TITLE: How small the probability $\Bbb{P}(X_1+X_2 +\dots +X_n < n + 1)$ can be if $\Bbb{E}(X_i) = 1$?
QUESTION [20 upvotes]: Let $X_1, X_2, \dots, X_n$ be $n$ nonnegative independent identically
  distributed random variables with the same expectation:  $$\forall 1
\le i \le n: \quad \Bbb{E}(X_i) = 1$$
How small the probability $\Bbb{P}(X_1+X_2 +\dots +X_n < n + 1)$ can
  be?

Ideally, for the fixed $n$, we need to find an infimum of this probability over all possible distributions of $X_i$ and then see if the lower bound can ever be achieved.
The natural idea is to rewrite the probability in question in the following form:
$$\Bbb{P}\left(\dfrac{X_1+X_2 +\dots +X_n}{n} < 1 + \dfrac{1}{n}\right)$$
Here appears the average of $X_1, X_2, \dots, X_n$, but I'm not sure what can be done next.
I'm also interested in solving the generalized version of the problem for random variables that are not necessarily identically distributed.
Any suggestions, hints on how to approach this problem, and useful comments would be greatly appreciated.

REPLY [5 votes]: I believe the following is a counter-example to Jason's claim that $E_n$ is a convex function.  Consider $n=2$. Define: 
\begin{align}
X &=  \left\{ \begin{array}{ll}
1/2 &\mbox{ with prob $\frac{1}{2}$} \\
3/2  & \mbox{ with prob $\frac{1}{2}$} 
\end{array}
\right.\\
Y &= \left\{ \begin{array}{ll}
0 &\mbox{ with prob $1-\frac{1}{\theta}$} \\
\theta  & \mbox{ with prob $\frac{1}{\theta}$} 
\end{array}
\right.
\end{align}
where $\theta$ is chosen so that $\theta>1$ and  $(1-\frac{1}{\theta})^2 = 3/4$, that is, $\theta = 2(2+\sqrt{3})\approx 7.46$. Note that $X$ and $Y$ are nonnegative random variables with $E[X]=E[Y]=1$. Let $X_1, X_2$ be independent copies of $X$ and $Y_1, Y_2$ be independent copies of $Y$. Then: 
$$ P[X_1+X_2<3] = 1-P[X_1=X_2=3/2] = 3/4$$
$$ P[Y_1 + Y_2 < 3] = P[Y_1=Y_2=0] = (1-1/\theta)^2 = 3/4 $$
To show convexity fails, it suffices to define a random variable $Z$ that is a mixture of $X$ and $Y$, yet $P[Z_1+Z_2<3]>3/4$ for $Z_1, Z_2$ independent copies of $Z$. 
Define $Z$ by independently flipping a fair coin and choosing $Z=X$ if heads; $Z=Y$ else: 
$$ Z = \left\{ \begin{array}{ll}
0 &\mbox{ with prob $\frac{1}{2}(1-\frac{1}{\theta})$} \\
1/2  & \mbox{ with prob $\frac{1}{4}$} \\
3/2 & \mbox{ with prob $\frac{1}{4}$} \\
\theta & \mbox{with prob $\frac{1}{2}\frac{1}{\theta}$}
\end{array}
\right.$$
Let $Z_1, Z_2$ be independent copies of $Z$.  Then
\begin{align}
P[Z_1+Z_2<3] &= P[Z_1=0]P[Z_2 \in \{0, 1/2, 3/2\}]  \\
&+ P[Z_1=1/2]P[Z_2 \in \{0, 1/2, 3/2\}] +P[Z_1=3/2]P[Z_2 \in \{0, 1/2\}] \\
&=\frac{3}{8}+ \frac{\sqrt{3}}{4} \\
&> 3/4
\end{align}