Expectation of a random variable which is the minimum of a random variable and a constant Given $X \sim exp(\lambda)$ and a constant $c \ge 0$. Let $Y=\min(X,c)$.
I would like to find $E[Y]$ and tried looking for possible approaches in link1 and link2.
However, the accepted answers in both URLs give very different results. For example, according to the answer in link1
\begin{align*}
\mathsf E(Y)
 & = \mathsf E(Y\mid X\leq c)\,\mathsf P(X\leq c)+\mathsf E(Y\mid X>c)\,\mathsf P(X> c)
\\
 & = \mathsf E(X\mid X\leq c)\,\mathsf P(X\leq c) +\mathsf E(c\mid X>c)\,\mathsf P(X>c)
\\
 & = \mathsf E(X\mid X\leq c)\,\mathsf P(X\leq c) +c\,\mathsf P(X>c) \\
 & = (1-e^{-\lambda c})\int_0^c t\lambda e^{-\lambda t}dt + ce^{-\lambda c}
\end{align*}
and according to the accepted answer in link2:
\begin{align*}
E(Y)&= \int_0^\infty P( Y > t) dt \\
       &= \int_0^\infty P( X > t, c > t) dt \\
       &= \int_0^\infty P( X > t)1\{ c > t\} dt \\
       &= \int_0^c P(X > t)dt \\
       &= \int_0^c e^{-\lambda t}dt
\end{align*}
What even confused me more is another direct solution to this problem shown in the second answer in link1.
Which one is correct? Honestly I am not sure how to start with this problem, but the first answer seems to make more sense to me.
 A: The first one is wrong. The first term there should be $\int_0^{c} t\lambda e^{-\lambda t} dt$ (without the factor $1-e^{-\lambda c})$ and if you integrate by parts this gives the same answer as in Link 2.
The method used by Graham Kemp in Link2 is not applicable here since $X$ is a continuous random variable.
A: I will give you a slightly easier way to think about the second method without indicators. Let's just split the integral domain. Kavis comment above is correct by the way.
\begin{align*}
E(Y)&= \int_0^\infty P( Y > t) dt \\
       &= 1) \int_0^c P( Y > t)dt + \int_c^{\infty}P(Y>t) dt \\ 
       &= 2) \int_0^c P( X > t)dt + \int_c^{\infty}0 dt \\
       &= 3) \int_0^c P(X > t)dt \\
       &= 4) \int_0^c e^{-\lambda t}dt
\end{align*}
Step $1)$ comes from simply splitting the integral over $(0,\infty)$ to $(0,c) \cup (c,\infty)$
Step $2)$ comes from thinking about what $Y$ is and how it can be in certain ranges. The only way it can be in between $0$ and $c$ is if it is $X$ taking this value. It also cannot be above $c$ because it is the minimum of something and $c$
Step $3)$ Just notes the integral of $0$ is $0$
Step $4)$ Just uses the CDF of exponential random variables.
