Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $a=\pm1$

share|improve this question
4  
You forgot the square around the numerator in the exponential defining $f_X$. Once you will have corrected this typo, everything will go fine. –  Did Nov 12 '11 at 19:48
    
How embarrassing. Thanks :) –  maliky0_o Nov 12 '11 at 19:50
2  
@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 and here. –  Zev Chonoles Dec 13 '11 at 4:05

2 Answers 2

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".

share|improve this answer
1  
Maybe I am tired but I fail to see how this is on topic. –  Did Nov 12 '11 at 20:04
1  
@Didier $aX$ is a linear combination of $X$, which is independent :-). –  whuber Nov 12 '11 at 21:56
    
@Didier Maybe you are tired. I think whuber's comment covers it. Although your comment under the question also does it. –  Michael Hardy Nov 13 '11 at 13:23
    
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). –  Did Nov 13 '11 at 14:20
1  
I think @MichaelHardy's answer is not off topic, but it may not be the most suitable. –  Tunococ Aug 15 '12 at 0:00

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.