I'm wondering whether I can use Markov's inequality to reach the following statement:

Given Markov's inequality on a non-negative random variable X:

$ P[X\geq a] \leq \frac{E[X]}{a}$

We can do the following:

$ P[X<a] = 1 - P[X\geq a]$

Thus, we can say:

$ P[X<a] \geq 1 - \frac{E[X]}{a}$

Now, since we know that $ 0\leq P[X<a] \leq 1 $, we can conclude that:

$E[X] \leq a$

What do you guys think?

Thanks in advance for your thoughts and ideas.

  • $\begingroup$ Well, I forgot to say $a\geq0$ (Given by Markov's inequality ) $\endgroup$ – pkenneth81 Apr 3 at 19:51
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    $\begingroup$ Let $a\downarrow 0$. This "proves" that $X$ is identically zero. You mixed up the direction of the inequality. The only meaningful bound you can extract from the third line is $\mathbb{E}X \ge 0$, which is already implied by non-negativity. $\endgroup$ – jth Apr 3 at 19:51
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    $\begingroup$ @htennek2k "Now, since we know that 0≤P[X<a]≤1, we can conclude that" I don´t see how you come to the next line with this reasoning. $\endgroup$ – callculus Apr 3 at 20:01
  • $\begingroup$ Thanks for your reply. Any probability is lower bounded by zero and upper bounded by 1. Then, since we have $P[X<a] \geq 1- B$, where $B=\frac{E[X]}{a}$, we can observe that B cannot be greater than 1 (otherwise we could get negative probability). Thus, with all the steps: we have $ 1 \geq 1- \frac{E[X]}{a}$. Then solving the inequality for $E[X]$ yields the result I claim. $\endgroup$ – pkenneth81 Apr 3 at 23:04
  • $\begingroup$ By writing my comment above, I realized that the weakness of my reasoning, is that I don't have equality, but a simple lower bound on $P[X<a]$. Thus, as it naturally is, can be greater than a negative value with no problem. $\endgroup$ – pkenneth81 Apr 3 at 23:07

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