The relation between two Random Variables $X a n d Y i s Y = a^{X}$ where X is uniformly distributed between [0, 1]. Then mean of Y will be

(1 - a) l o g_{e} a

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(a - 1) / l o g_{e} a

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a l o g_{e} a

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(a - 1) l o g_{e} a

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Solution

The correct option is B $(a - 1) / l o g_{e} a$

$E [Y] = E [a^{x}] = \int_{0}^{1} a^{x} d x$

$[a^{x} / l o g_{e} a]_{0}^{1}$

$(a - 1) / l o g_{e} a$

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