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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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    @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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    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.