Probability: Use Markov's inequality to prove a probability is above some bound?

Question

I have a random variable $X$ that arose from a Graph Theory problem, but I believe the details of the problem are not essential to this part, since the instructions explicitly say to use Markov's inequality. What might be important is that $0\leq X \leq m$ for some $m>1$. We have shown earlier that $E[X]\geq m/2$ and now are asked to use Markov's inequality to demonstrate that $P(X\geq 49m/100)\geq 1/51$.

I can't see how Markov's Inequality could be used to give a lower bound on a probability. I saw this question and answer: A lower bound on the probability that a variable is 3/2 times the expected value

I didn't follow some of the steps, but even so, in the end it appeared to me that the probability that was lower-bounded was just for the complement of the event in question. So as far as I can tell that solution would not help here.

Just for completeness I will write the inequality and do my best to try to use it:

$$P(X\geq 49m/100) \leq E[X]/(49m/100) \Rightarrow$$ $$49mp/100 \leq E[X]$$

where $p=P(X\geq 49m/100)$. I have heard rumor of a reverse Markov inequality in which I could compute

$$p \leq E[m-X]/(m-49m/100)$$ $$51mp/100 \leq E[m-X]$$

and I would imagine this further implies

$$51mp/100\leq m-E[X]$$ $$49mp/100\leq - E[X]$$ $$E[X]\leq 49mp/100$$

But clearly this result is paradoxical so I must not be fully understanding what the reverse Markov inequality really says, if it even is helpful here.

Answer 1

Let $Y=m-X$ so $Y$ is non-negative and $E[Y] \le \frac m2$

so can say from Markov's inequality that $P\left(Y \gt\frac{51m}{100}\right) \lt \frac{E[Y]}{51m/100} \le \frac{m/2}{51m/100}= \frac{50}{51}$

and thus $P\left(Y \le\frac{51m}{100}\right) \ge \frac{1}{51}$ and $P\left(X \ge\frac{49m}{100}\right) \ge \frac{1}{51}$

Answer 2

Let $A=\{X\geqslant x\}$ with $x\geqslant0$, then the pointwise inequality between random variables $$X\leqslant x\mathbf 1_{A^c}+m\mathbf 1_A$$ implies the inequality between expectations $$E(X)\leqslant xP(A^c)+mP(A)=x+(m-x)P(A)$$ Assuming furthermore that $x<m$ and $E(X)\geqslant y$, one gets $$P(A)\geqslant\frac{y-x}{m-x}$$ In your case, $x=0.49m$ and $y=0.5m$ hence, indeed, $$\frac{y-x}{m-x}=\frac1{51}$$