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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.

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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.

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