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Assume that we have a random variable $X$ that has a CDF $F$ (can be assumed to be continuous if required), and finite expectation, $\mathbb{E}(X) < \infty$. Then, does $$ \int_M^\infty (1-F(x))\,dx \rightarrow 0 $$ as $M\rightarrow\infty$?

This seems to be trivially clear (otherwise, the probability of $X$ exceeding $M$ would not go to 0), but I don't know how to show this formally. As the expectation is finite, the integral $\int_0^\infty (1-F(x))\,dx$ should be finite, but I am not sure if that really helps.

Did
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  • Yes; this follows from $x\mapsto 1-F(x)$ being monotone decreasing on $(0,\infty)$ and $\lim_{n\to\infty}(1-F(x))=0$. – Math1000 Aug 04 '16 at 09:20
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    Should the limit read $\lim_{x\rightarrow\infty}$? – user3825755 Aug 04 '16 at 09:30
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    Much simpler still: For every nonnegative function $G$ such that $\int_0^\infty G$ converges, one knows that $\int_M^\infty G\to0$ when $M\to\infty$ (this is almost the definition of the convergence of the integral). And you know that $\int_0^\infty(1-F)$ converges since you assumed that $E(X)$ is finite and, for every nonnegative $X$, $E(X)=\int_0^\infty (1-F)$. – Did Aug 04 '16 at 09:55
  • @Did Indeed, thanks. X is, however, not non-negative, but the argument should still hold as that integral should still be finite, right? – user3825755 Aug 04 '16 at 10:03
  • @user3825755 Yes, although you could also write the limit as $\lim_{n\to\infty}(1-F_(x_n))$ for an increasing sequence ${x_n}$ with $\sup_n x_n=+\infty$. – Math1000 Aug 04 '16 at 13:27
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    For $X$ real valued, check the convergence using $\int_0^\infty(1-F)=E(\max(X,0))\leqslant E(|X|)$. – Did Aug 04 '16 at 13:47
  • Ok, thanks, I see. I realized that one can also write the integral as $\int \mathbb{P}(X \geq x) dx = \mathbb{E}(X - M)I(X \geq M)$, where $I$ denotes the indicator function, which converges to 0 by the finiteness of the expectation of $X$. – user3825755 Aug 04 '16 at 13:54
  • Yes, same argument. Well done. Post this as an answer? – Did Aug 04 '16 at 13:55

1 Answers1

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Following from some comments to the question, note that the integral is equal to $\int_M^\infty\mathbb{P}(X\geq x)dx = \mathbb{E}((X-M)I(X\geq M))$ where $I$ denotes the indicator function. The last expression converges to $0$ by the assumed finiteness of the integral. Hence the result holds.

Did
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