Expectation when x>5 Exam question was:
Suppose $X$ is an exponential random variable with the expectation of $1$, find $E(X|X>5)$.  
My answer was that first find
$$P(X<x | X > 5)$$ 
Which I got to be $1-e^{5-x}$ and so
$$ E(X|X>5) = \int_5^{\infty} P(X>x) = 1 $$
I lost almost all point because the answer is said to be 6 in the key but I showed a lot of work why did I only get 1 out of 10 points? 
 A: There is a difference between the two notations
$\mathbb E(X \textbf{1}_{[X>5]})$ and and $\mathbb E(X|X>5)$. The former is defined by
$$
\mathbb E(X \textbf{1}_{[X>5]})=
\int_0^\infty f_X(x) x \textbf{1}_{[x>5]} \, \mathrm d x
=
\int_5^\infty f_X(x) x \, \mathrm d x
=
\int_5^{\infty } e^{-x} x \, dx
=\frac{6}{e^5},
$$
where $f_X(x)=e^{-x}$ is the density of $X$.
And the latter is
$$
\mathbb E(X|X>5) = 
\frac{\mathbb E(X \textbf{1}_{[X>5]})}{\mathbb P(X>5)}
=
\frac{\frac{6}{e^5}}{\frac{1}{e^5}}
=
6.
$$
A: It is a correct route to find $F(x)=P(X\leq x\mid X>5)$ and then to go for:$$\mathbb E(X\mid X>5)=\int x\;dF(x)$$
Going that route you find that $F(x)=1-e^{5-x}$ for $x\geq5$ and $F(x)=0$ otherwise.
So that would lead to:$$\mathbb E(X\mid X>5)=\int_5^{\infty} x\;d(1-e^{5-x})=\int_5^{\infty} xe^{5-x}\;dx=\int_0^{\infty} (y+5)e^{-y}dy=$$$$5+\int_0^{\infty} ye^{-y}dy=5+1=6$$

Also it is correct to go for:$$\mathbb E(X\mid X>5)=\int_0^{\infty}P(X>x\mid X>5)dx=\int_0^{5}1dx+\int_5^{\infty}e^{5-x}dx=5+1=6$$

Also you can take the route shown in the answer of @ablmf.

The most handsome route is realizing that exponential distribution is "memoryless" so that the distribution under condition $X>5$ is the same as the distribution of $5+X$ (and consequently has expectation $\mathbb E(5+X)=5+\mathbb EX=5+1=6$).
