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Let $x$ in $\mathbb{R}$, I want to compute $x^²$, but I only have a random variable $X$ such that with very high probability, say $1-(1/C)$ for $C > 0$, $|X-x| \leq \epsilon$ holds.

Can anything be said about the value of $X \cdot X$?

GregS
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1 Answers1

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Note $|X^2-x^2|=|X-x||X+x|$. Hence if $|X-x| \leq \epsilon$ then $|X^2-x^2| \leq \epsilon(2|x|+\epsilon)$.

Ian
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  • Thanks! So if I assume that I have an one-sided error, i.e., $X-x \leq \epsilon$, can I also just multiply $(x+\epsilon)(x+\epsilon)$. For some reason, I think I made an independence assumption... – GregS Aug 15 '16 at 16:57
  • Actually, then the situation can be more complicated because $f(x)=x^2$ is not a monotone function. If $X-x \leq \epsilon$ then $X^2-x^2=(X-x)(X+x)$. So you get an inequality comparing $X^2-x^2$ to $\epsilon(X+x)$, but the direction of the inequality depends on the sign of $X+x$, which is not necessarily determined by the given inequality. (Of course it is if $x+\epsilon<0$.) – Ian Aug 15 '16 at 17:00