Construction of non-negative random variables with fixed expectation to maximize probability

Question

I want to construct independent non-negative random variables $X_1,X_2,X_3$ and $X_4$ such that $\mathbb{E}(X_n)=n$ and then maximize the probability: $$\mathbb{P}\left(X_1+X_2+X_3+X_4 \geq 11\right)$$

By Markov's inequality this must be smaller or equal to 10/11 but I would not know how to actually maximize this.

Answer 1

Thanks to Axolotl's comment, we see this is related to Feige's Conjecture. In Feige's Paper he references a much older conjecture (Conjecture 2) due to Samuels that would answer your problem. It says

Let $X_1\ldots X_n$ be independent non-negative random variables with means $\mu_1\le\mu_2\le\ldots \le \mu_n.$ Then for every $\lambda>\sum_k\mu_k$ there is some $i$ with $1\le i\le n$ such that $P(\sum_kX_k\ge \lambda)$ is maximized when the $X_j$ are distributed as follows: 1) For $j<i,$ $X_j=\mu_j$ with probability $1$. 2) For $j\ge i$, $X_j$ takes the value $\lambda - \sum_{k=1}^{i-1} \mu_k$ with probability $\frac{\mu_j}{\lambda - \sum_{k=1}^{i-1} \mu_k}$ and $0$ otherwise.

Fortunately, he also claims that Samuels has proven this for $n\le 4$ in these two papers. So it appears this you can get the answer for finding the best value of $i$ above, though the proof might take a bit to work through.