Optimization of a Convex Function involving Standard Normal CDF and PDF Could someone provide closed form solutions, if any, and steps to get there for the following optimization problem? Please note this function has been shown to be a convex function and hence a minimum and unique solution exists for this optimization.
\begin{eqnarray*}
\underset{\left\{ x\right\} }{\min}\left\{ \left[\theta x^{2}+\sigma\frac{\phi\left(\frac{\theta x}{\sigma}\right)}{\Phi\left(\frac{\theta x}{\sigma}\right)}\right]+\left[\theta\left(W-x\right)^{2}+\frac{\phi\left(\frac{\theta\left(W-x\right)}{\sigma}\right)}{\Phi\left(\frac{\theta\left(W-x\right)}{\sigma}\right)}\right]\right\} 
\end{eqnarray*}
Here, $\phi$ and $\mathbf{\Phi}$ are the standard normal PDF and CDF, respectively. 
\begin{eqnarray*}
\text{Also, }W\geq x,\; x\geq 0,\;\theta>0,\:\sigma^{2}\text{ is the variance of a Normal Distribution.}
\end{eqnarray*}
FOC of this function  seems to give below, which we can see is zero at $x=W/2$. With $\sigma >0 $ Am I missing something?
\begin{eqnarray*}
4x-2W-\frac{\theta x\phi\left(\frac{\theta x}{\sigma}\right)}{\sigma\Phi\left(\frac{\theta x}{\sigma}\right)}-\frac{1}{\sigma}\left[\frac{\phi\left(\frac{\theta x}{\sigma}\right)}{\Phi\left(\frac{\theta x}{\sigma}\right)}\right]^{2}+\frac{\theta\left(W-x\right)\phi\left(\frac{\theta\left(W-x\right)}{\sigma}\right)}{\sigma\Phi\left(\frac{\theta\left(W-x\right)}{\sigma}\right)}+\frac{1}{\sigma}\left[\frac{\phi\left(\frac{\theta\left(W-x\right)}{\sigma}\right)}{\Phi\left(\frac{\theta\left(W-x\right)}{\sigma}\right)}\right]^{2}
\end{eqnarray*}
How would this change, if the function (convex again) were as follows?
\begin{eqnarray*}
\underset{\left\{ x\right\} }{\min}\left\{ \left[\theta x^{2}+\frac{\sigma x\phi\left(\frac{\theta x}{\sigma}\right)}{\Phi\left(\frac{\theta x}{\sigma}\right)}\right]+\left[\theta\left(W-x\right)^{2}+\frac{\sigma\left(W-x\right)\phi\left(\frac{\theta\left(W-x\right)}{\sigma}\right)}{\Phi\left(\frac{\theta\left(W-x\right)}{\sigma}\right)}\right]\right\} 
\end{eqnarray*}
 A: I agree with @Dougal that there is no general closed-form solution.  However, for the special case of $\sigma=1$ there is.  Using Mathematica (v 10.1) define the following functions:
\[CapitalPhi][z_] := CDF[NormalDistribution[0, 1], z]
\[Phi][z_] := PDF[NormalDistribution[0, 1], z]
f[x_, \[Theta]_, \[Sigma]_, 
  w_] := \[Theta] x^2 + \[Theta] (w - 
      x)^2 + \[Sigma] \[Phi][\[Theta] x/\[Sigma]]/\[CapitalPhi][\
\[Theta] x/\[Sigma]] + \[Phi][\[Theta] (w - 
        x)/\[Sigma]]/\[CapitalPhi][\[Theta] (w - x)/\[Sigma]]

So f is your function of interest.  Differentiate it with respect to x:
g = D[f[x, \[Theta], \[Sigma], w], x]

If we set $x = w/2$ for that derivative we get

We see that when $\sigma=1$, the derivative is zero for all values of $\theta$.  There might be other special cases.
Addition:  Just another observation:  If $w$ and $\theta$ are kept fixed and $\sigma\to\infty$, then the result is $x = w/2 + 1/(2\pi)$.  If $w$ and $\sigma$ are kept fixed and $\theta\to\infty$, then $x = w/2$.  So it seems that $w/2 \le x \le w/2+1/(2\pi)$ for any values of $\theta$ and $\sigma$.
