Minimum of $F(\alpha x + \beta) + F(-\alpha x + \beta)$where $F$ is the cdf of a standard normal

Question

Let $F$ be the cdf of a standard normal distribution and $\alpha, \beta$ be real numbers with $\beta \leq 0$. Prove that $x\mapsto F(\alpha x + \beta) + F(-\alpha x + \beta)$ reaches its global minimum at $x=0$.

This question originates from the theory of statistical tests. I want to know if there's an elegant way to prove this. Of course the function is even, it is therefore sufficient to prove that it's increasing for positive $x$. To that end, one could study the sign of the derivative $$\alpha[f(\alpha x + \beta) - f(-\alpha x + \beta)]$$ where $f$ is the pdf of a standard normal, but that looks tedious.

Answer 1

It's really not that tedious. Assume without loss of generality that $\alpha>0$. Then we have $$f(\alpha x + \beta)-f(-\alpha x + \beta)=\frac1{\sqrt{2\pi}}(e^{-(\alpha x + \beta)^2/2}-e^{-(-\alpha x + \beta)^2/2})=\frac{e^{-(\alpha^2x^2+\beta^2)/2}}{\sqrt{2\pi}}(e^{-\alpha\beta x}-e^{\alpha\beta x})=-\frac{2e^{-(\alpha^2x^2+\beta^2)}}{\sqrt{2\pi}}\sinh(\alpha\beta x).$$ Since $\beta\le0$, this expression is $\ge0$ for all $x\ge0$.

One should probably also point out that the function is constant if $\alpha=0$, so the result becomes trivially true.

Answer 2

The most succinct observation is $f(u)=f(v)$ iff $u=\pm v$. We can't solve $\alpha x +\beta =\alpha x -\beta$ unless $\beta=0$, but $\alpha x+\beta=-\alpha x+\beta$ implies $x=0$.

Answer 3

Let $G(\alpha, \beta, x) = F(\alpha x+\beta) + F(-\alpha x +\beta)$. Notice that

$$G(\alpha, \beta, x) = G(\alpha, \beta, -x) = G(-\alpha, \beta, x).$$

We may hence assume without loss of generality that $\alpha, x\geq 0$. Now, we have

\begin{align} G(\alpha, \beta, 0) - G(\alpha, \beta, x) &= \Big(F(\beta) - F(-\alpha x +\beta) \Big) - \Big( F(\alpha x+\beta) - F(\beta) \Big) \\&= \int_{-\alpha x+\beta}^\beta\,f(t)\,dt - \int_{\beta}^{\alpha x+\beta}\,f(t)\,dt \end{align}

From the fact that $f$ is even and decreasing in $[0,+\infty)$, one can deduce that

$$ G(\alpha, \beta, 0) - G(\alpha, \beta, x) = \int_{|\beta|}^{\alpha x+|\beta|}\,f(t)\,dt - \int_{-\alpha x + |\beta|}^{|\beta|}\,f(t)\,dt\tag{$*$} $$

satisfies:

• $(*) = 0$ when $x=0$ or $\alpha = 0$ or $\beta = 0$; and
• $(*) < 0$ when $\alpha, x > 0$ and $\beta < 0$.

Hence, $(*)\leq 0$ so that in particular, $H(0)$ is a global minimum for $H(x) = G(\alpha, \beta, x)$. Moreover, the global minimum is strict whenever $\alpha \neq 0$ and $\beta < 0$.