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Consider the function:

$$f_a(x) = [\phi(a+x)(a+x) + \phi(a-x)(a-x)][\Phi(a-x)-\Phi(-a-x)]+[\phi(a+x)-\phi(a-x)]^2$$

where $a>0$ is a real number and $\phi(\cdot), \Phi(\cdot)$ denote the probability density function and the cumulative distribution function of a standard normal random variable, respectively. I would like to prove that the function is non negative. The graph (blue) of $f_3$ is shown below. The claim looks right, and it is easy to check it for the interval $x \in [-a,a]$. Any help to prove the remaining part of the real line will be of great help. Note that by the symmetry of the function we only need to check it for $x>0$, or as I have already done a little, for $x>a$.

Edit 1: We also see that the function is such that $f_a(x) \to 0$ as $x \to \infty$. I am thinking we could check that $f_a''(0)>0$ and that for $x>0$ the function has a unique critical point (or not more than one). Then it will follow our claim, since if we prove this then there can not be some $x^*>0$ such that the function changes sign since then there would be another critical point. But of course to analyse the derivative and to prove doesn't look easy.

Edit 2: Changed the first sign, had a typo.

Edit 3: Graph (red) of the function $\phi(a+x)(a+x) + \phi(a-x)(a-x)$ for $a=3$ again.

enter image description here Edit 4: In black the same function divided by $[\Phi(a-x)-\Phi(-a-x)]^2$ is plotted. Since this is a non negative function, clearly the original is non negative iff this quotient is. This graph seems to indicate that the $f_a(x)$ is indeed non negative.

enter image description here

Barreto
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1 Answers1

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Each "component" of the above function is positive like so: $$ \phi(a-x)(a+x) + \phi(a-x)(a-x) = 2a\phi(a-x) > 0 $$ because $a > 0$ and $\phi(\cdot)$ is positive given that it is a probability density. Next notice that as $a > 0$ we must have $a - x > -a - x$ for any $x \in \mathbb R$. So $$ \Phi(a-x) > \Phi(-a-x) \implies [\Phi(a-x)-\Phi(-a-x)] > 0 $$ by monotonicity of cumulative distribution functions. Finally $$ [\phi(a+x)-\phi(a-x)]^2 \geq 0 $$ being a squared value. So when you combine them in the above manner you get a positive outcome.

balddraz
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    I am very sorry, I have adjusted one sign. Please see edit. – Barreto May 23 '23 at 17:13
  • Although the function for $x<0$ is greater than what was before, I mean, $\phi(a+x) > \phi(a-x)$ in this case and hence the inequality, right?. Then I guess I can just conclude using your answer. – Barreto May 23 '23 at 17:18
  • @Sarah It is no longer that simple; $\phi(a + x) > \phi(a - x)$ is not true for instance when $x = a$, $\phi(2a) < \phi(0)$ because the maximum of $\phi$ occurs at $0$. – balddraz May 23 '23 at 17:21
  • I mean for $x<0$, the inequality holds, doesn't it? Then the symmetry of the function helps us conclude the overall non negativity. – Barreto May 23 '23 at 17:22
  • @Sarah Yes at least for $x < 0$ you can in fact claim that. – balddraz May 23 '23 at 17:28
  • Yes! And then observe that $f(x)=f(-x)$ since $\Phi(x) = 1 - \Phi(-x)$ – Barreto May 23 '23 at 17:31
  • There are still issues: even though $\phi(a + x) > \phi(a - x)$ for $x < 0$, the inequality flips for large negative $x$: if $a + x < 0$ we still get $\phi(a + x)(a + x) < \phi(a - x)(a + x)$ – balddraz May 23 '23 at 17:39
  • I mean that for $x<0$, $\phi(a+x) > \phi(a-x)$. Hence for $x<0$ the first term is less than $\phi(a-x)(a+x) + \phi(a-x)(a-x) = 2\phi(a-x)a > 0$. Never mind, I see your point. – Barreto May 23 '23 at 17:44
  • @Sarah That's what I am saying is problematic. You want to essentially say $\phi(a + x)(a + x) + \phi(a - x)(a - x) > \phi(a - x)(a + x) + \phi(a - x)(a - x) = 2a\phi(a - x)$ right? For that you need the crucial step $\phi(a + x)(a + x) > \phi(a - x)(a + x)$ but you can't get that even with $\phi(a + x) > \phi(a - x)$. That seems odd until you realize that for $x < 0$, $a + x < 0$ is a possibility and multiplying a negative value to an inequality flips it. – balddraz May 23 '23 at 17:49
  • I understand. Added a plot of that term. – Barreto May 23 '23 at 17:53
  • Computations suggest the there is in fact an interval where the function takes negative values... – Barreto May 23 '23 at 18:11
  • @Sarah You can make an answer to your own question showing the computations as counterexamples. You can accept your own answer after 48 hours as per the rules. I can upvote it. In your answer, you can mention for the benefit of others that my answer was for a previous edit. – balddraz May 23 '23 at 18:17
  • I think that maybe it is just some error in the programming language. This is because if I plot it under the same range, but now I take it and divide it by a nonnegative function, I get something that is nowhere negative. – Barreto May 23 '23 at 20:42