Let $X\sim\text{Norm}(0,1)$ be a random variable which is standard normally distributed. I have an exercise where I need to compute the limit \begin{equation} \lim_{x\rightarrow\infty}\frac{x}{\overline{F}_{X}(x)}\int_{x}^{\infty}\overline{F}_{X}(s)ds \end{equation} So I know that of course $\overline{F}_{X}(x)=1-F_{X}(x)$, where $F_{X}(x)=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}dx$. Also I know that $F_{X}(-x)=1-F_{X}(x)$. Plugging this into my limit equation would give \begin{equation} \lim_{x\rightarrow\infty}\frac{x}{\int_{-\infty}^{-x}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}dx}\int_{x}^{\infty}\int_{-\infty}^{-x}\frac{1}{\sqrt{2\pi}}e^{-\frac{s^{2}}{2}}dsdx \end{equation} I have been trying to figure out how to solve this for a while now. The best I can come up with is applying L'Hopital's rule twice to remove the integrals but I am not sure if that is simply just allowed. Could someone help me in showing how to start solving this exercise?

  • $\begingroup$ Why would applying L'Hopital's rule not be allowed? $\endgroup$ Commented May 24, 2021 at 19:30
  • $\begingroup$ I am unsure due to the boundaries of the integral. While taking the derivative I believe they would just be disregarded. Furthermore, due to $x$ being in the numerator as well wouldn't applying L'Hopital's rule not require the use of the product rule and complicate the limit even more? $\endgroup$
    – TK99
    Commented May 24, 2021 at 19:34
  • $\begingroup$ I understand how to approach the question now. I like both answers given but I believe that Paresseux Nguyen shows the desired method as in the first step of the exercise I had to prove the first limit property that is used. The only thing I don't understand is the integration calculations done in the last two steps of the answer. $\endgroup$
    – TK99
    Commented May 25, 2021 at 9:17
  • 1
    $\begingroup$ @TK99: In the answer you are referring to, the last two steps are (1) change variables $t\mapsto t-\tfrac{x^2}{2}$ (2) dominated convergence $\frac{x^2}{x^2+t}\xrightarrow{\xrightarrow\infty}1$, so in the limit you get $\int^\infty_0e^{-t}\,dt=1$. I just did not know your background, and since you already tried l'Hospital, I approach the problem the same way. I only had the secret ingredient (the Asymptotics of the tail of the normal distribution) that simplyfy things a lot. $\endgroup$
    – Mittens
    Commented May 25, 2021 at 11:29

2 Answers 2


Let $Z$ be a Gaussian (0,1) random variable, and denote by $\phi(x)=\frac{e^{-x^2/2}}{\sqrt{2\pi}}$ it’s density. The following inequalities (see Feller, W, Introduciton to Probability, Vol 1, 1968, Secion 7.1) will be very useful: $$ \Big(\frac1x-\frac{1}{x^3}\Big)\phi(x)\leq 1-F(x)\leq \frac{1}{x}\phi(x),\qquad x>0 $$ From these inequalities, we have that $$\frac{x(1-F(x))}{\phi(x)}\xrightarrow{x\rightarrow\infty}1$$

  1. By l'hospital rule $G(x):=\frac{1-F(x)}{x}\sim-\frac{\phi(x)}{1}\xrightarrow{x\rightarrow\infty}0$.
  2. Also, since $\int^\infty_0(1-F(s))\,ds=\frac{1}{2}\int^\infty_0P(|Z|>s)\,ds=\frac12 E[|Z|]<\infty$, we have that $H(x):=\int^\infty_x(1-F(s))\,ds\xrightarrow{x\rightarrow\infty}0$.

We will use L'Hospital tule to estimate the desired limit. From $$\frac{H(x)}{G(x)}\sim \frac{H'(x)}{G'(x)}=\frac{x^2(1-F(x))}{x\phi(x)+(1-F(x))}=\frac{x(1-F(x))\phi^{-1}(x)}{1+(1-F(x))x^{-1}\phi^{-1}(x)},$$ it suffices to check that $\lim_{x\rightarrow\infty}\frac{H'(x)}{G'(x)}$ exists. Since $1-F(x)\xrightarrow{x\rightarrow\infty}0$ and $x\phi(x)\xrightarrow{x\rightarrow\infty}0$,
$$\frac{1-F(x)}{x\phi(x)}\sim\frac{-\phi(x)}{\phi(x)-x^2\phi(x)}=-\frac{1}{1-x^2}\xrightarrow{x\rightarrow\infty}0$$ by l' Hospital rule. Putting things together, we obtain $$\frac{H'(x)}{G'(x)}=\frac{x(1-F(x))\phi^{-1}(x)}{1+(1-F(x))x^{-1}\phi^{-1}(x)}\xrightarrow{x\rightarrow\infty}1$$ Consequently $$\frac{x}{1-F(x)}\int^\infty_x (1-F(s))ds = \frac{H(x)}{G(x)}\xrightarrow{x\rightarrow\infty}1$$


Define $\phi (x) :=\frac{1}{\sqrt{2\pi}} e^{-x^2/2}$

Based on some standard inequalities for normal distribution, we have that : $$\lim_{x \rightarrow +\infty} \frac{ \overline{F}(x)}{ \frac{1}{x} \phi (x) }=1$$ So according to L'Hopital's rule, one can imply that : $$\lim_{x \rightarrow +\infty} \frac{ \int_{x}^{\infty} \overline{F}(s)ds}{ \int_{x}^{\infty} \frac{1}{s} \phi (s)ds }=1$$ So based on two previous limits, the desired limit converges if the following limit converges $$\lim_{x\rightarrow+\infty} \frac{x^2}{ \phi (x)} \int_{x}^{\infty} \frac{1}{s}\phi(s)ds$$ And this limit is indeed convergent and equal to $1$.

Indeed, we have: $$\begin{align} &\lim_{x\rightarrow+\infty} \frac{x^2}{ \phi (x)} \int_{x}^{\infty} \frac{1}{s}\phi(s)ds= \lim_{x\rightarrow+\infty} x^2e^{x^2/2} \int_{x}^{\infty} \frac{1}{s}e^{-s^2/2} ds \\ &=\lim_{x\rightarrow+\infty} x^2e^{x^2/2} \int_{x^2/2}^{\infty} \frac{1}{2t}e^{-t} dt = \lim_{x\rightarrow+\infty} \int_{0}^{\infty} \frac{x^2}{x^2+2t}e^{-t} dt \\ &=1 \end{align}$$

  • $\begingroup$ How were the last two steps taken? $\endgroup$
    – TK99
    Commented May 25, 2021 at 8:27

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