Conditional expected value $Z_n$ Let $(X_n)_{n\geq0}$, following an exponential law of paramater 1. We define for each $n\geq0$, $Y_n = \min(X_1,X_2,...,X_n)$  and $Z_n=\mathbb{E}[Y_{n+1}|Y_n]$.
Show that $Z_n=1-e^{-Y_n}$
I tried using some independence properties linked to $(X_n)_{n\geq0}$ or even using the formula $\mathbb{E}[Y_{n+1}|Y_n]=\phi(Y_n)$.
I know that the conditional expected value is a variable but I can't find a way to end up with the same result.
 A: A most useful fact is that the expected value (when it exists) of a non-negative random variable $U$
\begin{equation}
    \mathbb{E}[U] = \int_0^\infty \mathbb{P}\{U > t\} dt.
\end{equation}
See Lemma 1.2.1 in this book for a quick proof of this fact. Applying this fact with $U = Y_{n+1}$ conditioned on $Y_n$ one has
\begin{align}
    \mathbb{E}[Y_{n+1} \ | \ Y_n] &= \int_0^\infty \mathbb{P}\{Y_{n+1} > t \ | \ Y_n\} dt\\
    &= \int_0^{Y_n} \mathbb{P}\{Y_{n+1} > t \ | \ Y_n \} dt + \int_{Y_n}^\infty \mathbb{P}\{Y_{n+1} > t| \ Y_n \} dt \\
    &= \int_0^{Y_n} 1 - \mathbb{P}\{Y_{n+1} \leq t  \ | \ Y_n \} dt + \int_{Y_n}^\infty 0 \ dt \\
    &= \int_0^{Y_n} 1 - \mathbb{P}\{Y_{n+1} \leq t  \ | \ Y_n \} dt
\end{align}
where we have used the integral identity in the first line, broken the integral into two parts in the second, and in the third used that for $t \geq Y_n$ we always have
\begin{equation}
    Y_{n+1} = \min(X_1,...,X_{n+1}) \leq \min(X_1,...,X_n) = Y_n \leq t
\end{equation}
which implies for all $t \geq Y_n$ that $\mathbb{P}\{Y_{n+1} > t| \ Y_n \} = 0$. Now we need to handle the remaining term.
Note that for all $t < Y_n$ we have $Y_{n+1} \leq t \iff X_{n+1} \leq t$, this follows from the fact that $Y_{n+1}$ is less than $Y_n$ if and only if $X_{n+1} < Y_n$. Therefore we have the identity for all $t < Y_{n}$ that
\begin{equation}
    \mathbb{P}\{Y_{n+1} \leq t \ | \ Y_n \} = \mathbb{P}\{X_{n+1} \leq t \}.
\end{equation}
and this expression we can deal with because we know that $X_{n+1}$ has an exponential distribution with parameter 1 and therefore
\begin{equation}
    \mathbb{P}\{X_{n+1} \leq t \} = 1 - e^{-t}
\end{equation}
(this is just the CDF of an exponential random variable). Now plugging this in above we have
\begin{align}
    \int_0^{Y_n} 1 - \mathbb{P}\{Y_{n+1} \leq t  \ | \ Y_n \} dt &= \int_0^{Y_n} 1 - \mathbb{P}\{X_{n+1} \leq t \} dt \\
   &= \int_0^{Y_n} 1 - (1 - e^{-t}) dt \\
   &= \int_0^{Y_n} e^{-t} dt \\
   &= -e^{-t} \bigg |_0^{Y_n} \\
   &= (-e^{-Y_n}) - (-e^{-0}) = 1 - e^{-Y_n} 
\end{align}
which when combined with the expression for the conditional expectation completes the proof. Note that we are taking advantage of the fact that we only consider values of $t$ less than $Y_n$.
Overall the main trick is to use the integral lemma and then consider the two possible cases: When $Y_{n+1} = Y_n$ and when $Y_{n+1} < Y_n$. This is why we chose to break up the integral at the point $Y_n$.
A: I assume $(X_n)$ is an i.i.d sequence.
Note that $Y_{n+1}=\min\{Y_n,X_{n+1}\}$ where $Y_n$ and $X_{n+1}$ are independently distributed.
Therefore, almost surely,
\begin{align}
E\left[Y_{n+1}\mid Y_n\right]&=E\left[\min\{Y_n,X_{n+1}\}\mid Y_n \right]
\\&=\int \min\{Y_n,x\}f_{X_{n+1}}(x)\,dx
\\&=\int_0^\infty \min\{Y_n,x\}e^{-x}\,dx
\end{align}
