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Let $X_{1}$, $X_{2}$, $X_{3}$, $X_{4}$, $X_{5}$ and $X_{6}$ be real-valued random variables that have the same probability distribution with finite moments, and they are independent. Does anyone know how to apply Markov inequality to show that $$ P\left(X_{1}+X_{2}+X_{3}+X_{4}+X_{5}+X_{6}\le3\right)\le2P\left(X_{1}\le1\right)? $$

Thanks for any helpful answers!

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All that's given is that the X's have the same distribution; however, we don't know if the first (absolute) moment of X exists, nor do we know whether the X's are independent. I fail to see how you can even apply Markov's inequality in this context. –  Mico Sep 24 '12 at 0:54
    
Dear Mico, I have added the assumptions accordingly. –  h636 Sep 24 '12 at 1:05
    
@h636: And you should not have. –  Did Sep 24 '12 at 7:32

2 Answers 2

Edit: This answer targets an earlier version of the question.

If $X_1 > 1$ and $X_2 > 1$ then $X_1+X_2+X_3+X_4 > 2$. Hence $$\Pr[X_1+X_2+X_3+X_4 > 2] \geq \Pr[X_1>1 \text{ and } X_2>1]. $$ The union bound shows that the negation of the event "$X_1 > 1$ and $X_2 > 1$" has probability at most $$\Pr[X_1 \leq 1] + \Pr[X_2 \leq 1] = 2\Pr[X_1 \leq 1].$$ Therefore $$\Pr[X_1+X_2+X_3+X_4 \leq 2] \leq 2\Pr[X_1 \leq 1]. $$ Note we didn't need to use the fact that $X_1$ and $X_2$ are independent.

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Dear Yuval, your proof was correct for 4 variables, but I was expecting a proof by Markov inequality. –  h636 Sep 24 '12 at 2:50
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Well, I guess you're on your own now... –  Yuval Filmus Sep 24 '12 at 4:19

The main task here is to guess what are the correct hypotheses of the statement to be proven... Let us assume that the random variables $X_k$ are (1) identically distributed, (2) almost surely nonnegative, (3) not necessarily independent.

Introduce the event $A=[X_1+\cdots+X_6\leqslant3]$ and the random variable $N=\sum\limits_{k=1}^6\mathbf 1_{X_k\leqslant1}$.

Then $\mathrm E(N)=6\mathrm P(X_1\leqslant1)$ and $A\subseteq[N\geqslant3]$ since $X_1+X_2+\cdots+X_6\geqslant N$ pointwise. Thus, $\mathrm P(A)\leqslant\mathrm P(N\geqslant3)\leqslant\frac13\mathrm E(N)=2\mathrm P(X_1\leqslant1)$, where $\mathrm P(N\geqslant3)\leqslant\frac13\mathrm E(N)$ is Markov inequality.

This proves the desired inequality. The same approach proves more generally that $$ \mathrm P(X_1+\cdots+X_n\leqslant k)\leqslant\frac{n}{n-k}\,\mathrm P(X_1\leqslant1). $$ One sees that neither independence nor finite moments are required. On the other hand, almost sure nonnegativity is crucial. To see this, assume for instance that $\mathrm P(X_k=-7)=p=1-\mathrm P(X_k=2)$ for every $k$ and that the random variables $X_k$ are independent.

Then $[N\geqslant1]\subseteq A$ hence $\mathrm P(A)\geqslant1-\mathrm P(N=0)$. Furthermore, $\mathrm P(X_1\leqslant1)=p$ and $\mathrm P(N=0)=(1-p)^6=1-6p+o(p)$ when $p\to0^+$, hence $\mathrm P(A)\geqslant1-(1-p)^6=6p+o(p)\gt3p=3\mathrm P(X_1\leqslant1)$ for every $p$ small enough.

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