= Solution
{c}
It suffices that the <moment-generating function> be finite on some open <interval> containing zero:
$$
\boxed{M_Y(t)<\infty\text{ for every }|t|<\delta\text{, for some }\delta>0.}
$$
Equivalently, some two-sided <exponential moment> exists: $\mathbb E e^{r|Y|}<\infty$ for some $r>0$. For the forward implication, choose $0<r<\delta$ and use $e^{r|Y|}\leq e^{rY}+e^{-rY}$; the converse follows from $e^{tY}\leq e^{r|Y|}$ for $|t|\leq r$.
Under this weaker hypothesis, all <moments> exist. For $|t|<r$, <dominated convergence> justifies the <Taylor series> argument in (i). Vanishing odd <moments> gives $M_Y(t)=M_{-Y}(t)$ near zero, and the local <uniqueness theorem for moment-generating functions> gives the <symmetric distribution>. The reverse implication follows directly from equality of the two <probability distributions>.
More generally, it suffices that all <moments> exist and the <probability distribution> have <moment determinacy>: $Y$ and $-Y$ then have identical <moments> and hence identical <probability distributions>. Existence of all <moments> by itself does not provide that uniqueness.
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