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$X$ and $Y$ are independent normal random variables and have the same moment-generating function and are thus identically distributed. Find the distribution of $Z$ where $Z=aX+bY$.

I have the MGF for $X$ and $Y$ :$M_X(t)=e^{{(t\mu_X)+t^2\sigma_X^2}/2}$ and $M_Y(t)=e^{{(t\mu_Y)+t^2\sigma_Y^2}/2}$.

Since they are identically distributed, can't I just replace $Y$ by $X$ and have $Z=(a+b)X$? That would make calculations a lot simpler.

Edit: Thanks for the pointers guys. So would the distribution for $Z$ then be $f_Z(x)=\frac{1}{\sqrt{2\pi}}(a\sigma_X+b\sigma_Y)e^{-x-(a\mu_X+b\mu_Y)^2/2(a^2\sigma_X^2+b^2\sigma_Y^2)}$?

Edit 2: Am I using the right assumption for my answer to the distribution of $Z$? That is, that I can just linearly combine the distributions for $X$ and $Y$?

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    No, identically distributed variables are not necessarily identical (their distributions are). Use, for independent $X$ and $Y$, $M_{aX+bY}(t)=M_X(at)M_Y(bt)$.2012-02-23
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    This question has been asked and answered several times before on math.SE. See for example [this answer](http://math.stackexchange.com/a/65871/15941)2012-02-23
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    Your distribution for $Z$ looks like you are aiming to write down a normal distribution but there are so many typographical and mathematical errors in what you have that it makes no sense whatsoever. Do you believe, for example, that $\sqrt{\alpha+\beta} = \sqrt{\alpha} + \sqrt{\beta}$? That is what you seem to be using....2012-02-23
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    Almost the same as [this other question](http://math.stackexchange.com/q/112305/15941). I recommend this one be closed.2012-02-23
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    @Dilip: they are indeed very similar. But I am not 100% sure that they are exact duplicates. As my vote would be binding, I'll refrain from casting it now.2012-02-24

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