I need to find a Random variable X such that $\lim_{t\to\infty} \frac{t^2}{\text{Var}X} \mathbb{P}(|X-\mathbb{E}X|\geq t) = 0$. I do understand that I can, for simplicity look at the case where the variance is 1 and the expected value is 0, simplifying it to $\lim_{t\to\infty} t² \mathbb{P}(|X|\geq t) = 0$. But I can't seem to find an example. Does anyone here know a solution?
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Concentration inequalities may help you : https://ai.stanford.edu/~gwthomas/notes/concentration.html â jvc Apr 22 '23 at 16:52
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Or take a gaussian rv : https://math.stackexchange.com/questions/74151/proof-of-an-estimate-for-the-tail-of-a-normal-distribution â jvc Apr 22 '23 at 16:54
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If you assume $\mathbb EX=0$ then all you'll need to find is a random variable $X$ having moments of order greater than $2$. Markov's inequality tells you that $$\mathbb P[|X|\geq t]\leq \frac{\mathbb E|X|^{2+\alpha}}{t^{2+\alpha}}$$ and hence $$\frac{t^2}{\text{Var}(X)}\mathbb P[|X|\geq t] \leq \frac{t^2}{\text{Var}(X)}\cdot\frac{\mathbb E|X|^{2+\alpha}}{t^{2+\alpha}}=C\cdot t^{-\alpha}\to 0$$ as $t\to\infty$.
You will probably know many such random variables.
Small Deviation
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For instance, any bounded random variable (with positive variance) will do. â PhoemueX Apr 23 '23 at 13:10