# Using Markov's Inequality to Derive a Conclusion about random variable

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

Thus, we can say:

$$P[X

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

$$E[X] \leq a$$

What do you guys think?

• Well, I forgot to say $a\geq0$ (Given by Markov's inequality ) – pkenneth81 Apr 3 at 19:51
• 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. – jth Apr 3 at 19:51
• @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. – callculus Apr 3 at 20:01
• 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. – pkenneth81 Apr 3 at 23:04
• 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. – pkenneth81 Apr 3 at 23:07