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$\begingroup$

I assumed it was true, and then found

$f_Y(y) = f_x(\frac{y}{a}).a^{-1} = a^{-1}\frac{1}{\sqrt{2\pi\sigma^2}}.\exp\{-\frac{(\frac{y}{a}-\mu)}{2\sigma^2}\} = \frac{1}{\sqrt{a^2.2\pi\sigma^2}}.\exp\{\frac{(y-\frac{\mu}{a})}{2a\sigma^2}\} $

which is almost of the form of a normal pdf, but there isn't a consistent value for $\sigma^2_Y$, unless $a=\pm1$

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    You forgot the square around the numerator in the exponential defining $f_X$. Once you will have corrected this typo, everything will go fine.2011-11-12
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    How embarrassing. Thanks :)2011-11-12
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    @maliky0_o: You can write up the final answer to your question and accept it, so that the whole web can get the benefit of the answer. This is explicitly encouraged by the SE network of sites; see [here](http://meta.stackexchange.com/questions/12513/should-i-not-answer-my-own-questions) and [here](http://blog.stackoverflow.com/2011/07/its-ok-to-ask-and-answer-your-own-questions/).2011-12-13

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