Expectation of Truncated Random Variable So this is a pretty basic question but it is tripping me up. 
Suppose we have a random variable $X$ so that $X\sim F$. Now, consider $Y$ where $Y=X$ if $X<T$ but $Y=T$ if $X\ge T$. What is $E[Y]$?
I have 
$E[Y]=E[X|X<T] \cdot Pr(X<T) + T \cdot Pr(X\ge T) = F(T) \cdot \int _0^T x dF(x)+T\cdot \bar{F}(T)$, which I believe should be right (as $T\to \infty$ for example this works out).
However, looking at this wikipedia article, it says that 
$E[X|X>y] = \frac{\int_y^\infty xdF(x)}{\bar F (x)}$ which instead has some normalizing thing in the denominator which would change my answer. Is this even correct?
 A: I think you're indeed missing a term when dealing with the first conditional expectation when calculating $\mathbb{E}[Y]$.
For the conditional expectation (in agreement with the link that you posted, but you can also check out 
https://en.wikipedia.org/wiki/Conditional_expectation 
for a more general discussion) for an event $A$ with $\mathbb{P}[A] > 0$ we have
\begin{align*}
\mathbb{E}[X | X \in A] = \frac{\mathbb{E}[X\mathbb{1}_A(X)]}{\mathbb{P}[A]} = \frac{\int_A x dF_X(x)}{\mathbb{P}[A]}.
\end{align*}
The intuition why there is a need to divide by $\mathbb{P}[A]$ is that by conditioning on $A$, or by restricting if you like, you need to consider a new probability measure $\overline{\mathbb{P}}[\cdot] = \frac{\mathbb{P}[\cdot]}{\mathbb{P}[A]}$.
In other words, restricting yourself to $A$ forces you to re-normalize by $\mathbb{P}[A]$ to make sure $\overline{\mathbb{P}}[A] = 1$.
You can also see it from another point of view, namely by computing the distribution function $Y$.
This leads to
\begin{align*}
F_Y(y) = \begin{cases}
F_X(y), \mbox{ if } y < T,\\
1, \mbox{ if } y \geq T.
\end{cases}
\end{align*}
The expectation of $Y$ can now be calculated as
\begin{align*}
\mathbb{E}[Y] = \int_0^{\infty}ydF_Y(y) = \int_0^T y dF_X(y) + T \overline{F}_X(T),
\end{align*}
which is exactly what you get when you take the missing term into account.
I hope this makes it more clear what's going on.

Just as an aside, the limit $\lim_{T \to \infty} T \overline{F}_X(T)$ is not necessarily well behaved, think of a Pareto distribution for example...
