# 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