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

Prove that $E\left( {{Y^2}} \right) = 2E(X)$.

P(Y) $\frac{1}{5}$ $\frac{3}{{10}}$ $\frac{2}{5}$ $\frac{1}{{10}}$

Since, we have to prove that, $E\left( {{Y^2}} \right) = 2E(X)$

$\therefore$ $E(X) = \Sigma XP(X)$
$= 0 \cdot \frac{1}{5} + 1 \cdot \frac{2}{5} + 2 \cdot \frac{1}{5} + 3 \cdot \frac{1}{5} = \frac{7}{5}$
$\Rightarrow$ $2E(X) = \frac{{14}}{5}$ …….(i)

$E{(Y)^2} = \Sigma {Y^2}P(Y)$

$= 0 \cdot \frac{1}{5} + 1 \cdot \frac{3}{{10}} + 4 \cdot \frac{2}{5} + 9 \cdot \frac{1}{{10}}$

$= \frac{3}{{10}} + \frac{8}{5} + \frac{9}{{10}} = \frac{{28}}{{10}} = \frac{{14}}{5}$

$\Rightarrow$ $E\left( {{Y^2}} \right) = \frac{{14}}{5}$

From Eqs. (i) and (ii),
$E\left( {{Y^2}} \right) = 2E(X)$