# Solving equation that contains cdf and pdf of standard normal distribution

I have the following equation: $x = \frac{1-\Phi(x)}{a\phi(x)}$, where $\Phi$ is the cdf and $\phi$ is the pdf of the standard normal distribution.

How can one solve for $x$? Is there an analytical approach? Or can this only be done numerically?

These might be rather stupid questions (it's quite a while ago that I last was exposed to this), but any suggestions are appreciated. Thanks a lot!

• First of all, have you tried to plot for example the 2 curves $y=a x \phi(x)$ and $y=1-\Phi(x)$ and see where they intersect in order, at least, to have an idea of where the root/roots is/are situated? A very coarse sketch shows for example that there are no negative roots Jun 21 '17 at 15:28
• I am almost certain that there are no analytical approaches yielding an exact solution. Jun 21 '17 at 15:52
• @JeanMarie" How fast time goes by ... Mar 1 '20 at 13:46
• @callculus As said by Heinrich Heine: "Ich hab' euch im besten Juli verlassen / Und find' euch wieder im Januar ; / Ihr sasset damals so recht in der Hitze / Jetzt sei ihr gekhült und kalt sogar..." (Die Heimkehr, 1823) Mar 1 '20 at 14:18
• @JeanMarie Wow. I didn´t expect that. C´est incroyable. I´ve to admit I have no idea about a similar lyric in french. It gives my the opportunity to concern with german literature and maybe with french literature-but translated. Mar 1 '20 at 14:55

Use the series expansions. You can see here how they can be derived. Let $a=1$.

The equation is $x\cdot \phi(x)=1-\Phi(x)$. The equation can be multipied by $\sqrt{2\cdot \pi}$. It becomes

$$x\cdot e^{-x^2/2}=\sqrt{2\cdot \pi}-\left(0.5\cdot \sqrt{2\cdot \pi}+\int_0^x e^{-t^2/2} \,dt \right)$$

Using the series expansion the approximated equation is

$$x\cdot \left(1-\frac{x^2}{2}+\frac{x^4}{8}-\frac{x^6}{48}+\frac{x^8}{384}\right)$$ $$=\sqrt{2\cdot \pi}-\left(0.5\cdot \sqrt{2\cdot \pi}+\frac{x}{1\cdot 1}-\frac{x^3}{2\cdot 3}+\frac{x^5}{8\cdot 5}-\frac{x^7}{48\cdot 7}+\frac{x^9}{384\cdot 9}\right)$$

This equation can be solved with Wolfram alpha. The result is $\boxed{x=0.751781}$

To check the result use calculators for the pdf and the cdf. I get

$$x\cdot \phi(x)=0.751781\cdot 0.30073521=0.226087...\approx 22.609\%$$

$$1-\Phi(x)=1-0.77391=0.22609=22.609\%$$

It looks like that the approximation is fine.

• Thank you very much for your fast help. This is very interesting, and I will have a closer look at this. Quick follow-up question: The factors in the series are $1,2,8,48,384,...$ rather than $1,2,8,48,484,...$, or am I confused? (The context of my original question is a simple economic model, in which x is quantity in a profit function.)
– Carl
Jun 22 '17 at 7:12
• @Carl No, you´re not confused. Good catch. Luckely the typo was in the last summand. So it doesn´t change the result much. I´ve edited the answer, including the WA-Link. .If you find the answer helpful you can accept it Jun 22 '17 at 10:09