random variable conditioned on its own value When I search online for conditional distributions, usually I find tutorials that one random variable is conditioned on another. I want to find some tutorial like the following exercise where there is a random variable that's conditioned on its own value.
X ∼ Exp(1)
Calculate the conditional PDF and CDF of X given X>1
Calculate E[X|X>1] 
Find Var(X|X>1)

What's the mathematical term for this kind of conditional density? Thanks in advance!
Edit:
Thanks for the discussions. There are some materials on "Conditioning a Random Variable by an Event" like the following one http://netsys.kaist.ac.kr/lecture/EE210_2017/lecture_notes/ch07_student.pdf. I want to make a summary about the following cases
X discrete
P(X in A1|X in A2)

X continuous
f(x|X in A1)

X discrete, Y discrete
P(X in A|Y in B)

X discrete, Y continuous
P(X in A|Y in B)

X continuous, Y discrete
f(x|Y in B)

X continuous, Y continuous
f(x|y) 

 A: You may be looking at a truncated distribution. 
E.g.: If $X \sim \mathcal{N}(0,1)$, then a $X$ conditional on $a<X<b$, follows a truncated normal distribution. 
https://en.wikipedia.org/wiki/Truncated_normal_distribution
A: If $X\sim\mathrm{Expo}(\lambda)$ then by definition of conditional expectation on an event,
\begin{align}
\mathbb E[X\mid X>1] &= \frac{\mathbb E[X\mathsf 1_{\{X>1\}}]}{\mathbb E[\mathsf 1_{\{X>1\}}]}.
\end{align}
Since $\mathbb E[\mathsf 1_{\{X>1\}}]=\mathbb P(X>1) = e^{-\lambda}$ and
$$
\mathbb E[X\mathsf 1_{\{X>1\}}] = \int_1^\infty \lambda x e^{-\lambda x}\ \mathsf dx = \frac{e^{-\lambda}(1+\lambda)}\lambda ,
$$
we have
$$
\mathbb E[X\mid X>1] = \frac{e^{-\lambda}(1+\lambda)}\lambda\cdot e^\lambda = \frac1\lambda + 1.
$$
Alternatively, by the memoryless property of the exponential distribution, we have
$$
\mathbb E[X\mid X>1] = \mathbb E[X] + 1 = \frac1\lambda + 1.
$$
Similarly,
$$
\mathbb E\left[X^2\mathsf 1_{\{X>1\}}\right] = \int_1^\infty \lambda x^2 e^{-\lambda x}\ \mathsf dx = \frac{e^{-\lambda } (\lambda  (\lambda +2)+2)}{\lambda ^2},
$$
and so 
$$
\mathbb E[X^2\mid X>1] = \frac{e^{-\lambda } (\lambda  (\lambda +2)+2)}{\lambda^2}\cdot\lambda = \frac{e^{-\lambda } (\lambda  (\lambda +2)+2)}{\lambda }.
$$
It follows that
\begin{align}
\operatorname{Var}(X\mid X>1) &= \mathbb E[X^2\mid X>1] - \mathbb E[X\mid X>1]^2\\
&= \frac{e^{-\lambda } (\lambda  (\lambda +2)+2)}{\lambda } - \left(\frac1\lambda + 1\right)^2\\
&= \frac{e^{-\lambda } (\lambda +1)^2 (\lambda  (\lambda +2)+2)}{\lambda ^3}.
\end{align}
