determine whether an integral is positive

Question

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 $\tau$.

Answer 1

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 $\tau$

Answer 2

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.