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

$X$ $0$ $1$ $2$ 3

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

$Y$ $0$ $1$ $2$ $3$

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

E (Y^{2}) = 2 E (X)

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E (Y^{2}) = E (X)

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E (Y) = E (X)

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E (X^{2}) = 2 E (Y)

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Solution

The correct option is C $E (Y^{2}) = 2 E (X)$
$E (X) = \sum X P (X) = 0 * \frac{1}{5} + 1 * \frac{2}{5} + 2 * \frac{1}{5} + 3 * \frac{1}{5} = 7 / 5 E (Y^{2}) = \sum y^{2} P (Y) = 0^{2} * \frac{1}{5} + 1^{2} * \frac{3}{10} + 2^{2} * \frac{4}{10} + 3^{2} * \frac{1}{10} = 28 / 10 = 14 / 5 E (Y^{2}) = 2 E (X)$

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Similar questions

X

$X = x$	$- 2$	$- 1$	$0$	$1$	$2$	$3$
$P (x)$	$\frac{1}{10}$	$K$	$\frac{1}{5}$	$2 K$	$\frac{3}{10}$	$K$

X, Y

P (X) = \frac{1}{2}, P (X | Y) = \frac{2}{3}, P (Y | X) = \frac{5}{6}

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$X = x$	$1$	$2$	$3$	$4$	$5$
$P (X = x)$	$k$	$2 k$	$3 k$	$4 k$	$5 k$

P (X \leq 4)

X

Y

P (X) = \frac{1}{3}, P (X | Y) = \frac{1}{2}

P (Y | X) = \frac{2}{5}

$X = x$	0	1	2	3	4	5	6	7
$P (X = x)$	$0$	$K$	$2 K$	$2 k$	$3 K$	$K^{2}$	$2 K^{2}$	$7 K^{2} + k$

P (0 < X < 5) =

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

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