determine whether an integral is positive Given a standardized normal variable $X\sim N\left(0,1\right)$, and constants $ \kappa \in \left[0,1\right)$  and  $\tau \in \mathbb{R}$, I want to sign the following expression:
\begin{equation}
\int_{-\infty}^\tau \left(X-\kappa \tau \right) \phi(X)\text{d}X
\end{equation}
where $\phi$ is the PDF of $X$.  Any comment will be appreciated. I would at least want to know if the sign of the expression can be determined given the information, or whether it hinges on the value of $\tau$.
 A: Note that 
$$ \int_{-\infty}^\infty (X-\kappa\tau)\phi(X)\,\mathrm dX=E[X-\kappa\tau]=-\kappa\tau$$
and for $\tau>0$
$$ \int_{\tau}^\infty (X-\kappa\tau)\phi(X)\,\mathrm dX\ge\int_{\tau}^\infty (1-\kappa)\tau\phi(X)\,\mathrm dX>0.$$
So for $\tau>0$ your expression is positive.
A: Let the function be $g$. Evaluating $g(\kappa =1)$, we to obtain 
\begin{equation*}
g=\int_{-\infty }^{\tau }\left( X-\tau
\right) \phi \left( X\right) \text{d}X \leq 0
\end{equation*}
the inequality follows since $X-\tau \leq 0$ for all $X \in (-\infty,\tau)$. Similarly, for $\kappa =0$, 
\begin{equation*}
g=\int_{-\infty }^{\tau }X\phi \left(X\right) \text{d}X \leq 0
\end{equation*}
since
\begin{equation*}
\int_{-\infty }^{\tau }X\phi \left( X\right) dX<\int_{-\infty }^{\infty
}X\phi \left( X\right) \text{d}X=0
\end{equation*}
and finally taking the derivative of $g$ with respect to $\kappa$
\begin{equation*}
\frac{\partial g}{\partial \kappa }=-\tau \int_{-\infty }^{\tau }\phi \left(
X\right) \text{d}X \leq 0
\end{equation*}
since, if it depends on $\tau$, then we will take $g\left( \kappa
=1\right) $ to find that it is negative. Therefore, 
\begin{equation*}
g = \int_{-\infty }^{\tau } \left( X-\kappa \tau \right) \phi \left(X\right) \text{d}X \leq 0
\end{equation*} 
and most importantly, this inequality does not depend on $\tau$
