In statistics, Stein's unbiased risk estimate (SURE) is an unbiased estimator of the mean-squared error of "a nearly arbitrary, nonlinear biased estimator."^[1] In other words, it provides an indication of the accuracy of a given estimator. This is important since the true mean-squared error of an estimator is a function of the unknown parameter to be estimated, and thus cannot be determined exactly.

The technique is named after its discoverer, Charles Stein.^[2]

Formal statement

Let ${\displaystyle \mu \in {\mathbb {R} }^{d}}$ be an unknown parameter and let ${\displaystyle x\in {\mathbb {R} }^{d}}$ be a measurement vector whose components are independent and distributed normally with mean ${\displaystyle \mu _{i},i=1,...,d,}$ and variance ${\displaystyle \sigma ^{2}}$. Suppose ${\displaystyle h(x)}$ is an estimator of ${\displaystyle \mu }$ from ${\displaystyle x}$, and can be written ${\displaystyle h(x)=x+g(x)}$, where ${\displaystyle g}$ is weakly differentiable. Then, Stein's unbiased risk estimate is given by^[3]

${\displaystyle \operatorname {SURE} (h)=d\sigma ^{2}+\|g(x)\|^{2}+2\sigma ^{2}\sum _{i=1}^{d}{\frac {\partial }{\partial x_{i}}}g_{i}(x)=-d\sigma ^{2}+\|g(x)\|^{2}+2\sigma ^{2}\sum _{i=1}^{d}{\frac {\partial }{\partial x_{i}}}h_{i}(x),}$

where ${\displaystyle g_{i}(x)}$ is the ${\displaystyle i}$th component of the function ${\displaystyle g(x)}$, and ${\displaystyle \|\cdot \|}$ is the Euclidean norm.

The importance of SURE is that it is an unbiased estimate of the mean-squared error (or squared error risk) of ${\displaystyle h(x)}$, i.e.

${\displaystyle \operatorname {E} _{\mu }\{\operatorname {SURE} (h)\}=\operatorname {MSE} (h),\,\!}$

with

${\displaystyle \operatorname {MSE} (h)=\operatorname {E} _{\mu }\|h(x)-\mu \|^{2}.}$

Thus, minimizing SURE can act as a surrogate for minimizing the MSE. Note that there is no dependence on the unknown parameter ${\displaystyle \mu }$ in the expression for SURE above. Thus, it can be manipulated (e.g., to determine optimal estimation settings) without knowledge of ${\displaystyle \mu }$.

Proof

We wish to show that

${\displaystyle \operatorname {E} _{\mu }\|h(x)-\mu \|^{2}=\operatorname {E} _{\mu }\{\operatorname {SURE} (h)\}.}$

We start by expanding the MSE as

${\displaystyle {\begin{aligned}\operatorname {E} _{\mu }\|h(x)-\mu \|^{2}&=\operatorname {E} _{\mu }\|g(x)+x-\mu \|^{2}\\&=\operatorname {E} _{\mu }\|g(x)\|^{2}+\operatorname {E} _{\mu }\|x-\mu \|^{2}+2\operatorname {E} _{\mu }g(x)^{T}(x-\mu )\\&=\operatorname {E} _{\mu }\|g(x)\|^{2}+d\sigma ^{2}+2\operatorname {E} _{\mu }g(x)^{T}(x-\mu ).\end{aligned}}}$

Now we use integration by parts to rewrite the last term:

${\displaystyle {\begin{aligned}\operatorname {E} _{\mu }g(x)^{T}(x-\mu )&=\int _{{\mathbb {R} }^{d}}{\frac {1}{\sqrt {2\pi \sigma ^{2d}}}}\exp \left(-{\frac {\|x-\mu \|^{2}}{2\sigma ^{2}}}\right)\sum _{i=1}^{d}g_{i}(x)(x_{i}-\mu _{i})d^{d}x\\&=\sigma ^{2}\sum _{i=1}^{d}\int _{{\mathbb {R} }^{d}}{\frac {1}{\sqrt {2\pi \sigma ^{2d}}}}\exp \left(-{\frac {\|x-\mu \|^{2}}{2\sigma ^{2}}}\right){\frac {dg_{i}}{dx_{i}}}d^{d}x\\&=\sigma ^{2}\sum _{i=1}^{d}\operatorname {E} _{\mu }{\frac {dg_{i}}{dx_{i}}}.\end{aligned}}}$

Substituting this into the expression for the MSE, we arrive at

${\displaystyle \operatorname {E} _{\mu }\|h(x)-\mu \|^{2}=\operatorname {E} _{\mu }\left(d\sigma ^{2}+\|g(x)\|^{2}+2\sigma ^{2}\sum _{i=1}^{d}{\frac {dg_{i}}{dx_{i}}}\right).}$

Applications

A standard application of SURE is to choose a parametric form for an estimator, and then optimize the values of the parameters to minimize the risk estimate. This technique has been applied in several settings. For example, a variant of the James–Stein estimator can be derived by finding the optimal shrinkage estimator.^[2] The technique has also been used by Donoho and Johnstone to determine the optimal shrinkage factor in a wavelet denoising setting.^[1]

References