Consider an auction in which two independent buyers bid for an antique object. If each of them draws their valuations from $\mathcal{U}[0,1]$ and the auction lets the highest bidder take the object by paying the lowest bid, what's the expected revenue of the auction?
Denote the two buyers' bids by the r.v.s $X, Y \sim \mathcal{U}[0,1]$.
I think we have to calculate $\mathbb{E}(X | X \geq Y) + \mathbb{E}(Y | Y \geq X)$. When I calculate it, I get $$2 \left(\int_{0}^{1} \int_{0}^{x} x \cdot \frac{1}{1-y} dydx\right) = 1.5$$
This is incorrect. Where did I go wrong?
Please note that I know a possible way to solve this is to evaluate $\mathbb{E}(\min(X,Y))$. But I am trying to figure out where I went wrong with my above method.
This is a funny auction but why not. The expected revenue is \begin{align} \mathbb E\big[\min(X,Y)\big]&=\textstyle\int_0^1\int_0^1\min(x,y)\,dx\,dy=\int_0^1\int_0^yx\,dx\,dy+\int_0^1\int_y^1y\,dx\,dy\\ &=\textstyle\int_0^1\frac{y^2}{2}\,dy+\int_0^1y-y^2\,dy=\int_0^1y-\frac{y^2}{2}\,dy\\ &=\textstyle\frac{1}{2}-\frac{1}{6}\\[2mm]&=\frac{1}{3}\,. \end{align} Alternatively one could use the law of total expectation by which \begin{align} \mathbb E\big[\min(X,Y)\big]&= \mathbb E\big[Y1_{\{X\ge Y\}}\big]+\mathbb E\big[X1_{\{Y>X\}}\big]\\[2mm] &=\underbrace{\mathbb E\big[Y\big|X\ge Y\big]}_{\textstyle=\frac{1}{3}}\,\underbrace{\mathbb P\big\{X\ge Y\big\}}_{\textstyle=\frac{1}{2}}+\underbrace{\mathbb E\big[X\big|Y> X\big]}_{\textstyle=\frac{1}{3}}\,\underbrace{\mathbb P\big\{Y>X\big\}}_{\textstyle=\frac{1}{2}}\\ &=\frac{1}{3}\,. \end{align} Proof.
Let's calculate the conditional expectations $$\mathbb E[Y|X\ge Y],\quad\mathbb E[Y|Y> X],\quad\mathbb E[X|X\ge Y],\quad\mathbb E[X|Y> X]\,$$ that were discussed in the comments. From $\mathbb P\{X\ge Y\}=\mathbb P\{Y>X\}=\frac{1}{2}$ we get \begin{align} \mathbb P\{Y\le z|X\ge Y\}&=\frac{\int_0^z\int_y^1\,dx\,dy}{1/2} =2\textstyle\int_0^z1-y\,dy=2z-z^2\,,\\ \mathbb P\{Y\le z|Y>X\}&=\frac{\int_0^z\int_0^y\,dx\,dy}{1/2} =2\textstyle\int_0^zy\,dy=z^2\,. \end{align} Taking the derivative w.r.t. $z$ the conditional PDFs of $Y$ are therefore, \begin{align} p(y|X\ge Y)=2-2y\,,\quad p(y|Y>X)=2y\,. \end{align} This leads to
\begin{align} \mathbb E[Y|X\ge Y]&=\textstyle\int_0^1y(2-2y)\,dy=2-\frac{2}{3}=\displaystyle\frac{1}{3}\,,\tag{1}\\ \mathbb E[Y|Y>X]&=\textstyle\int_0^12y^2\,dy=\displaystyle\frac{2}{3}\,,\\ \mathbb E[X|X\ge Y]&=\frac{2}{3}\,,\quad\text{ (by symmetry) }\\ \mathbb E[X|Y>X]&=\frac{1}{3}\,.\quad\text{ (by symmetry) }\tag{2}\\ \end{align} $$\tag*{$\Box$} \quad $$ Remarks.
The proposed solutions in the edited OP, resp. in a recent comment \begin{align} &\mathbb E[X|X\ge Y]+\mathbb E[Y|Y> X]=\frac{4}{3}\,,\\ &\mathbb E[Y|X\ge Y]+\mathbb E[X|Y> X]=\frac{2}{3} \end{align} are both wrong because we add expectations that are conditioned on mutually exclusive events. They are also different from $1/3$ which was mentioned as the correct answer in the OP that got overridden by some edits.
The other proposed solution (1) in a comment equals the correct solution (as well as (2) does) but I think this is a coincidence that is based on the independence of $X$ and $Y$ and on the uniformity of their distributions.