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I am reading the 2008 annals of statistics paper "Ranking and empirical minimisation of U-statistics" by Clémençon et. al, and read a statement which I do not know why is true. In order to accurately frame my question, I will give a brief account of the context. It is not necessary to read this to understand the question, so if you trust that I have reasoned correctly, you can skip to the bottom.

Assume that we observe two independent pairs $(Y_1, X_1),$ $(Y_2, X_2)$ from the same distribution, with support $\{-1, 1\}\times\mathcal{X}.$ We seek a ranking rule $r:\mathcal{X}\times\mathcal{X}\rightarrow \{-1, 1\}$ that minimises the ranking risk $$ R(r) = Pr((Y_1 - Y_2)\cdot r(X_1, X_2) < 0 ). $$ One can show that the minimum ranking risk in this case is $$ R^* = \mathbb{E}(\min(\eta(X_1), \eta(X_2)) - (\mathbb{E}(\eta(X)))^2, $$ where $\eta(x) = Pr(Y=1\vert X=x).$

So far so good. The authors then state, and this is where the basis of my question lies, that we also have $$ R^* = \text{Var}\left(\frac{Y+1}{2}\right) - \frac{1}{2}\mathbb{E}\vert \eta(X_1) - \eta(X_2)\vert. $$ We obviously have that for $W = (Y+1)/2,$ $W\vert X\sim\text{Bernoulli}(\eta(X)),$ so by using the double rule of variance we can see that $$ \text{Var}((Y+1)/2) = \mathbb{E}\{\eta(X)\} - (\mathbb{E}\{\eta(X)\})^2. $$ By inserting this, and equating the two supposed equivalent expressions for $R^*,$ we can see that both these statements are true if $$ \mathbb{E}\{\min(\eta(X_1), \eta(X_2))\} = \mathbb{E}\{\eta(X)\} - \mathbb{E}\{\vert\eta(X_1) - \eta(X_2)\vert\}. $$

My question:

The above statement may be rephrased in the following way. Given $X_1, X_2$ independent and identically distributed according to a distribution function $F$ whose support is a subset of $[0, 1],$ we have that $$ \mathbb{E}(\min(X_1, X_2)) = \mathbb{E}(X_1) - \frac{1}{2}\mathbb{E}(\vert X_1 - X_2\vert), $$ or equivalently that $$ \mathbb{E}(\min(X_1, X_2)) = 2\mathbb{E}(X_1) - \mathbb{E}(\max(X_1, X_2)). $$ This seems quite reasonable, and I have checked some examples to verify that it is true for a couple of distributions. I have tried searching for a result that states something along the lines of this, asked around if anyone knew, but have so far been unsuccessful. Does anyone know of such a result, what the proof is, and what the most general conditions for when it holds are?

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    $\begingroup$ The last result follows easily from the iid assumption and the observation $X_1+X_2=\min(X_1,X_2) + \max(X_1,X_2).$ $\endgroup$
    – whuber
    Feb 1, 2019 at 17:51

1 Answer 1

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You want to show that $ \mathbb{E}(\min(X_1, X_2)) = \mathbb{E}(X_1) - \frac{1}{2}\mathbb{E}(\vert X_1 - X_2\vert), $ following the comment from @whuber, we have that $\DeclareMathOperator{\E}{\mathbb{E}} X_1 + X_2 = \min(X_1, X_2) + \max(X_1, X_2)$ as our starting point. $$ X_1 + X_2 = \min(X_1, X_2) + \max(X_1, X_2) \\ = \min(X_1, X_2) + \left(\min(X_1, X_2) + \mid X_1 - X_2 \mid \right) \\ = 2 \min(X_1, X_2) + \mid X_1-X_2 \mid $$ reorganizing $$ \min(X_1, X_2) = \frac12 \left( X_1 + X_2 - \mid X_1 - X_2 \mid \right) $$ taking expectations and using that $X_1$ and $X_2$ have the same expectation $$ \E \min(X_1, X_2) = \E X_1 - \frac12 \E \mid X_1-X_2 \mid $$

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