{5}$ $\frac{2}{5}$ $\frac{1}{5}$ $\frac{1}{5}$ Y 0 1 2 3 P(Y) $\frac{1}{5}$ $\frac{3}{10}$ $\frac{2}{5}$ $\frac{1}{10}$ Prove that $E(Y^2)=2E(X)$.

Toolbox:
• $E(x)$=$\sum$p$_i$X$_i$
• $E(y)^2$=$\sum$p$_i$X$_i$$^2$
$E(x)$=$\sum$p$_i$X$_i$=$\Large0\times\;\frac{1}{5}\;+1\times\;\frac{2}{5}\;+2\times\;\frac{1}{5}\;+3\times\;\frac{1}{5}$
=$\Large\frac{3}{10}\;+\;\frac{8}{5}\;+\;\frac{9}{10}\;=\;\frac{28}{10}\;=\;\frac{14}{5}$
$E(y)^2$=$2\;E(x)$ proved

edited Jun 4, 2013