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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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Definitely linear combinations of independent normals are normally distributed. Also linear combiations of of jointly normally distributed random variables are normally distributed. Look on Wikipedia under "normal distribution" and "multivariate normal distribution".

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    Maybe I am tired but I fail to see how this is on topic.2011-11-12
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    @Didier $aX$ is a linear combination of $X$, which is independent :-).2011-11-12
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    @Didier Maybe you are tired. I think whuber's comment covers it. Although your comment under the question also does it.2011-11-13
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    Maybe I am tired but I fail to see what my comment and your post have in common. I spot the exact small mistake the OP did. You introduce *families* of normal random variables and the *independence* property in a context where both are off-topic since exactly one random variable is concerned. To answer this question by suggesting to look for *multivariate normal distribution*... this beats me (but maybe I am tired).2011-11-13
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    I was simply pointing out that this is an instance of a well-known phenomenon.2011-11-14
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    I think @MichaelHardy's answer is not off topic, but it may not be the most suitable.2012-08-15
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Answered by did's comment:

You forgot the square around the numerator in the exponential defining $f_X$. Once you will have corrected this typo, everything will go fine.