$X$ 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 $Y$?

  • 1
    $\begingroup$ No, identically distributed variables are not necessarily identical (their distributions are). Use, for independent $X$ and $Y$, $M_{aX+bY}(t)=M_X(at)M_Y(bt)$. $\endgroup$ – David Mitra Feb 23 '12 at 16:44
  • $\begingroup$ This question has been asked and answered several times before on math.SE. See for example this answer $\endgroup$ – Dilip Sarwate Feb 23 '12 at 19:17
  • $\begingroup$ 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.... $\endgroup$ – Dilip Sarwate Feb 23 '12 at 19:32
  • 1
    $\begingroup$ Almost the same as this other question. I recommend this one be closed. $\endgroup$ – Dilip Sarwate Feb 23 '12 at 23:51
  • $\begingroup$ @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. $\endgroup$ – 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 $a^2+b^2 + 2ab$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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