If a random variable X takes values $(- 1)^{k} 2^{k} / k; k = 1, 2, 3, . . . .$ with probabilities $P (X = k) = \frac{1}{2^{k}}$ then E(X)=

l o g 2

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l o g e

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l o g (\frac{1}{2})

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l o g (\frac{1}{4})

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Solution

The correct option is B $l o g (\frac{1}{2})$
$E [X] = \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k} 2^{k}}{k} \frac{1}{2^{k}} = \frac{{(- 1)}^{k}}{k} = \frac{- 1}{1} + \frac{1}{2} + \frac{- 1}{3} + \frac{1}{4} + . . . . . . . . = - 1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + . . . . . . . . . = - log 2 = log \frac{1}{2}$

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Similar questions

1, 2, 3

P (X = 1) = K, P (X = 2) = 2 K, P (X = 3) = 3 K,

X

$X$	$0$	$1$	$2$	$3$	$4$
$P (X)$	$0.1$	$k$	$2 k$	$2 k$	$k$

k

P (X \geq 2)

x

p (x = k) = \frac{(k + 1) C}{2^{k}}, k = 0, 1, 2, 3, . . . . . . . . . . . .

C .

X = x : 1, 2, 3

p (X = x) : \frac{1}{k}, \frac{2}{k}, \frac{3}{k}

X

10 k

$X$	$- 2$	$- 1$	$0$	$1$	$2$	$3$
$P (X)$	$0.1$	$k$	$0.2$	$2 k$	$0.3$	$k$

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