14

It holds $$ \begin{array}{rcl} \operatorname V(x) &= &\dfrac1N\left\| x-\dfrac{e^\top x}{N} e \right\|^2 \\ & = & \dfrac1N\left(x^\top x+\dfrac{(e^\top x)^2 e^\top e}{N^2}-2\dfrac{(e^\top x)^2}N\right) \\ & = & \dfrac{x^\top x}{N} - \dfrac{(e^\top x)^2}{N^2}. \end{array} $$ So you are minimizing the $\ell^2$-norm of an ...


7

Since $\hat q_{N'}(\hat x)\approx\Bbb E[Q(\hat x,\xi)]$ and $\Bbb V[\hat q_{N'}(\hat x)]=\Bbb V[Q(\hat x,\xi)]/N'$, we have \begin{align}\hat\sigma_{N'}^2(\hat x)=\Bbb V[\hat q_{N'}(\hat x)]&=\frac1{N'}\cdot\frac1{N'-1}\sum_{j=1}^{N'}[Q(\hat x,\xi^j)-\Bbb E[Q(\hat x,\xi)]]^2\\&\approx\frac1{N'(N'-1)}\sum_{j=1}^{N'}[Q(\hat x,\xi^j)-\hat q_{N'}(\hat x)]...


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