Expectation of maximum between zero and a function of a random variable Let $v$ be a random variable uniformly distributed over [0,1]. Let $\alpha$ be a constant, where $0<\alpha<1/2$, so that $0<2\alpha<1$.
How can I express: $\mathbb{E}\left(Max\{2\alpha, v\}-v\right)$?

Here's what I've tried "intuitively":

*

*$$\mathbb{E}(Max\{2\alpha, v\}-v) = \mathbb{E}(Max\{2\alpha-v, 0\})\\
=\mathbb{P}(2\alpha-v \geq 0)\cdot (2\alpha-v)+ [1-\mathbb{P}(2\alpha-v \geq 0)]\cdot 0 \\
= \mathbb{P}(v \leq 2\alpha) \cdot (2\alpha-v) + 0 = 2\alpha \cdot (2\alpha-v)$$


*Which seems incorrect given this very similar question here. Instead,
$$\mathbb{E}(Max\{2\alpha, v\}-v) = \mathbb{E}(Max\{2\alpha-v, 0\})\\
=\mathbb{P}(2\alpha-v \geq 0)\cdot \mathbb{E}(2\alpha-v|2\alpha-v\geq0)+ [1-\mathbb{P}(2\alpha-v \geq 0)]\cdot 0 \\
= \mathbb{P}(2\alpha-v \geq 0)\cdot \mathbb{E}(2\alpha-v|2\alpha-v\geq 0) \\
= \mathbb{P}(v \leq 2\alpha) \cdot \mathbb{E}(2\alpha-v|2\alpha-v\geq 0)\\
= 2\alpha \cdot \mathbb{E}(2\alpha-v|2\alpha-v\geq 0)$$
However I am not sure about $\mathbb{E}(2\alpha-v|2\alpha-v>0)$, in turn because I can't find the distribution of $2\alpha-v$, is it a Uniform over [-1, 1]? so that we would have: $\mathbb{E}(2\alpha-v|2\alpha-v> 0)=1/2$, and $\mathbb{E}\left(Max\{2\alpha, v\}-v\right) = \alpha$?
Thanks,
 A: Distribution of $2\alpha -v$ is not needed. Use the fact that if $X$ has density function $f$ then $Eg(X)=\int g(x)f(x)dx$
$E(2\alpha -v|2\alpha -v \geq 0)=\frac 1 {2\alpha} \int_0^{2\alpha} (2\alpha -t)dt=\alpha$.
[The denominator is $P(v \leq 2\alpha)$].
A: You are right in that your "intuitive" method (1) is not correct - notice that your random variable $v$ still appears at the end of your expected value, which should be a number.
Your method (2) following the linked post uses a wise strategy (the "law of total expectation" to decompose it into two manageable parts). Your calculations up to the end look correct, so it suffices to compute $\mathbb{E}[2\alpha - v \mid 2\alpha - v > 0]$ as you recognised.
Notice that $v \sim \mathrm{Uniform}(0, 1)$, so $2\alpha - v \sim \mathrm{Uniform}(2\alpha, 2\alpha - 1)$. You can prove that $2 \alpha - v \mid 2\alpha - v > 0$ is actually uniform over $[0, 2\alpha]$ - intuitively this is by symmetry, and more rigorously you can prove this by using the definition of conditional probability. Thus, the expected value is $\alpha$.
(Alternatively, the answer by @Kavi Rama Murthy is of course true in general as well, since it is essentially using the definition of expectation.)
