How do we go about solving equations of this sort, where we need to find $x$ satisfying the below? Here $K$ and $\xi$ are known constants. Also, $\phi$ and $\mathbf{\Phi}$ are the standard normal PDF and CDF, respectively.

\begin{eqnarray*} \frac{\xi\left(K-x\right)^{2}\phi\left(\xi\left\{ K-x\right\} \right)}{\Phi\left(\xi\left\{ K-x\right\} \right)}+\left(K-x\right)\left[\frac{\phi\left(\xi\left\{ K-x\right\} \right)}{\Phi\left(\xi\left\{ K-x\right\} \right)}\right]^{2} & & =\\ \left\{ K-2x\right\} +\frac{1}{\xi}\frac{\phi\left(\xi\left\{ K-x\right\} \right)}{\Phi\left(\xi\left\{ K-x\right\} \right)}+\frac{\xi Kx\phi\left(\xi x\right)}{\Phi\left(\xi x\right)}+K\left[\frac{\phi\left(\xi x\right)}{\Phi\left(\xi x\right)}\right]^{2} \end{eqnarray*}

This comes up during the minimization of this problem.

\begin{eqnarray*} \underset{\left\{ x\right\} }{\min}\left[K\left\{ \xi x+\frac{\phi\left(\xi x\right)}{\Phi\left(\xi x\right)}\right\} +\left(K-x\right)\left\{ \xi\left(K-x\right)+\frac{\phi\left(\xi\left(K-x\right)\right)}{\Phi\left(\xi\left(K-x\right)\right)}\right\} \right] \end{eqnarray*}

Please note this can be shown to be convex and there is a separate thread on this. Convexity of Function of PDF and CDF of Standard Normal Random Variable

  • $\begingroup$ Maybe the most strait method is to plug in $\xi(K-x)$ .(i.e treat it as a new variable) to pdf and cdf of standard normal and then simplify. $\endgroup$
    – Deep North
    Oct 16, 2015 at 11:41
  • $\begingroup$ Please note that this will introduce an extra term in those parts where there is only $\xi$$x$ right now. $\endgroup$
    – texmex
    Oct 16, 2015 at 11:44
  • $\begingroup$ Yes, it more complicated than I had though :( $\endgroup$
    – Deep North
    Oct 16, 2015 at 12:13
  • 2
    $\begingroup$ Could you give some context---how do this equation arise? It might be simpler to attack the original problem directly? $\endgroup$ Oct 16, 2015 at 13:03
  • $\begingroup$ @kjetilbhalvorsen This arises as a result of an optimization problem. Let me add it to the original question, if that would be helpful. Please let me know if you need any further details. $\endgroup$
    – texmex
    Oct 18, 2015 at 3:39

1 Answer 1


Denote by $\lambda(x)$ the Inverse Mill's ratio $$\lambda(x) := \frac{\phi(x)}{1 - \Phi(x)} = \frac{\phi(-x)}{\Phi(-x)}.$$

Your objective can be rephrased as

$$\min_{x} K\bigg( \xi x+\lambda(-\xi x) \bigg) +\left(K-x\right)\bigg( \xi\left(K-x\right)+\lambda(-\xi\left(K-x \right))\bigg).$$ I don't see how convexity follows from the question you linked, since that question restricts the domain to $x \geq 0$ and focuses on the right term.

But, assuming you've convinced yourself of the convexity, such a function is easy to minimize numerically. The convexity gives you that any stationary point is a global minimizer. Here's how to do the optimization in Python.

from scipy import optimize
from scipy import stats

def inv_mills(x):
    return scipy.stats.pdf(-x)/scipy.stats.cdf(-x)

#define your constants K and xi
xi = 1.0; K = 2.0;

result = optimize.minimize_scalar(lambda x: K*(xi*x + inv_mills(-xi*x)) + (K-x)*(xi*(K-x) + inv_mills(-xi*(K-x))))

My example has the following result:

    nfev: 14
     nit: 10
 success: True
       x: 1.3308029569427138

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