解答
1
$$
\begin{aligned}
\mathbb{E}\left(N_k\right)
&= \mathbb{E}\left(\sum_{h=1}^{n_0}Y_k^{(h)}\right) = \mathbb{E}\left(\sum_{h=1}^{n_0}\mathbb{I}\left(X^{(h)} = k\right)\right)\\
&= \sum_{h=1}^{n_0}\mathbb{E}\left(\sum_{h=1}^{n_0}\mathbb{I}\left(X^{(h)} = k\right)\right)\\
&= n_0q_k
\end{aligned}
$$