Limits regarding Cumulative Distribution Function when Expectancy is finite

Question

The exercise states: let ${F_X(x)}$ be the cumulative distribution function of a random variable. Prove that if ${X}$ has a finite expectancy (${E[X]<\infty}$), then: $${\lim_{x\to \infty}x(1-F_X(x))=0}$$ and $${\lim_{x\to -\infty}x(F_X(x))=0}$$

So I started using that for ${X}\in(-\infty,\infty)$

${E[X]} = \int_0^\infty (1-F_X(x))dx + \int_{-\infty}^0 F_X(x)dx$

I used integration by parts for both integrals, using ${u = 1-F_X(x)}$ in the first one and ${u = F_X(x)}$ in the second one.

$ = x(1-F_X(x))\bigg\rvert_0^\infty + \int_0^\infty xf_X(x)dx - xF_X(x)\bigg\rvert_{-\infty}^0+ \int_{-\infty}^0 xf_X(x)dx$

(Where ${f_X(x)}$ is the probability density function)

$ = x(1-F_X(x))\bigg\rvert_0^\infty - xF_X(x)\bigg\rvert_{-\infty}^0+ \int_{-\infty}^\infty xf_X(x)dx$

$ = x(1-F_X(x))\bigg\rvert_0^\infty - xF_X(x)\bigg\rvert_{-\infty}^0+ {E[X]} $

And since ${E[X]<\infty}$ then,

${(1-F_X(x))\bigg\rvert_0^\infty - xF_X(x)\bigg\rvert_{-\infty}^0 = 0}$

$${\lim_{x\to \infty}x(1-F_X(x))- \lim_{x\to -\infty}x(F_X(x))=0}$$

And there is where I'm stuck, I don't know how to conclude that both limits are zero. Is there a better way to prove it? Am I missing something?

Answer 1

Regarding your proof: at the end, note that when $x\to -\infty$, $x F_X(x)$ is negative, so the term $$-\lim_{x\to-\infty} xF_X(x)$$ is non-negative. As the first term is also non-negative, for the sum of both to be zero, both terms have to be zero, yielding the conclusion you seek. (I haven't checked the rest of your proof)

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A different argument.

Consider the random variable $Y=|X|\geq 0$. It is non-negative, and its cumulative distribution function is goven, for $y\geq 0$, by $$ F_Y(y) = \mathbb{P}\{ Y \leq y \} = \mathbb{P}\{ -y \leq X \leq y \} = F_X(y) - F_X(-y) \tag{1} $$ (assuming $X$ is continuous).

We also have by definition $\mathbb{E}[Y] = \mathbb{E}[|X|]<\infty$, and now we also can write $$ \mathbb{E}[Y] = \int_0^\infty \mathbb{P}\{Y > y\} dy = \int_0^\infty (1-F_Y(y)) dy\tag{2} $$ However, $y\mapsto (1-F_Y(y))$ is non-negative and non-increasing, so similarly to the case for series we can show that $$ \lim_{y\to\infty} y (1-F_Y(y)) = 0\tag{3} $$ Indeed, we can write $$ \int_0^{2y} (1-F_Y(t)) dt - \int_0^{y} (1-F_Y(t)) dt = \int_y^{2y} (1-F_Y(t)) dt \geq \int_y^{2y} (1-F_Y(y)) dt = y(1-F_Y(y)) \geq 0 $$ and conclude by the squeeze theorem, as the LHS converges to $\int_0^{\infty} (1-F_Y(t)) dt-\int_0^{\infty} (1-F_Y(t)) dt = 0$.

So we have, by (1) and (3), $$ 0 \leq y (1-F_X(y)) \leq y (1-F_X(y)) + yF_X(-y) = y(1-F_Y(y)) \xrightarrow[y\to\infty]{} 0 $$ showing that $$ \boxed{\lim_{y\to\infty} y (1-F_X(y)) = 0.} $$