Efron-Stein theorem

[[concept]]

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Efron-Stein Theorem

Let $x_1,\dots ,x_N$ be independent random variables and $f$ any function of them. Then $\text{Var}(f)\le {\sum }_{i=1}^N{\mathbb{E}}_{x_1,\dots ,x_{i-1},x_{i+1},\dots x_N}\left[{\text{Var}}_{x_i}(f(x_1,\dots ,x_{i-1},x_i,x_{i+1},\dots ,x_N))\right]$

Remark

The only assumption that we make in the above is that $f:\underset{x_1\in }{\underbrace{{\Sigma }_1}}\times \cdots \times \underset{x_N\in }{\underbrace{{\Sigma }_N}}\to \mathbb{R}$ And the $x_i$ are independent.

NOTE

We generally want something to avoid taking these expectations

Proof

Let’s denote ${\mathbb{E}}_i:={\mathbb{E}}_{x_i}$ and ${\mathbb{E}}_{\le i}:={\mathbb{E}}_{x_1,\dots ,x_i}$

\text{Var}(f) &= \mathbb{E}\left[f-\mathbb{E}\left[f\right]^2 \right] \\ &= \mathbb{E}\left[ (f-\cancel{ \mathbb{E}_{\leq 1}\left[f\right] } ) + \underbrace{ (\cancel{ \mathbb{E}_{\leq 1}\left[f\right] } - \cancel{ \mathbb{E}_{\leq 2}\left[f\right] } ) }_{ \mathbb{E}_{y_{1}}\left[f(y_{1},x_{2},\dots,x_{N})\right] -\mathbb{E}_{y_{1},y_{2}}\left[y_{1},y_{2},x_{3},\dots,x_{N}\right] }+\cancel{ \dots }+(\cancel{ \mathbb{E}_{\leq N-1}\left[f\right] } -\mathbb{E}_{\leq N}\left[f\right] )^2\right] \\ & \end{align}$$ Expanding for the cross terms, if $i \leq j$, we get $$\begin{align} \mathbb{E}_{x_{1},\dots,x_{N}}&\left[\,(\mathbb{E}_{y_{1},\dots,y_{i}}\left[y_{1},\dots,y_{i},x_{i+1},\dots,x_{j},\dots,x_{N}\right]) \times(\mathbb{E}_{z_{1},\dots,z_{j}}\left[z_{1},\dots,z_{j},x_{j+1},\dots,x_{N}\right] ) \,\right] \\ &=\mathbb{E}\left[\,\mathbb{E}_{\leq j}\left[f\right] ^2\,\right] \end{align}$$ Generally, $\mathbb{E}\left[\,\mathbb{E}_{\leq i}\left[f\right]\,\,\mathbb{E}_{\leq j}\left[f\right]\,\right]=\mathbb{E}_{}\left[(\mathbb{E}_{\leq\max\{ i,j \}}\left[f\right])^2\right]$ $$\begin{align} &\mathbb{E}_{}\left[(\mathbb{E}_{\leq i}\left[f\right] -\mathbb{E}_{\leq i+1}\left[f\right] )\times(\mathbb{E}_{\leq j}\left[f\right] -\mathbb{E}_{\leq j+1}\left[f\right] )\right] \\ &= \mathbb{E}_{}\left[(\mathbb{E}_{\leq j}\left[f\right]) ^2\right] - \mathbb{E}_{}\left[(\mathbb{E}_{\leq j}\left[f\right])^2 \right] +\mathbb{E}_{}\left[(\mathbb{E}_{\leq j+1}\left[f\right] )^2\right] -\mathbb{E}_{}\left[(\mathbb{E}_{\leq j+1}\left[f\right] )^2\right] \\ &=0 \end{align}$$ By [[Jensen's inequality]]. So we get $$\begin{align} \text{Var}(f) &= \sum_{i=1}^N \mathbb{E}_{}\left[(\mathbb{E}_{\leq i-1}\left[f\right] -\mathbb{E}_{\leq i}\left[f\right] )^2\right] \\ &= \sum_{i=1}^N \mathbb{E}_{}\left[\mathbb{E}_{y_{1},\dots,y_{i-1}}\left[f(y_{1},\dots,y_{i-1},x_{i},\dots,x_{N})\right] -\mathbb{E}_{y_{i}}\left[f(y_{1},\dots,y_{i-1}, y_{i}, x_{i+1},\dots,x_{N})\right] \right] \\ (*)&\leq \sum_{i=1}^N \underbrace{ \mathbb{E}_{}\left[\mathbb{E}_{x_{i}}\left[f(x_{1},\dots,x_{N})\right] -\mathbb{E}_{y_{i}}\left[f(x_{1},\dots,x_{i-1}, y_{i}, x_{i+1},\dots x_{N})\right] \right] }_{ \text{Var}_{x_{i}}f(x) } \\ &= \sum_{i=1}^N \text{Var}_{x_{i}}(f(x)) \end{align}$$ Where we apply the inequality at $(*)$, and note that the $y_{i}$ are iid copies of the $x_i$ (and therefore can be interchanged freely, as convenient) $$\tag*{$\blacksquare$}$$

References

References

See Also

Mentions

Mentions

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