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I have been trying to prove that $$\small \left(\phi\left( x \right)\Phi\left( x-y \right)-\Phi\left( x \right)\phi\left( x-y \right)\right)\times \phi\left( x+y \right)-\left(\phi\left( x+y \right)\Phi\left( x \right)-\Phi\left( x+y \right)\phi\left( x \right)\right)\times\phi\left( x-y \right)>0,$$ where $\phi$ is the standard normal density function and $\Phi$ is the standard normal CDF. The following R code suggests that it holds true for all $x,y \in \mathbb{R}$, but I cannot find a way to show it formally:

library(plotly)
f <- function(x, y) (dnorm(x) * pnorm(x - y) - pnorm(x) * dnorm(x - y)) * dnorm(x + y) - (dnorm(x + y) * pnorm(x) - pnorm(x + y) * dnorm(x)) * dnorm(x - y)
x <- seq(-5, 5, length = 1000)
y <- seq(-5, 5, length = 1000)
z <- outer(x, y, f)
plot_ly(x = x, y = y, z = z, type = "surface")  # 3D plot
sum(z <= 0) # any nonpositive function values?
# [1] 0

Can someone maybe help to prove/ disprove it or point to some insights/ results for the normal distribution that may be of help in that regard?

Many thanks and best wishes

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  • $\begingroup$ I am curious: why would you need that? $\endgroup$
    – Snoop
    Jun 28, 2022 at 12:43
  • $\begingroup$ @Snoop It is needed for establishing an (econometric) identification result, where I try to examine whether the determinant of the Jacobian of some function is always positive (which is the inequality stated above). $\endgroup$
    – cliu55
    Jun 29, 2022 at 16:56

1 Answer 1

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Notice that $$ \Phi(x+y) = \int_{-\infty}^{x+y} \phi(z) \mathrm{d}z = \int_{-\infty}^x \phi(z+y) \mathrm{d}z.$$

Further, $\phi(x+y) = e^{-y^2/2 - xy} \phi(x). $ Therefore, $$ \phi(x+y)(\phi(x) \Phi(x-y) - \Phi(x) \phi(x-y)) = \int_{-\infty}^x \phi^2(x) \phi(z) e^{-y^2} \left( e^{zy-xy} - e^{xy - xy} \right) \mathrm{d}z, \\ \phi(x-y)(\Phi(x) \phi(x+y) - \phi(x)\Phi(x+y)) = \int_{-\infty}^x \phi^2(x) \phi(z)e^{-y^2} ( e^{xy - xy} - e^{xy -zy}) \mathrm{d}z.$$ So, it suffices to argue that $$ \int_{-\infty}^x \phi^2(x) \phi(z) e^{-y^2}( e^{zy -xy} + e^{xy -zy} -2) \mathrm{d}z \overset{?}> 0.$$

But, using the AM-GM inequality, we have that $$ \exp(a) + \exp(-a) \ge 2\sqrt{\exp(a) \cdot \exp(-a)} = 2, $$ with strict inequality so long as $a \neq 0.$ It follows that if $y \neq 0,$ then the integrand is stricly positive for $z \in (-\infty, x)$. Thus, the the conclusion follows for $y \neq 0$ (and of course, for $y = 0,$ the difference equals $0$).

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  • $\begingroup$ Thanks a lot for the brilliant answer! $\endgroup$
    – cliu55
    Jul 20, 2022 at 9:47

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