= Solution
For each <random variable>, the <probability mass function> gives
$$
\mathbb E[X_i]=\frac{-1+0+1}{3}=0,
\qquad \mathbb E[X_i^2]=\frac{1+0+1}{3}=\frac23,
\qquad\operatorname{var}(X_i)=\frac23.
$$
Let $S_m=\sum_{i=1}^mX_i$. By <linearity of expectation> and <variance additivity for independent random variables>,
$$
\boxed{\mathbb E[S_m]=0,\qquad\operatorname{var}(S_m)=\frac{2m}{3}.}
$$
The <independence> is needed for the <variance> calculation, but not for the <mean>.
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