Let $X,Y$ be IID random variable that are uniformly distributed on $[0,1]$. Does the following equation hold?
$$\text{E}[X\vert X\geq Y] = \int_{-\infty}^{\infty} \text{E}[X\vert X\geq y]f_Y(y)\,\text{d}y$$
This would then imply that
$$\text{E}[X\vert X\geq Y] = \int_0^1 \text{E}[X\vert X\geq y]\,\text{d}y$$
where
$$f_{X\vert X\geq y}(x) = \begin{cases}\frac{1}{1-y}&y\leq x\leq 1\\0&\text{otherwise}
\end{cases}$$
such that
$$\text{E}[X\vert X\geq y] = \int_y^1 x\,\frac{1}{1-y}\,\text{d}x = \frac{1+y}{2}$$
Finally, we obtain
$$\text{E}[X\vert X\geq Y] = \int_0^1 \frac{1+y}{2}\,\text{d}y = \frac{3}{4}$$
which is not correct. Hence, there must be an error somewhere. I appreciate your help!
For completeness, the (presumably) correct solution is given below.
$$\text{E}[X\vert X\geq Y] = \int_{-\infty}^{\infty} x f_{X\vert X\geq Y}(x)\,\text{d}x$$
Using Bayes' rule, we obtain
\begin{align}
f_{X\vert X\geq Y}(x) &= \frac{\text{Pr}[X\geq Y\vert X = x] f_X(x)}{\text{Pr}[X\geq Y]}\\
&= \frac{\text{Pr}[Y\leq x] f_X(x)}{1/2}\\
&= \begin{cases}
2x& x\in[0,1]\\
0&\text{otherwise}
\end{cases}
\end{align}
such that
$$\text{E}[X\vert X\geq Y] = \int_{-\infty}^{\infty} x \cdot 2x\,\text{d}x = \frac{2}{3}$$
Answer
All credit goes to @Thomas who found the mistake in my initial approach. Thank you @Thomas and thanks to those who added helpful answers or comments.
The main issue lies in my very first equation, namely
$$\text{E}[X\vert X\geq Y] = \int_{-\infty}^{\infty} \text{E}[X\vert X\geq y]f_Y(y)\,\text{d}y$$
As @Thomas suggested, the density $f_Y(y)$ should also be conditioned on $X\geq Y$, that is, $f_{Y\vert X\geq Y}(y)$. The corrected equation is given as
$$\text{E}[X\vert X\geq Y] = \int_{-\infty}^{\infty} \text{E}[X\vert X\geq y]f_{Y\vert X\geq Y}(y)\,\text{d}y$$
with
\begin{align}
f_{Y\vert X\geq Y}(y) &= \frac{P(X\geq Y\vert Y = y)f_Y(y)}{P(X\geq Y)}\\
&= \frac{P(X\geq y)f_Y(y)}{P(X\geq Y)}\\
&= \begin{cases}
\frac{1-y}{1/2} &0\leq y \leq 1\\
0 &\text{otherwise}
\end{cases}\\
&= \begin{cases}
2(1-y) &0\leq y \leq 1\\
0 &\text{otherwise}
\end{cases}
\end{align}
This results in the following solution.
\begin{align}
\text{E}[X\vert X\geq Y] &= \int_{-\infty}^{\infty} \text{E}[X\vert X\geq y]f_{Y\vert X\geq Y}(y)\,\text{d}y\\
&= \int_0^1 \text{E}[X\vert X\geq y]\,2(1-y)\,\text{d}y\\
&= \int_0^1 \frac{1+y}{2}\,2(1-y)\,\text{d}y\\
&= \int_0^1 (1-y^2)\,\text{d}y\\
&= \left[y – \frac{1}{3}\,y^3\right]_0^1\\
&= \frac{2}{3}
\end{align}
where $\text{E}[X\vert X\geq y] = \frac{1+y}{2}$ follows from the computations above (initial approach).
Best Answer
First of all we can recall the tower property. If we have a partition of the probability space $A_i$, with $P(A_i)>0$, than:
$E[X]=\sum_i E[X|A_i]P(A_i)$
If we have an event $B$ this reads:
$E[X|B]=\sum_i E[X|A_i,B]P(A_i|B)$ [1]
Now consider $B=\{X>Y\}$ and $A_y=\{Y=y\}$. We would be tempted to guess from [1]:
$E[X|X>Y]=\int dy E[X|Y=y,X>Y]P(Y=y|X>Y)$ [2]
This is similar to your formula in spirit, but note that we have a conditional density on P.
But the problem here is that also this formula is undefined because we are conditioning on null events, something that is undefined. Nevertheless, if we consider a $y_1=0,...,y_n=1$ equidistant with distant $\Delta$ and consider the events $A_i=\{Y \in [y_i,y_{i+i}]\}$ we have a well defined expression, similar to 2, but with summations instead of integrals. We can than let $\Delta \rightarrow 0$.
If you do that, you see that you need to substitute :
, which leads to the correct result (2/3). Something that one could have also 'guessed'.
Of course there is a more direct way to solve the exercise but here I wanted to show how to change your first approach so that you get the correct result.