I am looking to prove that the following function $G: [0, \infty) \to \mathbb{R}$ is decreasing over it's domain:
$$G(x) := \frac{x \phi(x)}{2 \Phi(x) – 1}$$ for $x \geq 0$. Here $\phi(x)$, and $\Phi(x)$ denote the PDF, and CDF (respectively) of a $\mathcal{N}(0, 1)$ random variable. As noted here, $G(0) = \frac{1}{2}$ by L'Hospital's rule.
Based on numerical plots, it does appear to be decreasing for all $x \in [0, \infty)$. Formally, I tried showing that $G^{\prime}(x) \leq 0$, but got the following complicated expression for $G^{\prime}(x)$:
\begin{equation}
G^{\prime}(x) = \frac{(2 \Phi(x) – 1)(\phi(x) + x \phi^{\prime}(x)) – 2 x \phi^{2}(x)}{(2 \Phi(x) – 1)^{2}}
\end{equation}
but I wasn't able to show this is non-negative. Could anyone please show how to obtain this using elementary methods and properties of the standard normal random variable?
Best Answer
Approach 1: to prove $G'(x) < 0$ for all $x > 0$:
Here is a proof using @fedja's idea:
(Actually, it is not difficult without using fedja's idea. But fedja's idea is better.):
We have $G(x) > 0$ for all $x > 0$. Let $$f(x) := \ln G(x) = \ln x + \ln \phi(x) - \ln(2\Phi(x) - 1).$$
We have $$f'(x) = \frac{1}{x} - x - \frac{1}{2\Phi(x) - 1}\cdot 2\phi(x) = \frac{1 - x^2}{x} - \frac{\mathrm{e}^{-x^2/2}}{\int_0^x \mathrm{e}^{-t^2/2}\,\mathrm{d} t}$$ where we have used $2\Phi(x) - 1 = 2 \int_0^x \phi(t)\, \mathrm{d} t$.
Using $\mathrm{e}^{-u} \ge 1 - u$ for all $u \ge 0$, we have $\mathrm{e}^{-x^2/2} \ge 1 - x^2/2 > 1 - x^2$ for all $x > 0$. Also, we have $\int_0^x \mathrm{e}^{-t^2/2}\,\mathrm{d} t < \int_0^x 1 \,\mathrm{d} t = x$ for all $x > 0$. Thus, we have, for all $x > 0$, $$f'(x) < \frac{\mathrm{e}^{-x^2/2}}{x} - \frac{\mathrm{e}^{-x^2/2}}{x} = 0.$$
Thus, we have $G'(x) < 0$ for all $x > 0$.
Dealing with $G'(x)$ for the case: $x = 0$:
Now, let us deal with $x = 0$. Since $\lim_{x\to 0} G(x) = 1/2$, we define $G(0) = 1/2$. Using L'Hopital rule, we have $$\lim_{x\to 0} \frac{G(x) - G(0)}{x - 0} = 0$$ that is $G'(0) = 0$ (see Remarks at the end for details).
Approach 2: to prove $G'(x) < 0$ for all $x > 0$:
We have $G(x) > 0$ for all $x > 0$. Let $$f(x) := \ln G(x) = \ln x + \ln \phi(x) - \ln(2\Phi(x) - 1).$$
We have $$f'(x) = \frac{1}{x} - x - \frac{1}{2\Phi(x) - 1}\cdot 2\phi(x).$$
(1) If $x \ge 1$, we have $f'(x) < 1/x - x \le 0$.
(2) If $x\in (0, 1)$, let $$g(x) := \frac{2\Phi(x) - 1}{1/x - x}f'(x) = 2\Phi(x) - 1 - \frac{x}{1 - x^2}\cdot 2\phi(x).$$ We have $$g'(x) = 2\phi(x) - \frac{x^2 + 1}{(1 - x^2)^2}\cdot 2\phi(x) - \frac{x}{1 - x^2}\cdot 2\phi(x) \cdot (-x) = - \frac{2x^2}{(1-x^2)^2}\cdot 2\phi(x) < 0.$$ Also, $\lim_{x\to 0} g(x) = 0$. Thus, we have $g(x) < 0$ for all $x \in (0, 1)$. Thus, we have $f'(x) < 0$ for all $x \in (0, 1)$.
Thus, $f'(x) < 0$ for all $x > 0$. Thus, $G'(x) < 0$ for all $x > 0$.
Remarks: Dealing with $G'(x)$ for the case: $x = 0$::
We have $$\frac{G(x) - G(0)}{x - 0} = \frac{\frac{x \phi(x)}{2 \Phi(x) - 1} - \frac12}{x} = \frac{2x\phi(x) - (2\Phi(x) - 1)}{2x(2\Phi(x) - 1)}.$$ Let $N(x) := 2x\phi(x) - (2\Phi(x) - 1)$ and $D(x) := 2x(2\Phi(x) - 1)$. We have $$N'(x) = 2\phi(x) + 2x\phi'(x) - 2\phi(x) = 2x\phi'(x)$$ and $$D'(x) = 2(2\Phi(x) - 1) + 2x \cdot 2\phi(x).$$ Note that $\lim_{x\to 0} N'(x) = 0$ and $\lim_{x\to 0} D'(x) = 0$. We have $$N''(x) = 2\phi'(x) + 2x\phi''(x)$$ and $$D''(x) = 4\phi(x) + 4\phi(x) + 4x \phi'(x) = 8\phi(x) + 4x\phi'(x).$$ We have $\lim_{x\to 0} N''(x) = 0$ and $\lim_{x\to 0} D''(x) = \frac{8}{\sqrt{2\pi}}$. We have $\lim_{x\to 0} \frac{N''(x)}{D''(x)} = 0$. Using L'Hoptial rule, we have $\lim_{x\to 0} \frac{N(x)}{D(x)} = \lim_{x\to 0} \frac{N''(x)}{D''(x)} = 0$.