Let $X$ be an exponetial r.v. with parameter $1$, and $c$ be a nonnegative coonstant. I'm trying to guess $E[X|X\land c]$ and $E[X|X\lor c].$ Once I know what they are, I should be able to verify them using the $\pi-\lambda$ theorem, but if someone could help with the initial guess, that'd be great.
As Ben noted below, $\lor$ means max and $\land$ means min.
Here's an intuition which should help guide you to an initial guess: we can think of $X$ as measuring the time of failure of some system, and $X\wedge c$ measures the information that we have $X$ if we observe the system up until time $c$: if the system fails before time $c$, i.e. if $X < c$, then $X\wedge c< c$, and based on $X\wedge c$ we will know exactly what $X$ is, i.e. exactly what time the system failed; namely, it must be $X\wedge c$. On the other hand, if $X\geq c$, then $X\wedge c = c$, and $X\wedge c$ gives us no further information about the lifetime of the system other than that it is at least $c$. If you put these two cases together, and recall the memoryless property of the exponential distribution, then you should be able to write down a guess for $E(X\mid X\wedge c)$ as a linear combination of the indicator functions defining the two cases. The other part, with $X \vee c$ can be handled similarly.
Suppose that $X$ a random variable defined on a probability space $(\Omega, \mathcal{F}, P)$. Further suppose that $\mathcal{G} \subset \mathcal{F}$ is a sub-$\sigma$ algebra and that $P$ is the unit-exponential probability measure.
Recall that if $A \in \mathcal{G}$ is an atom ($P(A) > 0$ and $B \subset A \implies B = \emptyset$), then $\mathbb{E}[X | \mathcal{G}]$ is constant on $A$. Further, one can prove $\mathbb{E}[X | \mathcal{G}](\omega) = \frac{1}{P(A)} \int_A X dP(\omega) \; $ for all $\omega \in A$. This result can be found in Resnick as an exercise (additionally this problem is also an exercise in Resnick).
Suppose that you fix $t > 0$. Then, $\mathcal{G} := \sigma(t \wedge X) = \sigma(\mathcal{B}(0,t)\cup[t,\infty))$ where $\mathcal{B}(0,t)$ is the Borel $\sigma-$algebra generated by $(0,t)$. It should be clear that $[t,\infty)$ is an atom of $\mathcal{G}$ and $\mathcal{B}(0,t)$ has no atoms.
Without loss of generality, we may assume $X(\omega) = \omega$ and $\Omega = \mathbb{R}^{+}$. Then, from the result above: $$\mathbb{E}[X | \mathcal{G}](\omega) = \frac{1}{P([t,\infty))} \int_t^{\infty} X(\omega) dP(\omega) = \frac{1}{e^{-t}} \int_t^{\infty} x e^{-x} dx = e^{t} \big[ e^{-t}(t + 1)\big] = t+ 1$$
for $\omega \in [t, \infty)$
You can prove that if $G \in \mathcal{G}$ does not have an atom and is such that $P(G) > 0$ then $\mathbb{E}[X | \mathcal{G}](\omega) = X(\omega)\;$ almost surely (or use some other intuitive explanation for why this must be the case).
Therefore,
$\mathbb{E}[X | X \wedge t](\omega) = \left\{ \begin{array}{ll} X(\omega) & \omega \in (0,t) \\ t + 1 & \omega \in [t,\infty)\\ \end{array} \right.$
So you can actually find an explicit functional form.
