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Let $X_1,X_2,X_3,X_4$ be four indipendent random variable with normal distribution of mean 0 and variance 1. The exercise asks me to calculate the moment-generating function of $X_1X_2$. I was able to do it and I found that $$M_{X_1X_2}(z) = \displaystyle\frac{1}{\sqrt{1-z^2}}$$ I believe that this result is correct since I found a similar question here. The exercise goes on and ask to prove that the moment-generating function of $Z:=X_1X_2+X_3X_4$ is $$M_Z(t)= \displaystyle\frac{1}{1+t^2}$$ What I write was: $$M_Z(t)=\Bbb E[e^{tZ}]=\Bbb E[e^{tX_1X_2+tX_3X_4}]=\Bbb E[e^{tX_1X_2}e^{tX_3X_4}] \overset{indip.}{=} \Bbb E[e^{tX_1X_2}]\Bbb E[e^{tX_3X_4}]=\displaystyle \left(\frac{1}{\sqrt{1-t^2}}\right)^2=\frac{1}{1-t^2}$$ Am I missing something or there is an error in the exercise's text?

TEuler27
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    Result is certainly true : https://stats.stackexchange.com/q/71126/119261. – StubbornAtom Jan 27 '19 at 11:45
  • Thank you @StubbornAtom. I wasn't able to find a reference of this – TEuler27 Jan 27 '19 at 12:54
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    @StubbornAtom I have look that answer but it seems that they are calculating the characteristic function instead of the moment-generating function. In fact $M_Z(t)=\phi(-it)$ (where $\phi$ is the characteristic function) seems to confirm that my calculation where good(I have manually found $M_Z(t)$). – TEuler27 Jan 27 '19 at 13:44
  • You are right, sorry. Yes the stated answer is wrong (if it was characteristic function it would have been true). It should be $M_Z(t)=1/(1-t^2)$, as confirmed here. – StubbornAtom Jan 27 '19 at 13:57
  • @StubbornAtom Thank you for your time. If you want to make your comments into an answer I will gladly accept it! – TEuler27 Jan 27 '19 at 13:59
  • I would ask you to post an answer and accept it. – StubbornAtom Feb 21 '20 at 06:53

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