What choices of $c, d$ would make $\int_{x=-\infty}^{x_0} (cx - d) \mathcal{N}(x; \mu, \sigma^2)\,dx $ positive?

Question

Consider the following integration: $$ I(x_0|c, d) = \int_{x=-\infty}^{x_0} (cx - d) \mathcal{N}(x; \mu, \sigma^2) \,dx $$

In this notation, $\mathcal{N}(x; \mu, \sigma^2)$ is a normal distribution: $$ \mathcal{N}(x; \mu, \sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2} } e^{ -\frac{(x-\mu)^2}{2\sigma^2} } $$

For a fixed and finite $x_0$ what choices of $d$ and $c$ would result in positive $I$?

Answer 1

So $\mathcal{N}(x; \mu, \sigma^2)$ is the density of a normal distribution. Let $F(x; \mu,\sigma^2)$ be the distribution function for this normal distribution.

First $$I(x_0\ |\ c,d) = c\int_{x=-\infty}^{x_0} x \mathcal{N}(x; \mu, \sigma^2)dx - d\int_{x=-\infty}^{x_0} \mathcal{N}(x; \mu, \sigma^2)dx.$$

You can check that

$$\frac{d}{dx}\mathcal{N}(x; \mu, \sigma^2) = -\frac{x-\mu}{\sigma^2}\mathcal{N}(x; \mu, \sigma^2).$$ Thus

\begin{align*} I(x_0\ |\ c,d) &= -c\sigma^2\int_{x=-\infty}^{x_0}-\frac{x-\mu}{\sigma^2} \mathcal{N}(x; \mu, \sigma^2)dx + (c\mu + d)\int_{x=-\infty}^{x_0} \mathcal{N}(x; \mu, \sigma^2)dx \\ &=-c\sigma^2\int_{x=-\infty}^{x_0} \frac{d}{dx}\mathcal{N}(x; \mu, \sigma^2)dx + (c\mu + d)F(x_0;\mu,\sigma^2)\\ &= -c\sigma^2 \mathcal{N}(x_0; \mu, \sigma^2) + (c\mu + d)F(x_0;\mu,\sigma^2). \end{align*} Both $\mathcal{N}(x; \mu, \sigma^2)$ and $F(x; \mu,\sigma^2)$ are known. For $I(x_0\ |\ c,d) > 0$ to hold, you want $$(c\mu + d)\frac{F(x_0;\mu,\sigma^2)}{\mathcal{N}(x_0; \mu, \sigma^2)} > c\sigma^2.$$ At this point, I would split into a couple of cases. If $c\mu+d>0$, then you want to choose $c, d$ such that $$\frac{F(x_0;\mu,\sigma^2)}{\mathcal{N}(x_0; \mu, \sigma^2)} > \frac{c\sigma^2}{(c\mu + d)},$$ and with the reverse inequality if $c\mu+d<0$.