Moments of a uniformly sampled fixed-size client set

Assume $N>1$ and define $I_i={\mathbf{1}}_{\{i\in \mathcal{S}\}}$ and [given] $p=S\mathrm{/}N$. Counting equally likely subsets gives $\mathbb{E}[I_i]=\left(\genfrac{}{}{0px}{}{N-1}{S-1}\right)\mathrm{/}\left(\genfrac{}{}{0px}{}{N}{S}\right)=p$. For distinct $i,j$, $\mathbb{E}[I_i{I}_j]=S(S-1)\mathrm{/}(N(N-1))$: this is zero for $S=1$, and for $S\ge 2$ it is $\left(\genfrac{}{}{0px}{}{N-2}{S-2}\right)\mathrm{/}\left(\genfrac{}{}{0px}{}{N}{S}\right)$. Thus $\mathbb{E}[(I_i-p{)}^2]=\frac{S(N-S)}{N^2},\mathbb{E}[(I_i-p)(I_j-p)]=-\frac{S(N-S)}{N^2(N-1)}(i\mathrm{\ne }j)\mathrm{.}$ These expectations are finite weighted sums by [F2].

Since ${\sum }_i{I}_i=S$, we have [given] ${\bar{v}}_{\mathcal{S}}-\bar{v}=S^{-1}{\sum }_i(I_i-p)(v_i-\bar{v})$. Also ${\sum }_i(v_i-\bar{v})=0$, so ${\sum }_{i\mathrm{\ne }j}⟨v_i-\bar{v},v_j-\bar{v}⟩=-{\sum }_i\mathrm{\parallel }{v}_i-\bar{v}{\mathrm{\parallel }}_2^2$. Expanding the squared norm [F3] and using step 1.1 yields $\begin{array}{rl}{\displaystyle \mathbb{E}\mathrm{\parallel }{\bar{v}}_{\mathcal{S}}-\bar{v}{\mathrm{\parallel }}_2^2}& {\displaystyle =\frac{1}{S^2}[\frac{S(N-S)}{N^2}\sum _i\mathrm{\parallel }{v}_i-\bar{v}{\mathrm{\parallel }}_2^2-\frac{S(N-S)}{N^2(N-1)}\sum _{i\mathrm{\ne }j}⟨v_i-\bar{v},v_j-\bar{v}⟩]}\\ {\displaystyle }& {\displaystyle =\frac{N-S}{SN(N-1)}\sum _i\mathrm{\parallel }{v}_i-\bar{v}{\mathrm{\parallel }}_2^2\mathrm{.}}\end{array}$ The first-moment identity follows from $\mathbb{E}[I_i]=S\mathrm{/}N$ in step 1.1.

The identity $N^{-1}{\sum }_i\mathrm{\parallel }{v}_i-\bar{v}{\mathrm{\parallel }}_2^2=N^{-1}{\sum }_i\mathrm{\parallel }{v}_i{\mathrm{\parallel }}_2^2-\mathrm{\parallel }\bar{v}{\mathrm{\parallel }}_2^2$ and $N\mathrm{/}(N-1)\le 2$ give the stated upper bound. If $N=1$, the only size-one subset is the full set, so its sample mean equals $\bar{v}$ and its variance is zero.