Let 
\begin{equation}
	N:=\inf\{n\ge2\colon X_{n-1}>X_n\}, 
\end{equation}
where $X_1,X_2,\dots$ are independent random variables uniformly distributed on $[0,1]$. 
We want to find 
\begin{equation}
	EX_N=\sum_{n=2}^\infty EX_n\,1(X_1\le\cdots\le X_{n-1}>X_n).
\end{equation}

We have 
\begin{equation}
\begin{aligned}
	&EX_n\,1(X_1\le\cdots\le X_{n-1}>X_n) \\
	&=EX_n\,1(X_1\le\cdots\le X_{n-1}) \\ 
	&-EX_n\,1(X_1\le\cdots\le X_{n-1}\le X_n), 
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
	&EX_n\,1(X_1\le\cdots\le X_{n-1}) \\
	&=EX_n\,P(X_1\le\cdots\le X_{n-1})=\frac12\,\frac1{(n-1)!},
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
	&E(1-X_n)\,1(X_1\le\cdots\le X_{n-1}\le X_n) \\ 
	&=E\int_0^1 dx\,1(X_1\le\cdots\le X_{n-1}\le X_n\le x) \\ 
	&=\int_0^1 dx\,P(X_1\le\cdots\le X_{n-1}\le X_n\le x) \\ 
	&=\int_0^1 dx\,x^n\frac1{n!}= \frac1{(n+1)!},
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
	&EX_n\,1(X_1\le\cdots\le X_{n-1}\le X_n) \\ 
	&=P(X_1\le\cdots\le X_{n-1}\le X_n)-\frac1{(n+1)!} \\ 
	&=\frac1{n!}-\frac1{(n+1)!},
\end{aligned}	
\end{equation}
\begin{equation}
\begin{aligned}
	EX_N&=\sum_{n=2}^\infty \Big(\frac12\,\frac1{(n-1)!}-\frac1{n!}+\frac1{(n+1)!}\Big) \\ 
	&=\frac e2-1\approx0.359.
\end{aligned}	
\end{equation}

---

One may also note that 
\begin{equation}
\begin{aligned}
	&EN=E\sum_{n=0}^\infty1(N>n)=\sum_{n=0}^\infty P(N>n) \\ 
	&=\sum_{n=0}^\infty P(X_1\le\cdots\le X_n) =\sum_{n=0}^\infty \frac1{n!}=e\approx2.72. 
\end{aligned}
\end{equation}

---

Simulation with 
Mathematica appears to confirm these results (click on the image below to enlarge it): 

[![enter image description here][1]][1]


  [1]: https://i.sstatic.net/1Rbqq.png