Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have been working on the following problem for some time now, to no avail, so any advice whatsoever is greatly appreciated.

I would like to optimize

$\min_{0 \leq r \leq 1} \ \ L = Prob(y>0)*(k) + (1-Prob(y>0))*(1-r)*l$


$L = (1-F_Y (0))*k+F_Y (0)*(1-r)*l$

with respect to $r$, where $F_Y (y)$ is the cdf of $y \sim \mbox{ i.i.d. } \mathcal{N}[\mu+r,\sigma^2]$

So in a sense, the minimization looks for the "optimal distribution" of $y$ out of a set of distributions which vary only with respect to the mean.

I have worked on finding the derivative of $L$ with respect to $r$, which (if I haven't mad a mistake) comes down to

$\frac{d L}{d r} = - \frac{l}{2} - \frac{1}{\sqrt{\pi}}e^{-(-\frac{\mu+r}{\sqrt{2 \sigma^2}})^2}(k-(1-r)l) - l*\frac{1}{\sqrt{\pi}}\int\limits_{-\infty}^{-\frac{\mu+r}{\sqrt{2 \sigma^2}}} {e^{-t^2}dt} $

Since there is no closed form of the error function's integral, I am kind of stuck with how to solve for $r$. Any kind of help would be greatly appreciated - please apologize in advance any obvious blunders...

Edit: I know that $I \equiv \int\limits_{-\infty}^{\infty} {e^{-t^2}dt}=\sqrt{\pi}$, and $\int\limits_{-\infty}^{0} {e^{-t^2}dt}=\frac{\sqrt{\pi}}{2}$. Would it be possible to somehow find a similar solution with $-\frac{\mu+r}{\sqrt{2 \sigma^2}}$ as the upper bound of the integral?

share|cite|improve this question

"Since there is no closed form of the error function's integral"

Yes there is, erf is in the newer C standards. You can't get much more closed form than that. It is right there alongside cos and exp.

share|cite|improve this answer
Thanks binn for your comment! But would that really help me in finding the optimal r? Or is it just a numerical solution? Note that I've got r in the integral's boundary. Could you please give me some more hints here? Thanks!! – jush Sep 3 '12 at 22:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.