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I encountered a question in a test:

If $F(x)$ is a cumulative distribution function of a continuous non-negative random variable $X$, then find $\int_0^\infty (1 - F(x))\,\mathrm{d}x$

After a bit of pondering, I thought that the answer should depend upon the density of the random variable, so I checked the "none of these" option , but the correct answer was $\,E(X)$.So later I tried to work out the question properly.

If the density of random variable $X$ is $f(x)$ then it is necessary that $f(x) > 0$ and $\int_0^\infty f(x)\,\mathrm{d}x = 1$ Doing the integration by parts $$\left. x\Big(1-F(x)\Big)\right|_0^\infty - \int_0^\infty x\left(-\frac{\mathrm{d}}{\mathrm{d}x}F(x)\right)\,\mathrm{d}x$$ which reduces to $$ \lim_{x\to \infty} x\big(1\,- F(x)\big)\;+ \int_0^\infty xf(x) \, \mathrm{d}x $$

Now the $\int_0^\infty xf(x)\,\mathrm{d}x$ is clearly $E(X)$ but the limit is where my problem arises. Applying L'Hopital rule in the limit we have $$\lim_{x\to\infty}\frac{1-F(x)}{\frac{1}{x}} = \lim_{x\to\infty}\;x^2f(x)$$. Now is there any way to further reduce that limit to $0$ so that $E(X)$ is the correct answer or am I doing something wrong?

Dominik
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Jatin
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1 Answers1

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There are various mistakes in your approach.

First of all, the random variable doesn't need to have a density, so your whole approach doesn't work. But even if this were the case, you can't just apply L'Hospital because the resulting limit doesn't need to exist.

Instead, this exercise can be solved easily with Fubini's theorem. Simply note $$\int_0^\infty 1 - F(x) \, dx = \int_0^\infty E[I\{X > x\}] \, dx = E\Big[ \int_0^\infty I\{X > x\} \, dx \Big] = E[X].$$

Dominik
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