Expectation exponential distribution for $X\leq c$ and $X>c$ Let $X \sim \text{Exp}(\lambda)$ and suppose we are allowed to know whether $X\leq c$ or $X>c$. What is our updated expectation for $X$? Let us take $(\Omega, \mathcal{F}, P) = (\mathbb{R}_+, \mathcal{B}(\mathbb{R}_+),\mu)$ with $\mu(dx) = \lambda e^{-\lambda x} dx$, the identity random variable $X(\omega)=\omega$, and the sub-$\sigma$-algebra $\mathcal{A}=\{\emptyset, \Omega, [0,c], (c,\infty)\}$. 
For example, for any $\omega \in (c, \infty)$ we would compute,
\begin{align}
\mathbb{E}[X\mid\mathcal{A}](\omega) &= \frac{\mathbb{E}[\mathbb{1}_{(c,\infty)}X]}{\mu(c,\infty)}\mathbb{1}_{(c,\infty)}(\omega) \\
&= \frac{1}{\mu(c,\infty)} \int_c^{\infty} x \lambda e^{-\lambda x}\ dx \\
&= e^{\lambda c}(c e^{-\lambda c} + \lambda^{-1}e^{-\lambda c})\\
&=c + \lambda^{-1}.
\end{align}
Now, for any $\omega \in [0, c]$ we would compute,
\begin{align}
\mathbb{E}[X\mid\mathcal{A}](\omega) &= \frac{1}{\mu[0,c]} \int_0^c x \lambda e^{-\lambda x}\ dx.
\end{align}
I found that $\int_0^c x \lambda e^{-\lambda x}\ dx = -e^{-\lambda c}(c+\lambda^{-1})+\lambda^{-1}$. But, maybe a silly question, how to determine $\mu[0,c]$? $\mu(c,\infty)$ has to be $e^{-\lambda c}$ but for me it is not clear how you can find this. In the end, of course, we want to find that $\mathbb{E}[X\mid\mathcal{A}] = \frac{1}{\lambda}$. 
Secondly, which strategy should you suggest to prove the identity $ \mathbb{E}[\mathbb{E}[X\mid\mathcal{A}]] = \mathbb{E}[X]$? And why would it be understandable that the expectation is constant on $[0,c]$ and $(c,\infty)$?  
In reaction:
\begin{align}
\mathbb{E}[X \mid \mathcal{A}] &= \frac{\mathbb{E}[\mathbb{1}_{[0,c]}X]}{\mu[0,c]}\mathbb{1}_{[0,c]}(\omega) + \frac{\mathbb{E}[\mathbb{1}_{(c,\infty)}X]}{\mu(c,\infty)}\mathbb{1}_{(c,\infty)}(\omega)\\
&=\mathbb{E}[\mathbb{1}_{[0,c]}X] + \mathbb{E}[\mathbb{1}_{(c,\infty)}X]\\
&= \int_0^c x \lambda e^{-\lambda x}\ dx + \int_c^\infty x \lambda e^{-\lambda x}\ dx\\
&= -e^{-\lambda c}(c+\lambda^{-1})+\lambda^{-1} + \big(c e^{-\lambda c} + \lambda^{-1}e^{-\lambda c}\big) \\
&=\lambda^{-1}
\end{align}
 A: To condition the expected value on $\mathcal A = \{ \varnothing, \Omega = [0,\infty), [0,c], (c,\infty) \}$ means you're finding the conditional expected value when you have information telling you which of those four sets $\omega$ is in.  That means you know either that it's in $[0,c]$ or that it's in $(c,\infty)$, but you don't know more than that.  If you know it's in $[0,c]$ but not where in that interval it is, then moving it to a different location within $[0,c]$ will not change the information you've got, so the conditional probability will still be the same.  And similarly for $(c,\infty)$.  Abstractly, the definition of conditional expectation given a $\sigma$-algebra tells you that it must be measurable with respect to that $\sigma$-algebra, and from that it follows that it's constant on $[0,c]$ and also on $(c,\infty)$.
First find four numbers:


*

*$a=\operatorname{E}(X\mid X\le c)$,

*$b=\operatorname{E}(X\mid X> c)$,

*$p=\Pr(X\le c)$,

*$q=\Pr(X > c)$.


That $b = c + \dfrac 1 \lambda$ follows immediately from the memorylessness of the exponential distribution.
$$
p= \mu([0,c]) = \int_0^c e^{-\lambda x} (\lambda \, dx) = \int_0^{\lambda c} e^{-u}\,du = 1 - e^{-\lambda c}.
$$
$$
q = \int_c^\infty e^{-\lambda x} (\lambda\,dx) = e^{-\lambda c}.
$$
Then the random variable $\operatorname{E}(X\mid A)$ is
$$
\operatorname{E}(X\mid A) = \left. \begin{cases} a & \text{if } X\le c, \\  b & \text{if } X > c, \end{cases} \right\} = \begin{cases} a & \text{with probability } p, \\ b & \text{with probability }q. \end{cases}
$$
In other words, conditioning the expected value on $A$ means finding conditional expected values when you are given information telling you which members of $A$ the outcome $\omega$ is in.
The expected value of that is $ap+bq$ and should be the same as the expected value of $X$.
