Was anyone able to explain why $$E(U_2) = 0$$

I don't quite understand what the relevance of the underlined statement - "by the symmetry of $U_1$" in determining $E(U_2)$ is enter image description here

edit: I get it now, was brain dead on the night of asking this question. thanks for the responses anyway

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    $\begingroup$ Could you please state the book from which this exercise is taken? Thanks. $\endgroup$ – COOLSerdash Nov 8 '14 at 18:30
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    $\begingroup$ The textbook is making a lot of fuss over nothing. Since $U_1$ and $U_2$ are independent, $E[U_2\mid U_1] = E[U_2]$ without the necessity of dragging in covariances (which might not even exist). The only issue here is why $E[U_2] = 0$ and the reason given for this, while perfectly correct in this simple situation, need not apply in general: symmetry of the density need not mean that $E[U] = 0$ as the well-known example of Cauchy random variables shows. $\endgroup$ – Dilip Sarwate Nov 8 '14 at 22:52
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    $\begingroup$ @whuber $U_2$ and $-U_2$ have the same distribution? $\endgroup$ – elbarto Nov 9 '14 at 2:29
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    $\begingroup$ The statement "... $E[U_2]=0$ by symmetry of $f_{U_1}$." has the obvious typo that it should have said "$f_{U_2}$." $\endgroup$ – Dilip Sarwate Nov 9 '14 at 5:04
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    $\begingroup$ FWIW, the duplicate was found as the first relevant hit in a site search for distribution symmetry expectation. The next hit I could find is stats.stackexchange.com/questions/46843, which directly addresses a generalization of the present question (to all odd moments). (Most of the other hits do not discuss symmetry per se, but only invoke it without further comment.) $\endgroup$ – whuber Nov 9 '14 at 17:38