# Leave-one-out cross validation: Relatively unbiased estimate of generalization performance?

I have read that leave-one-out cross-validation provides a relatively “unbiased estimate of the true generalization performance” (e.g. here) and that this is an advantageous property of the leave-one-out CV.

However, I don't see how this follows from the properties of leave-one-out CV. Why is the bias of this estimator low when compared to others?

### Update:

I keep investigating the topic, and I believe it has to do with the fact that this estimator is less pessimistic than, say, K-fold validation, since it uses all the data but one instance, but it would be great to read a mathematical derivation of this.

• Hi Amelio - I have seen several mathematical derivations of this in the papers cited in this answer stats.stackexchange.com/questions/280665/… - is there one in particular you want to see ? Commented Jul 24, 2018 at 12:07

I don't think there is a need for a mathematical derivation of the fact that in ML, with increasing training test size, the prediction error rates decrease. LOO -- compared to k-fold validation -- maximizes the training set size, as you have observed.

However, LOO can be sensitive to "twinning" -- when you have highly correlated samples, with LOO you have the guarantee that for each sample used as a test set, the remaining "twins" will be in the training set. This can be diagnosed by a rapid decrease in accuracy when LOO is replaced by, say, 10-fold crossvalidation (or a stratified validation, if for example the samples are paired). In my experience, this can lead to a disaster if generally your data set is small.

In a perfect world, you have also a validation set that you never use to train your model, not even in a CV setting. You keep it for the sole purpose of testing the final performance of a model before you send of the paper :-)

I am new to this topic but I think I can give you a concrete and elementary example of how least square cross-validation with the leave-one-out method produces an "unbiased" estimate of integrated mean square error ($$\mathrm{IMSE}$$ for short) in the context of kernel density estimation.

Recall a typical univariate kernel density estimator $$\widehat{f}$$: $$\widehat{f}(x)=\frac{1}{n h} \sum_{i=1}^n K\left(\frac{X_i-x}{h}\right),$$ where we have imposed the following assumptions:

• A.1 $$X_1, \cdots, X_n \stackrel{i . i . d .}{\sim} f$$.
• A.2 $$f^{\prime\prime}(x)$$ is continuous and bounded in the neighborhood of $$x$$.
• A.3 The kernel function $$K(\cdot)$$ is a symmetric pdf which maximized at $$0$$ and satisfies: (i)\begin{aligned}\int K(u) d u=1\end{aligned},\quad (ii)\begin{aligned}\nu_2:=\int K^2(u) d u<\infty\end{aligned},\quad (iii)\begin{aligned}\kappa_2:=\int u^2 K(u) d u \in(0, \infty)\end{aligned}
• A.4 $$h \rightarrow 0$$, $$n h \rightarrow \infty$$ as $$n \rightarrow \infty$$, which means $$n^{-1}=o\left((n h)^{-1}\right)$$.

Under the above assumptions, the integrated mean square error can be computed as: \begin{aligned} \mathrm{IMSE}\left(\widehat{f}\right) =& \int \operatorname{MSE} \widehat{f}(x) d x\\ =& \frac{1}{n h}\int K(u)^2 d u+\frac{h^4}{4}\left(\int u^2 K(u) d u\right)^2 \int\left[f^{\prime\prime}(x)\right]^2 d x+o\left(h^4+\frac{1}{n h}\right), \end{aligned} which can be easily derived from the bias rendered by KDE (the derivation can be referred to in this question).

Accompanied with this knowledge, let us see how least square cross-validation with the leave-one-out method produces an "unbiased" estimate of $$\mathrm{IMSE}\left(\widehat{f}\right)$$.

Let us define the objective function of least square cross-validation as follows: \begin{aligned} LSCV&=\int[\widehat{f}(x)-f(x)]^2 d x \\ &=\underbrace{\int[\widehat{f}(x)]^2 d x}_{=:\mathcal{I}_{1n}}-2 \underbrace{\int \widehat{f}(x) f(x) d x}_{=:\mathcal{I}_{2n}}+\int f^2(x) d x \\ &=\mathcal{I}_{1n} - 2\mathcal{I}_{2n}+\int f^2(x) d x\\ &={\mathcal{I}}_{1n} - 2\widehat{\mathcal{I}}_{2n}+\int f^2(x) d x + o_{\mathbb{P}}(1),\\ \end{aligned} where $$\widehat{\mathcal{I}}_{2n}$$ denotes the estimand of $${\mathcal{I}}_{2n}$$.

The leave-one-out method is applied to the computation of $$\widehat{\mathcal{I}}_{2n}$$. To ease of exposition, let me define two versions of $$\mathcal{I}_{2n}$$: \begin{aligned} \widehat{\mathcal{I}_{2n}}^{\text{LOO}} :=& \frac{1}{n} \sum_{i=1}^n \widehat{f}_{-i}\left(X_i\right) =\frac{1}{n^2 h} \sum_{i=1}^n \sum_{j=1, j \neq i}^n K\left(\frac{X_j-X_i}{h}\right),\\ \widehat{\mathcal{I}_{2n}}^{\text{without LOO}} :=& \frac{1}{n} \sum_{i=1}^n \widehat{f}\left(X_i\right)=\frac{1}{n^{2}h} \sum_{i=1}^n\sum_{j=1}^n K\left(\frac{X_j-X_i}{h}\right). \end{aligned} Notice that their only difference lies in: $$\widehat{\mathcal{I}_{2n}}^{\text{without LOO}} = \widehat{\mathcal{I}_{2n}}^{\text{LOO}} + \frac{1}{n h} K(0).$$ But as we will see later, this difference plays a key role in the conclusion that without leave-one-out, the objective function $$LSCV$$ is biased concerning $$\mathrm{IMSE}$$ and is minimized at $$h=0$$, which violates the condition of $$n h \rightarrow \infty$$ as $$n \rightarrow \infty$$.

Now, taking the expectation $$\mathbb{E}_{X}(\cdot)$$ to $$LSCV^{\text{LOO}}$$ gives: $$\mathbb{E}_{X}\left(LSCV^{\text{LOO}}\right) = \mathbb{E}_{X}\left[{\mathcal{I}}_{1 n}\right]-2 \mathbb{E}_{X}\left[\widehat{\mathcal{I}}_{2 n}^{\text{LOO}}\right]+\int f^2(x) d x.$$ The first term is computed as: \begin{aligned} \mathbb{E}_{X}\left[{\mathcal{I}}_{1 n}\right]=& \mathbb{E}_{X}\left[\int[\widehat{f}(x)]^2 d x\right] = \int\mathbb{E}_{X}\left[[\widehat{f}(x)]^2\right] d x\\ =& \int\mathrm{Var}\widehat{f}(x) d x + \int\mathbb{E}_{X}^2\left[\widehat{f}(x)\right] d x \\ =&\int \frac{f(x)}{n h} \nu_2+o\left(\frac{1}{n h}\right) + \left(f(x)+\frac{h^2}{2} f^{\prime \prime}(x) \kappa_2 + o\left(h^2\right)\right)^{2} dx \\ =& \frac{\nu_2}{n h}+o\left(\frac{1}{n h}\right) + \int f^{2}(x) dx + \frac{h^4}{4} \int u^2 K(u) d u \int\left[f^{\prime \prime}(x)\right]^2 d x+o\left(h^4\right)\\ & + \int h^{2}f(x)f^{\prime \prime}(x) \kappa_2 dx + 2 \int f(x)o\left(h^2\right) dx , \end{aligned} where the third line can be referred (again) to the link and $$\nu_2:=\int K^2(u) d u.$$ The second term is computed as: \begin{aligned} 2 \mathbb{E}_X\left[\widehat{\mathcal{I}}_{2 n}^{\mathrm{LOO}}\right] =& \mathbb{E}_{X}\left[\frac{1}{n^2 h} \sum_{i=1}^n \sum_{j=1, j \neq i}^n2 K\left(\frac{X_j-X_i}{h}\right)\right] = \mathbb{E}_{X}\left[\frac{2(n-1)}{n^{2}} \sum_{i=1}^n \widehat{f}_{-i}(X_{i})\right] \\ =& \frac{2(n-1)}{n^{2}} \sum_{i=1}^n \mathbb{E}_{X_{i}}\left[\mathbb{E}_{X_{-i}}\left[\widehat{f}_{-i}(X_{i})\right]\right] \\ =& \frac{2(n-1)}{n^{2}} \sum_{i=1}^n \mathbb{E}_{X_{i}}\left[f(X_{i})+\frac{h^2}{2} f^{\prime \prime}(X_{i}) \kappa_2+o\left(h^2\right)\right] \\ =& \frac{2(n-1)}{n} \int\left[f(X_{i})+\frac{h^2}{2} f^{\prime \prime}(X_{i}) \kappa_2+o\left(h^2\right)\right]f(X_{i}) d X_{i}\\ =& \frac{2(n-1)}{n} \int\left[f(x)+\frac{h^2}{2} f^{\prime \prime}(x) \kappa_2+o\left(h^2\right)\right]f(x) d x\\ =& \frac{2(n-1)}{n} \int f^{2}(x) dx +\frac{2(n-1)}{n} \int \frac{h^2}{2} f^{\prime \prime}(x)f(x) \kappa_2 dx + \frac{2(n-1)}{n} o\left(h^2\right) \\ \end{aligned} Inserting the two expressions back $$\mathbb{E}_{X}\left[{LSCV}^{\text{LOO}}(h)\right]$$ gives: \begin{aligned} & \mathbb{E}_{X}\left[{LSCV}^{\text{LOO}}(h)\right]\\ =& \frac{\nu_2}{n h}+o\left(\frac{1}{n h}\right) + \int f^{2}(x) dx + \frac{h^4}{4} \int u^2 K(u) d u \int\left[f^{\prime \prime}(x)\right]^2 d x+o\left(h^4\right) \\ &+ \int h^{2}f(x)f^{\prime \prime}(x) \kappa_2 dx + 2 \int f(x)o\left(h^2\right) dx \\ &-\left[\frac{2(n-1)}{n} \int f^{2}(x) dx +\frac{2(n-1)}{n} \int \frac{h^2}{2} f^{\prime \prime}(x)f(x) \kappa_2 dx + \frac{2(n-1)}{n} o\left(h^2\right)\right]+\int f^2(x) d x\\ =& \frac{\nu_2}{n h}+o\left(\frac{1}{n h}\right) + \frac{h^4}{4} \int u^2 K(u) d u \int\left[f^{\prime \prime}(x)\right]^2 d x+o\left(h^4\right) + O\left(\frac{1}{n}\right), \end{aligned}

Now, we can see that $$\mathbb{E}_{X}\left[{LSCV}^{\text{LOO}}(h)\right]\sim \operatorname{IMSE}(\widehat{f}).$$ And it is quite evident to see that $$\mathbb{E}_{X}\left[LSC V^{\text {without } L O O}(h)\right]=\mathbb{E}_{X}\left[LSC V^{L O O}(h)\right]-\frac{2}{n h} K(0).$$

To this end, one can be informed in the context of KDE that the leave-one-out cross-validation provides a relatively “unbiased estimate of the true generalization performance”.