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For the standard normal density function $f$, show that $f'(x) + xf(x) = 0$. Hence show that for $x>0$, $$\frac{1}{x}-\frac{1}{x^3} < \frac{1-F(x)}{f(x)} < \frac{1}{x}-\frac{1}{x^3}+\frac{3}{x^5}.$$

The first part is fine. The inequality is what's bugging me.

$F(x)$ is the CDF of the standard normal distribution where $f(x)$ is the PDF and $F^\prime(x)=f(x)$, so it's given by $$F(x) = \frac{1}{\sqrt{2\pi}}\int_{0}^{x}e^{-\frac{x^2}{2}}dx.$$

This is the integral of a Gaussian function and as far as I know it cannot be computed if there isn't an infinite limit somewhere (correct me if I'm wrong).

I have no idea where to start. I tried messing around with limits but to no avail. This probably has something to do with some kind of Analysis theorem, and Analysis is not my strong point.

Any help would be much appreciated.

sandbag66
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1 Answers1

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The question is wrong. The gvien lower-limit-is-$0$ definition of $F(x)$ implies $1-F(x)\approx\frac12$ for large $x>0$, whence $\lim_{x\to\infty}\frac{1-F(x)}{f(x)}=\infty$, contradicting the desired bounds. The question only makes sense if the definition uses a lower limit of $-\infty$, as per the CDF. My answer reflects that.

You want to bound $\frac{\int_x^\infty f(t)dt}{f(x)}$. The desired results are refinements of a famous upper bound viz.$$\int_x^\infty f(t)dt<\frac1x\int_x^\infty tf(t)dt=\frac{f(x)}{x}.$$But we'll adapt a more complicated proof of that result, which goes as$$f(x)\left(\frac1x\right)=\int_x^\infty f(t)\left(1+\frac{1}{t^2}\right)dt>\int_x^\infty f(t)dt.$$On the one hand,$$f(x)\left(\frac1x-\frac{1}{x^3}\right)=\int_x^\infty f(t)\left(1-\frac{3}{t^4}\right)dt<\int_x^\infty f(t)dt.$$On the other hand,$$f(x)\left(\frac1x-\frac{1}{x^3}+\frac{3}{x^5}\right)=\int_x^\infty f(t)\left(1+\frac{15}{t^6}\right)dt>\int_x^\infty f(t)dt.$$In fact, you can see each extra term as cancelling out a term in the integrand only to replace it with another. See if you can work out how the bounds continue (and you'll notice they alternate between being upper and lower bounds on $\frac{\int_x^\infty f(t)dt}{f(x)}$).

J.G.
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  • I understand your proof, but I don't understand why we're using it? And why have the limits of the integral changed? – sandbag66 Nov 02 '19 at 22:07
  • @sandbag66 Since $F(x)=\int_{-\infty}^xf(t)dt$, $1-F(x)=\int_x^\infty f(t)dt$. If you use an Ansatz $1-F(x)=f(x)\sum_{n\ge0}\frac{a_n}{x^{2n+1}}$, the partial sums give you these approximations. You can then use $f(x)g(x)=\int_x^\infty(-fg)^\prime dt=\int_x^\infty f(t)(tg(t)-g^\prime(t)) dt$ (provided $fg\pm\infty)=0$) to compare $fg$ to $\int_x^\infty fdt$, showing the partial sums alternate between over-/underestimating it. – J.G. Nov 02 '19 at 22:28
  • But I want limits between 0 and x? Sorry for being so obtuse lol – sandbag66 Nov 03 '19 at 00:11
  • @sandbag66 No you don't, because then it would be wrong, as the CDF starts at $-\infty$. I'll correct your post. – J.G. Nov 03 '19 at 07:20
  • It's for x>0.... read the quoted question please. – sandbag66 Nov 03 '19 at 11:31
  • @sandbag66 I've added a paragraph to my answer. – J.G. Nov 03 '19 at 11:38