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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 $Y$, $M_{aX+bY}(t)=M_X(at)M_Y(bt)$. – David Mitra Feb 23 '12 at 16:44
This question has been asked and answered several times before on math.SE. See for example this answer – Dilip Sarwate Feb 23 '12 at 19:17
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.... – Dilip Sarwate Feb 23 '12 at 19:32
Almost the same as this other question. I recommend this one be closed. – Dilip Sarwate Feb 23 '12 at 23:51
@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. – Willie Wong Feb 24 '12 at 8:58

Because in the expression $aX+bY$, the two random variables $X$ and $Y$ are independent, whereas in $aX+bX$, the two random variables $X$ and $X$ are as far from independent as any two random variables can get.

Notice that $$\operatorname{var}(aX+bY) = a^2 \operatorname{var}(X) + b^2 \operatorname{var}(Y), $$ but $$ \operatorname{var}(aX+bX) = (a+b)^2 \operatorname{var}(X), $$ and $a^2 + b^2$ differs from $(a+b)^2$, which is the same as $a^2+b^2 + 2ab$.

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