Where is the mistake with this conditional expectation calculation? Let $X$ and $Y$ be independent uniform random variables in $[0,1]$ and let $\alpha\geq 1$. I am interested in computing $E(\alpha)=\mathbb{E}(X\mid X\geq \alpha Y)$. Intuitively, I expect to have $E'(\alpha)>0$ as when $\alpha$ increases, conditional on $\{X\geq \alpha Y\}$, I know $X$ can only take higher values. I am stuck with the computations, however. This is what I've done:
We know that
$$\mathbb{E}(X\mid X\geq \alpha Y)=\frac{\mathbb{E}(X\mathbb{I}_{X\geq \alpha Y})}{\mathbb{P}(X\geq \alpha Y)}.$$
As $X\in[0,1]$,
$$\mathbb{P}(X\geq \alpha Y)=\int_{0}^{1/\alpha}\mathbb{P}(X\geq \alpha y)\,dy+\underbrace{\int_{1/\alpha}^{1}\mathbb{P}(X\geq \alpha y)\,dy}_{=0}=\int_{0}^{1/\alpha}[1-F_X(\alpha y)]\,dy=\frac{1}{2\alpha}.$$
This expression seems about right, as I know that $\mathbb{P}(X>Y)=\frac{1}{2}$, and it is decreasing in $\alpha$ as it should intuitively be. Likewise, I can compute
$$\mathbb{E}(X\mathbb{I}_{X\geq \alpha Y})=\mathbb{E}\left(X\mathbb{I}_{X\geq \alpha Y}\mathbb{I}_{Y\leq \frac{1}{\alpha}}\right)+\underbrace{\mathbb{E}\left(X\mathbb{I}_{X\geq \alpha Y}\mathbb{I}_{Y> \frac{1}{\alpha}}\right)}_{=0}=\int_0^{1/\alpha}\int_{\alpha y}^1x\,dx\,dy=\int_{0}^{1/\alpha}\left(\frac{1}{2}-\frac{\alpha^2 y^2}{2}\right)dy= \frac{1}{3\alpha}$$
Then, $\mathbb{E}(X\mid X\geq \alpha Y)=\frac{2}{3}$, which I know that it is true if $\alpha=1$, but doesn't make sense for other $\alpha>1$. Can someone point me where is the mistake in my computations?
EDIT: As someone already pointed out, the calculations are correct. Can someone come up with a nice intuitive explanation for it?
 A: Consider the square $$[0,1]^2 \subset \mathbb R^2.$$  The region in this square that satisfies $x \ge \alpha y$ is the triangle with vertices $$\{(0,0), (1,0), (1,1/\alpha)\}.$$  Therefore, given that $(X,Y)$ is a point in this triangle, the expected value of $X$ is simply $$\operatorname{E}[X \mid X \ge \alpha Y] = 2\alpha \int_{x=0}^1 \int_{y=0}^{x/\alpha} x \, dy \, dx = \frac{2}{3},$$ which is independent of the choice of $\alpha$ (so long as it is at least $1$).  But why?  Well, because the joint conditional expected value $\operatorname{E}[(X,Y) \mid X \ge \alpha Y]$ is the geometric centroid of the triangle, hence the desired conditional marginal is simply the $x$-coordinate of that centroid.  And since we know that the centroid of a triangle is the intersection of its medians, and we know that the medians divide each other in a $2:1$ ratio, the desired expectation is invariant to the height of the triangle (as it depends only on the length of the triangle's base, which is always $1$) and occurs at $x = 2/3$.

As a bonus, we immediately can see that $$\operatorname{E}[Y \mid X \ge \alpha Y] = \frac{1}{3\alpha}.$$
