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I would like to solve the indeterminacy presents in this limit

$\displaystyle \lim_{x \rightarrow -\infty} x F(x)$,

in which $F(x)$ is a distribution function of a random variable $X$. The answer is obviously zero, but how to show the steps formally for any continuous distribution? Is it easy?

Thank you!

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up vote 1 down vote accepted

This is actually false. Consider: $$F(x) = \begin{cases} \left|\frac{1}{x+2}\right| & x < -4 \\ 1 - \frac{1}{32} x^2 & -4 \le x \le 0 \\ 1 & x > 0 \end{cases}$$

$F(x)$ is a cumulative distribution function (since it is continuously differentiable and nondecreasing, with limit to $-\infty$ at $0$ and limit to $+\infty$ at $1$), but $$\lim_{x \to -\infty} xF(x) = -1 \ne 0$$

You can similarly construct counterexamples that converge to any negative number or that don't converge at all (try $\displaystyle \frac{\arctan x}{\pi} + \frac{1}{2}$ for another counterexample).

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I really appreciate your counterexamples. You are right! The second function is the cumulative distribution function of an standard Cauchy and its heavy tails are associated with this result. I will explore situations in which the limit holds for another distributions. Thank you! – Emmanuel Ferreira Apr 8 '13 at 1:17

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