Two probability distributions of the discrete random variable X a… - Vidyadip

Question

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

$\text{X}$	$0$	$1$	$2$	$3$
$\text{P}(\text{X})$	$\frac{1}{5}$	$\frac{2}{5}$	$\frac{1}{5}$	$\frac{1}{5}$

$\text{Y}$	$0$	$1$	$2$	$3$
$\text{P}(\text{Y})$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$	$\frac{1}{10}$

Prove that $E(Y^2) = 2E(X)$.

✓

Answer

Since, we have to prove that, $E(Y^2) = 2E(X)$ .......(i)
Taking L.H.S. of equation (i), we have
$\text{E}(\text{Y}^2)=\sum\text{Y}^2\text{P}(\text{Y})$
$=0\cdot\frac{1}{5}+1\cdot\frac{3}{10}+4\cdot\frac{2}{5}+9\cdot\frac{1}{10}$
$=\frac{3}{10}+\frac{8}{5}+\frac{9}{10}$
$=\frac{28}{10}=\frac{14}{5}$
$\Rightarrow\text{E}(\text{Y})^2=\frac{14}{5}\ ......(\text{ii})$
Now taking R.H.S. of equation (i), we get
$\text{E}(\text{X})=\sum\text{XP}(\text{X})$
$=0.\frac{1}{5}+1\cdot\frac{2}{5}+2\cdot\frac{1}{5}+3\cdot\frac{1}{5}=\frac{7}{5}$
$2\text{E}(\text{X})=\frac{14}{5}\ ......(\text{iii})$
Thus, from equations (ii) and (iii), we get
$E(Y^2) = 2E(X)$
Hence proved.

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