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I have a system of equations:

$$1/2 + {\rm Erf}(x) - {\rm Erf}(\frac{x+y}{2})=0$$ $$-1/2 + {\rm Erf}(y) - {\rm Erf}(\frac{x+y}{2})=0,$$ Where $x \le y$ and ${\rm Erf}$ is the Error Function.

By ansatz I know that one solution is $(x,y)=(-{\rm Erf}^{-1}(1/2),{\rm Erf}^{-1}(1/2))$, but I can not manage to prove that this is the only solution, even though I strongly suspect it. My question therefore is how to prove uniquness.

One re-arrangement I have tried results in: ${\rm Erf}(\frac{x+y}{2}) = \frac{{\rm Erf}(x)+{\rm Erf}(y)}{2}$, which I am pretty sure is equivalent to $y=-x$ (apparent by implicit plot), but again I can not prove this. Proving that $y=-x$ for all solutions would imply the desired uniqueness.

Any advice on manipulation tactics for the error function would be much appreciated..!

Tagged as probability theory because of the relation to the Normal distribution CDF.

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[Edited only to change each $Erf$ to ${\rm Erf}$.] – Noam D. Elkies Apr 19 '12 at 2:20
I have posed a follow-up question about a generalization of this problem:… – Johan Ugander Apr 25 '12 at 20:09
up vote 6 down vote accepted

This can be solved as a problem in inequalities. The desired implication $$ {\rm Erf}\Bigl(\frac12(x+y)\Bigr) = \frac12\bigl({\rm Erf}(x) + {\rm Erf}(y)\bigr) \Longleftrightarrow x + y = 0 $$ is not quite correct, because the hypothesis holds also for $x=y$; but these are the only possibilities, which is enough to yield the desired result. Because the error function is odd, it is enough to prove:

Proposition. If $x$ and $y$ are real numbers such that $x+y > 0$ then $$ {\rm Erf}\Bigl(\frac12(x+y)\Bigr) \geq \frac12\bigl({\rm Erf}(x) + {\rm Erf}(y)\bigr), $$ with equality if and only if $x=y$.

Proof: Let $a = (x+y)/2$ and $b = (x-y)/2$. Then $a>0$ and $x,y = a \pm b$, and we claim $$ \bigl({\rm Erf}(a+b) + {\rm Erf}(a-b)\bigr) - 2\phantom.{\rm Erf}(a) \leq 0, $$ with equality iff $b=0$. By symmetry we may assume $b\geq 0$. Equality clearly holds for $b=0$. We claim that the left-hand side (LHS) is a decreasing function of $b$ for $b \geq 0$, which will prove that it is negative for all $b>0$. Indeed derivative of the LHS is $$ \frac2{\sqrt{\pi}} \left( e^{-(a+b)^2} - e^{-(a-b)^2} \right) = \frac2{\sqrt{\pi}} e^{-(a^2+b^2)} (e^{-2ab} - e^{+2ab}) < 0 $$ for $b>0$, QED.

(This uses the normalization of ${\rm Erf}(t)$ that has derivative $(2/\sqrt\pi) \phantom. e^{-t^2}$. There are other normalizations, but the proof works in the same way.)

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Thank you. To clarify (for myself and potentially others), your approach breaks up the possibilities of (x,y) into three cases: x+y>0, x+y<0, x+y=0. Your proposition proves that x+y>0 => equality iff x=y. Oddness gives x+y<0 => equality iff x=y. x+y=0 clearly gives x=-y. So you proves that the statement I derived implies that x=y or x=-y. But x=y does not work with the other equations in the system, so x=-y, and thus (x,y)=(-InvErf(1/2),InvErf(1/2)). Thanks again! – Johan Ugander Apr 19 '12 at 3:07
Small typo: in the derivative of the LHS, first expression, the second exp() should be exp(-(a-b)^2), not exp(-(a+b)^2), but this is a typo only and you did not carry it forward (the result is correct). – Johan Ugander Apr 19 '12 at 20:31
right, thanks; I'll fix it. – Noam D. Elkies Apr 21 '12 at 17:09

Here is another proof, arguably more elementary in that it only uses basic properties of $e^{-t^2}$.

Notice that the two original equations reduce to: $$ \int_x^{(x+y)/2} e^{-t^2}dt = \frac{\sqrt{\pi}}{4} , \hspace{0.3in} \int_{(x+y)/2}^y e^{-t^2}dt = \frac{\sqrt{\pi}}{4}, $$ It is clear from this that we must have $y\ne x$.

With a variable substitution $a=(x+y)/2, b=(x-y)/2$, we also have: $$ \int_{a-b}^{a} e^{-t^2} dt - \int_{a}^{a+b} e^{-t^2}dt =0. $$ The following proposition then completes the proof.

Proposition. For $f(t)$ integrable, positive, symmetric at $t=0$, and increasing for $t<0$, and $b \ne 0$, $$ \int_{a-b}^{a} f(t)dt - \int_{a}^{a+b} f(t)dt =0 \hspace{0.2in} \Leftrightarrow \hspace{0.2in} a=0. $$ Proof.

The implication $\Leftarrow$ is straight forward.

To see the implication $\Rightarrow$, notice that a symmetry manipulation gives us (steps omitted): $$ \int_{a-b}^{a} f(t)dt - \int_{a}^{a+b} f(t)dt = \int_{-a}^{a} f(t)dt - \int_{b-a}^{b+a} f(t)dt. $$ For $a=0$ this clearly equals $0$.

For $a>0$, the increasing property and symmetry give us that: $$ \int_{-a}^{a} f(t)dt > \int_{b-a}^{b+a} f(t)dt. $$ for all $b \ne 0$. An analogous argument holds for $a<0$. QED.

I have not checked wether integrability and positivity are strictly necessary conditions on $f$ for the proposition to hold, but $f(t)=e^{-t^2}$ does have these properties, and generalizations I'm interested in relate to probability distributions, so these assumptions are fine for my purposes.

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