# Comparing the Markov and Chebyshev's inequalities

This question was given to me as a review for an upcoming exam.

Let X be a positive random variable with mean 3 and variance $$\frac{1}{4}$$. Use Chebyshev's and Markov's inequalities to obtain the probability that $$X \geq 6$$.

Also give an example of such an X

My work thus far:

Chebyshev's

$$P(|X-\overline{x}| \geq a) \leq \frac{\sigma^{2}}{a^2}$$

$$P(X -3 \geq6-3) \leq \frac{\frac{1}{4}}{3^2} = \frac{1}{36}$$

Markov's

$$P(X \gt a) \leq \frac{E[X]}{a}$$

$$P(X \gt 6) \leq \frac{3}{6}= \frac{1}{2}$$

I feel as though I've made an error somewhere, could someone help me figure out where?

Your expression for Chebyshev's two sided inequality perhaps should have been $$P(|X-\mu| \geq a) \leq \frac{\sigma^{2}}{a^2}$$ and would have led to an upper bound of $$\frac1{36}$$, slightly higher than you have. What you have actually used looks like a one sided version $$P(X-\mu \geq a) \leq \frac{\sigma^{2}}{\sigma^{2}+a^2}$$ for $$a \gt 0$$, and your result of $$\frac1{37}$$ as an upper bound would then appear to be correct. There would be equality if $$P(X=6)=\frac1{37}$$ and $$P\left(X=\frac{35}{12}\right)=\frac{36}{37}$$
Your use of Markov's inequality for a non-negative random variable and $$a \gt 0$$ and your result of $$\frac12$$ as an upper bound both appear to be correct. There would be equality if $$P(X=6)=\frac1{2}$$ and $$P\left(X=0\right)=\frac{1}{2}$$, though this would have a variance of $$9$$, rather higher than that given in the question, and showing that taking the variance into account here gives a tighter bound through Chebyshev's one sided inequality