$E[\overline{X}]$ of random samples from discrete uniform population

Question

Would $E[\overline{X}]$ of random samples from discrete uniform population be $\frac{N+1}{2}$ (X goes from 1 to N)?

Answer 1

Yes. If $X_i$ is sampled uniformly from $1,\dots,N$, then \begin{align*} \mathbb{E}[X_i] &= 1 \cdot \frac{1}{N} + 2\cdot \frac{1}{N} + \cdots + N\cdot \frac{1}{N}\\ &=\frac{1}{N}(1+2+\cdots+N)\\ &=\frac{1}{N}\cdot\frac{N(N+1)}{2}\\ &=\frac{N+1}{2}. \end{align*}

So if $X_1,\dots,X_n$ are IID sampled uniformly from $1,\dots,N$ and $\bar{X}=\frac{1}{n}(X_1+\cdots+X_n)$, then \begin{align*} \mathbb{E}[\bar{X}]&=\mathbb{E}\left[\frac{1}{n}(X_1+\cdots+X_n)\right]\\ &=\frac{1}{n}(\mathbb{E}[X_1]+\cdots+\mathbb{E}[X_n])\\ &=\frac{1}{n}\left(n\cdot \frac{N+1}{2}\right)\\ &=\frac{N+1}{2}. \end{align*}