Answer

1. No, since \(X_{i}\) is also uniform in \([-1,1]\)

2. Yes, \(E[Y_{i}] = 0\) by symmetry. For \(\epsilon > 0\),

\[ \begin{aligned} \lim_{i \to \infty}(P\vert Y_{i} - \mu_{i} \vert > \epsilon) &= \lim_{i \to \infty} P(\vert \frac{X_{i}}{i} - 0 \vert > \epsilon)\newline &= \lim_{i \to \infty} P(\frac{X_{i}}{i} > \epsilon \text{ and } \frac{X_{i}}{i} < -\epsilon)\newline &= \lim_{i \to \infty} [P(X_{i} > i\epsilon) + P(X_{i} < -i\epsilon)] = 0 \end{aligned} \]

3. Yes, \(E[Y_{i}] = 0\) by symmetry. For \(\epsilon > 0\),

\[ \begin{aligned} \lim_{i \to \infty}P(\vert Z_{i} - 0 \vert > \epsilon) &= \lim_{i \to \infty}P((X_{i})^{i} > \epsilon \text{ or } (X_{i})^{i} < -\epsilon)\newline &= \lim_{i \to \infty} [\frac{1}{2}(1 - \epsilon^{1/i}) + \frac{1}{2}(1 - \epsilon^{1/i})]\newline &= \lim_{i \to \infty}(1 - \epsilon^{1/i}) = 0 \end{aligned} \]