# A scenario where Markov's inequality give better bound than Chebyshev?

I encountered a very interesting problem, given $$E[X]=2$$ and $$Var(X)=9$$, Find a $$t$$ satisfy $$t>2$$ where Markov's inequality gives a better bound than Chebyshev's inequality for $$P(X\geq t)$$.

My intuition is that Chebyshev bounds the absolute value of the r.v minus its mean, in this case, that is $$P(|X-E[X]|\geq t)=P(-t\geq X-E[X]\geq t)=P(-t+E[X]\geq X\geq t+E[X])$$, and a part of the bound is "wasted" on the minus and plus $$E[X]$$ part, but I'm not sure if I'm right, please show me how can I find my $$t$$, any help is appreciated.

Let $$X$$ be a positive random variable with $$E(X) = \mu = 2$$ and $$Var(X) = 9$$. Let $$t = 2$$
Now from Chebyshev’s inequality we have $$P(|X-2|>2) = P(X>4) \leq \frac{9}{4}$$ which doesn’t help us at all since $$\frac{9}{4} > 1$$.
However from the Markov’s Inequality we have $$P(X > 4) \leq \frac{1}{2}$$.
In general, if the variance is too high, the added advantage of using the variance in Chebyshev’s inequality can only be realized for larger bounds. So you can choose a $$t$$ small enough has to to make Markov inequality work better than Chebyshev’s inequality.