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.
Best Answer
Define $c_n$ as the infimum value of $P[X_1 + ... + X_n<n+1]$ over all distributions for i.i.d. nonnegative random variables $\{X_i\}$ with $E[X_i]=1$. Here I prove the simple upper and lower bounds: $$ \frac{1}{n+1} \leq c_n \leq (1-\frac{1}{n+1})^n \quad, \forall n \in \{1, 2, 3, ...\} $$ Notice that the upper and lower bounds meet when $n=1$, so $c_1=1/2$.
Achievability (upper bound):
Consider the nonnegative random variables $$ X_i = \left\{ \begin{array}{ll} n+1 &\mbox{ , with prob $\frac{1}{n+1}$} \\ 0 & \mbox{ , with prob $1-\frac{1}{n+1}$} \end{array} \right.$$ These have $E[X_i]=1$ and: $$P[X_1 + ... + X_n<n+1] = P[\mbox{all $X_i$ are zero}] = (1-\frac{1}{n+1})^n$$ Hence, $c_n \leq (1-\frac{1}{n+1})^n$.
Lower bound:
Let $\{X_i\}$ be any (possibly dependent and non-identically distributed) nonnegative random variables with $E[X_i]=1$. By the Markov inequality: $$ P[X_1 + ... + X_n\geq n+1] \leq \frac{E[X_1+...+X_n]}{n+1} = \frac{n}{n+1}$$ and hence $P[X_1 + ... + X_n < n+1] \geq \frac{1}{n+1}$. Hence, $c_n \geq \frac{1}{n+1}$.