Expected value for $X, Y \sim \mathcal{U}[0,10]$ vs. expected value of $E(2x)$ $X, Y$ are independent events both $\sim \mathcal{U}[0,10]$.
I know that $$E[X+Y] = E[X] + E[Y] = 5 + 5 = 10$$
and $$E[2X] = 2 E[X] = 10 \Rightarrow E[X+Y] = E[2X].$$
However, is $E\left[(X+Y-10)_+\right] = E\left[(2X-10)_+\right]$ ? 
I believe the answer is no, but why?

The right one is
$$
\frac{1}{10} \int_0^{10} (2x-10)^+dx 
  = \frac{1}{10} \int_5^{10} (2x-10)dx = 2.5
$$
and the left one is
$$
\begin{split}
\frac{1}{100} \int_0^{10} \int_0^{10} (x+y-10)^+ dydx
 &= \frac{1}{100} \int_0^{10} \int_{10-x}^{10} (x+y-10) dydx \\
 &= \frac{1}{100} \int_0^{10} \left[ x(x-10) + 50 - (10-x)^2/2 \right] dx \\
 &= \frac{1}{100} \frac{500}{3} = \frac{5}{3}.
\end{split}
$$
the intuition is that in the $X+X$ case, knowing the left also fixes the right, but for $X+Y$ the left and the right are independent so there is more possibility for successful outcomes.
