Is $\frac{d}{dx} \ln(\phi(x))$ decreasing where $\phi$ is the Gaussian cumulative distribution function?

Question

I was plotting the function $x \mapsto \frac{d}{dx} \ln(\phi(x))$ and it seams like this is a decreasing function. But somehow I cannot prove this. Here, $\phi$ is the Gaussian cumulative distribution function, i.e. $$ \phi(x) = \int_{-\infty}^x \gamma(t) \, dt $$ and $\gamma$ is the Gaussian. Do you have any suggestion how to show this?

Answer 1

We have $\displaystyle\frac{d}{dx}\ln(\phi(x))= \frac{\phi'(x)}{\phi(x)} = \frac{\gamma(x)}{\phi(x)},$ where $\gamma$ is the standard normal PDF, $$\gamma(x) = \frac1{\sqrt{2\pi}}e^{-x^2/2}$$ Differentiating, we get

$$\gamma'(x) = -\frac1{\sqrt{2\pi}}xe^{-x^2/2}=-x\gamma(x)$$

Hence,

$$\left(\frac{\gamma}{\phi}\right)' = \frac{\phi(x)\gamma'(x)-\gamma(x) \phi'(x)}{\phi^2(x)}= \frac{\phi(x)\gamma'(x)-\gamma^2(x)}{\phi^2(x)}=-\frac{\gamma(x)}{\phi^2(x)}[x\phi(x)+\gamma(x)]$$

Let $F(x) = x\phi(x) + \gamma(x).$ Then $\lim_{x \rightarrow -\infty}F(x) = 0$ and

$$F'(x)=\phi(x)+x\phi'(x)+\gamma'(x)=\phi(x) +x\gamma(x)-x\gamma(x)= \phi(x) > 0 $$

Hence, $F$ is increasing and $F(x) > 0$ for all $x$.

Therefore, $\displaystyle\left(\frac{\gamma}{\phi}\right)' = -\frac{\gamma(x)}{\phi^2(x)}F(x)< 0,$ and $\displaystyle x \mapsto \frac{d}{dx}\ln(\phi(x))= \frac{\gamma(x)}{\phi(x)}$ is decreasing.

Answer 2

$(ln (\phi(x))'=\frac {\gamma (x)} {\phi(x)}$. In view of smoothness this function is decreasing iff its derivative is $\leq 0$. So we require $\phi \gamma' \leq \gamma^{2}$ or $\phi \leq \frac {\gamma^{2}} {\gamma'}$. Look at the behavior at $\infty$ to see that this is false.