# Is this ratio of normal PDFs and CDFs decreasing?

I am trying to show that the following function is decreasing in x:

$$\frac{\phi(x+a)-\phi(a)}{\Phi(x+a)-\Phi(a)},$$

where $\Phi(x)$ and $\phi(x)$ are CDF and PDF of the standard normal distribution and $a \in \mathbb{R}$.

Taking derivatives leads to an expression that I have problem signing:

$$\frac{-(x+a)\phi(x+a)[\Phi(x+a)-\Phi(a)] - \phi(x+a) [\phi(x+a)-\phi(a)] }{[\Phi(x+a)-\Phi(a)]^{2}}$$

This suggests that it is enough to show that

$$-(x+a)[\Phi(x+a)-\Phi(a)] - [\phi(x+a)-\phi(a)]<0$$

but was not able to establish this inequality.

I have tried using simple results about Mill's ratio to sign the derivative (or to prove the above inequality) but was not able to. However, numerically, it seems that this function is decreasing.

Any help would be much appreciated!

In other terms, we need to show that $$f_a(x)=\frac{\int_{a}^{a+x} z e^{-z^2/2} \,dz}{\int_{a}^{a+x} e^{-z^2/2}\,dz} \tag{1}$$ is an increasing function over $\mathbb{R}$ for any $a\in\mathbb{R}$. For such a purpose we may assume $y>x$ and prove that with such assumptions $$\int_{a}^{a+y}ze^{-z^2/2}\,dz \int_{a}^{a+x}e^{-z^2/2}\,dz \geq \int_{a}^{a+x}ze^{-z^2/2}\,dz \int_{a}^{a+y}e^{-z^2/2}\,dz \tag{2}$$ holds. That is equivalent to: $$\int_{0}^{y}(z+a)e^{-(z+a)^2/2}\,dz \int_{0}^{x}e^{-(z+a)^2/2}\,dz \geq \int_{0}^{x}(z+a)e^{-(z+a)^2/2}\,dz \int_{0}^{y}e^{-(z+a)^2/2}\,dz \tag{3}$$ or to:

$$\int_{0}^{y}\int_{0}^{x}(w-z) e^{-\frac{(z+a)^2+(w+a)^2}{2}}\,dz\,dw\geq 0 \tag{4}$$ that is geometrically trivial. The rectangle $[0,x]\times[0,y]$ in the $zw$ plane can be decomposed as the union of a square (over which the integrand function has a zero integral by symmetry) and a smaller rectangle over which $(w-z)$ is positive. This proves the claim.

• Thank you! This problem was driving me insane.
– Mdoc
Commented Jun 22, 2017 at 7:28

This is just a partial answer.

Considering $$\phi \left( x\right) =\frac{1}{\sqrt{2\pi }}e^{-\frac{x^{2}}{2}}\qquad \text{and}\qquad \Phi \left( x\right) =\int\limits_{-\infty }^{x}\phi \left( t\right) dt=\frac{1}{2} \left(1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right)$$ $$A=\frac{\phi(x+a)-\phi(a)}{\Phi(x+a)-\Phi(a)}=\sqrt{\frac{2}{\pi }}\frac{ e^{-\frac{(a+x)^2}{2} }-e^{-\frac{a^2}{2}}}{\text{erf}\left(\frac{a+x}{\sqrt{2}}\right)- \text{erf}\left(\frac{a}{\sqrt{2}}\right)}$$ Now using Taylor series built around $x=0$ and using $$\text{erf}(y+b)=\text{erf}(b)+\frac{2 e^{-b^2} }{\sqrt{\pi }}y-\frac{2 b e^{-b^2} }{\sqrt{\pi }}y^2+O\left(y^3\right)$$ then $$A=-a-\frac{1}{2}x+\frac{a }{12}x^2+O\left(x^3\right)$$ So, at least at the origin, $A$ is a decreasing function.

• Thanks! That's useful to know.
– Mdoc
Commented Jun 21, 2017 at 10:00
• @Mdoc. You are welcome ! Commented Jun 21, 2017 at 10:13

This is not an analytical answer, but you can see that indeed this ratio is monotone decreasing function. If you change $\alpha$ you just "shift" the function along the $x$ axis.

• Thanks! However, I am trying to establish this analytically...
– Mdoc
Commented Jun 21, 2017 at 10:04