Upper bound of the waiting time of a sum process Let $n \in \mathbb{N}$, $x_1, \ldots, x_n \in (0,1)$ fix but arbitrary, s.t. $\sum_{i=1}^n x_i = 1$. Let $X_i \sim \operatorname{Unif}(\{x_1, \ldots, x_n\})$ i.i.d., and $T_n = \min\{t \in \mathbb{N} \, : \, \sum_{i=1}^t X_i \geq 1\}$.
I think there exists a constant $C$, which is independent of $n$, s.t.
$$\mathbb{E}[T_n] \leq C \cdot n.$$
(At least, if we assume, that $x_1 \geq \frac{1}{2}$, then $C \leq 2$.)
My question is the following: What is the smallest possible $C$, s.t. the estimate holds? Do you know, if there is any work regarding this problem?
I could show, that in a special case $C \approx 1.36$, and I think it is the worst case, but couldn't prove it up to now.
 A: This looks like a "Wald equality" question. Define $Y=\sum_{i=1}^{T_n} X_i$.  Then: 
\begin{align} 
1 + x_{max} \geq Y =\sum_{i=1}^{\infty} X_i1\{T_n\geq i\}\\
\end{align} 
where $1\{T_n\geq i\}$ is an indicator function that is $1$ if $T_n\geq i$, and 0 else. 
Taking expectations of both sides gives: 
$$ 1 + x_{max} \geq \sum_{i=1}^{\infty} E[X_i]Pr[T_n\geq i] $$ 
where we used the fact that $X_i$ is independent of the event $\{T_n \geq i\}$. Using the fact that $E[X_i]=1/n$ for all $i$ gives: 
$$ 1 + x_{max} \geq (1/n)\sum_{i=1}^{\infty}Pr[T_n\geq i] = (1/n)E[T_n] $$
Thus $E[T_n] \leq (1+x_{max})n \leq 2n$.
So you can use $C=1+x_{max}$ if you have a bound on $x_{max}$, or you can use $C=2$ else.  

The James Martin observation looks interesting if you only require the condition to hold for large $n$.
The Lorden inequality (link in comment above) shows that $E[T_n]/n \leq 1 + \frac{E[X^2]}{E[X]}$. However, intuitively I would expect the overshoot for large $n$ to be $\approx \frac{E[X^2]}{2E[X]}$, which can be arbitrarily close to $1/2$ (so $C \approx 1.5$) by choosing $\{x_1, \ldots, x_n\} = \{\epsilon, \ldots, \epsilon, 1-(n-1)\epsilon\}$. [See comment below by James Martin on why this conjecture is not correct due to the threshold "1" being too small]
