# How small the probability $\Bbb{P}(X_1+X_2 +\dots +X_n < n + 1)$ can be if $\Bbb{E}(X_i) = 1$?

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.

• With the distribution $P[X=0]=1-\frac{1}{n+1}, P[X=n+1]=\frac{1}{n+1}$ we get $$P[X_1 + ... + X_n < n+1] = \left(1-\frac{1}{n+1}\right)^n$$ This bound is the lowest possible for $n=1$. Also $(1-1/(n+1))^n \approx 1/e$ for large $n$. – Michael May 4 '17 at 2:32
• In general you can argue that you can restrict attention to distributions for $X_i$ that have support over $[0, n+1]$. – Michael May 4 '17 at 2:38
• @Michael Thanks for your comment! Could you please elaborate more on why this expression is the lowest possible bound for $n=1$? Also, I didn't quite understand what did you mean in your second comment. – Ramil May 4 '17 at 14:58
• I expand my comments in my answer below. – Michael May 4 '17 at 17:36
• See the discussion in mathoverflow.net/questions/187938/lower-bound-for-prx-geq-ex/… The best lower bound for arbitrary $n$ is still unknown. – fedja May 10 '17 at 0:48

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, $c_n \geq \frac{1}{n+1}$.

• Thank you! Now it's more clear what you meant in comments. – Ramil May 4 '17 at 20:48
• The lower bound $P[X_1 + ... + X_n<n+1]\geq 1/(n+1)$ is tight if we consider all nonnegative random vectors $(X_1, ..., X_n)$ with $E[X_i]=1$ for all $i \in \{1, ..., n\}$, but likely is not tight if we impose the i.i.d. assumption. – Michael May 8 '17 at 18:13

EDIT: Michael's upper bound is locally optimal, not globally as I had originally stated. Specifically, there is some neighborhood $N$ of the distribution $\alpha:=\frac{n}{n+1}\delta_0+\frac1{n+1}\delta_{n+1}$ (in the weak topology) such that $\mathbb P(X_1+\ldots+X_n<n+1)<\mathbb P(Y_1+\ldots+Y_n<n+1)$ whenever $(X_i),(Y_i)$ are iid with $X_1\sim\alpha$ and $Y_1\sim\mu\in N\setminus\{\alpha\}$.

To see this, let $S$ be the set of probability measures $\mu$ on $[0,\infty)$ such that $\int_0^\infty x\,\mu(dx)=1$, and for $\mu\in S$ define

$$E_n[\mu]:=\int_{\sum_{i=1}^nx_i<n+1}\mu(dx_1)\ldots\mu(dx_n).$$

Note that if $(X_i)$ are iid nonnegative random variables with $\mathbb E[X_1]=1$, and $\mu$ is the law of $X_1$, then $E_n[\mu]=\mathbb P(X_1+\ldots+X_n<n+1)$. Let $\alpha=\frac{n}{n+1}\delta_0+\frac1{n+1}\delta_{n+1}$, i.e. $\alpha$ is the distribution of the random variable Michael defines for the upper bound.

Observe that $S$ is convex, so given arbitary $\mu\in S\setminus\{\alpha\}$ the function $\Phi_n(t):=E_n[(1-t)\alpha+t\mu]$ is well-defined for $t\in[0,1]$. The formula for $\Phi_n(t)$ is complicated, but we do not need much of it:

$$\Phi_n(t)=c_0+\left(\sum_{i=1}^n\int_{\sum_jx_j<n+1}\mu(dx_i)\prod_{k\neq i}\alpha(dx_k)-\int_{\sum_jx_j<n+1}n\prod_{k=1}^n\alpha(dx_k)\right)t+\sum_{i=2}^nc_it^i$$

where the $c_i$ are constants depending on $\mu$ but not $t$. This yields

\begin{align*} \Phi_n'(0) &=n\left(\int_{\sum_j x_j<n+1}\mu(dx_1)\prod_{k=2}^n\alpha(dx_k)-\int_{\sum_j x_j<n+1}\prod_{k=1}^n\alpha(dx_k)\right)\\ &=n\left[\mathbb P\left(Y_1+\sum_{i=2}^nX_i<n+1\right)-\mathbb P\left(\sum_{i=1}^nX_i<n+1\right)\right] \end{align*}

where $Y_1$ is a random variable with law $\mu$, independent of the iid random variables $X_i$ which have law $\alpha$. Note that since $\mu\neq\alpha$ and $Y_1\ge0$, we have $\mathbb P(Y_1<n+1)>1-\frac1{n+1}$ and hence

$$\mathbb P\left(Y_1+\sum_{i=2}^nX_i<n+1\right)=\mathbb P(Y_1<n+1)\mathbb P(X_1=0)^{n-1}>\left(1-\frac1{n+1}\right)^n$$ and thus $\Phi_n'(0)>0$. This implies that there exists $\delta>0$ such that $\Phi_n(0)<\Phi_n(t)$ for all $t\in(0,\delta)$; since $\mu$ was arbitrary, it follows that $E_n$ has a strict local minimum at $\alpha$.

EDIT: I had originally written here that $E_n$ is convex and conclude the result. However, as Michael points out in the comments, this assertion is not true. It may still be possible to continue with this argument and conclude that $\alpha$ is the global minimizer of $E_n$, but at this point the best I have is that $\alpha$ is a local minimizer.

After having seen fedja's comment and reading related material, I find it unlikely anyone will be able to solve it the generalized version here as it is an open research problem.

• Interesting approach. However, I am not sure that your function $E_n$ is convex, see my counter-example below for the case $n=2$. – Michael May 10 '17 at 17:25
• However, if you assume $\{X_i\}_{i=1}^n$ are mutually independent but not necessarily identically distributed, then $P[X_1 + ... + X_{n-1} + X_n < n+1] = \int_{y=0}^{\infty} P[X_n<n+1-y]dG(y)$ where $G$ is the CDF of $X_1 + ... + X_{n-1}$. This is linear in the distribution for $X_n$. To minimize a linear function over the convex set of all distributions on nonnegative random variables with mean 1 (and we can restrict to those with support over $[0,n+1]$), we can just consider the extreme distributions. I believe the only extreme distributions are those with 1 or 2 point masses. – Michael May 10 '17 at 18:17
• Ah, damn. The induction argument I was thinking of only shows $E_n[t\mu+(1-t)\nu]\le tE_{n-1}[\mu]+(1-t)E_{n-1}[\mu]$, which is not enough. I will leave up my answer since I think a variational approach could still be promising but as it stands this answer is not correct. – Jason May 10 '17 at 18:19
• That sounds promising, but without further analysis I think that only shows that for fixed $X_1,\ldots,X_{n-1}$, the minimizing $X_n$ is two point masses. Still, I think that argument is worth exploring. – Jason May 10 '17 at 18:22
• On your answer: It is valuable to prove local optimality even if not globally optimal. The intuition I had for my upper bound distribution was to take an extreme distribution and hope for some form of concavity (rather than convexity). In general I would be very interested to see more of the kind of analysis you give above, if you have examples where it works nicely let me know. For the current problem, another approach would be Lagrange multipliers and KKT conditions. – Michael May 10 '17 at 18:25

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}