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I am trying to solve this integral equation for $f$ (can be assumed to be positive)

$$ f(y) \int_0^1 \frac{e^{-\frac{(x-y)^2}{2}}}{\int_0^1 e^{-\frac{(x-z)^2}{2}} f(z) dz} dx =1, \quad y \in [0,1]$$

Any insight is welcome, I have no idea how this kind of equations is called and how to solve them. I thought about differentiating both side with regard to $y$, we get

$$f(y)\int_0^1 \frac{(x-y)e^{-\frac{(x-y)^2}{2}}}{\int_0^1 e^{-\frac{(x-z)^2}{2}} f(z) dz} dx + f'(y)\int_0^1 \frac{e^{-\frac{(x-y)^2}{2}}}{\int_0^1 e^{-\frac{(x-z)^2}{2}} f(z) dz} dx =0$$

Plugging in the first equation, we get

$$ f(y)\int_0^1 \frac{xe^{-\frac{(x-y)^2}{2}}}{\int_0^1 e^{-\frac{(x-z)^2}{2}} f(z) dz} dx -y + \frac{f'(y)}{f(y)} = 0$$

But it only seems to make the problem harder.

EDIT: As this question seems to interest some people, here is some more info. We can rewrite the problem as a system of two integral equations:

$$ f(y) \int_0^1 e^{-\frac{(x-y)^2}{2}} \hat f(x) dx =1, \quad y \in [0,1]$$

$$ \hat f(x) \int_0^1 e^{-\frac{(x-y)^2}{2}} f(y) dy =1, \quad x \in [0,1]$$

The existence of solutions $f,\hat f$ is a deep result in stochastic processes and probability theory. The trained eye recognized the heat kernel (Brownian transition density). The product $f(y)\hat f(x)$ is actually the density (not w.r.t Lebesgue, it's complicated) of a certain coupling of 2 given random variables. So $f,\hat f$ contain all the information about the dependence between those 2 random variables. They are unique up to multiplicative constant (when you multiply one by $c$, you divide the other one by $c$)

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  • $\begingroup$ The issue is that differentiating usually only helps when $y$ is one of the endpoints of the integrals. It's also worth pointing out that your final line isn't correct, you can't make those fraction cancellations. Can I ask where this problem came up? Is this a problem you've been given as is or did you arrive at this problem in the process of doing something else? I want to know how tractable this problem might be, and what you might have been studying when you came across this problem. $\endgroup$ – Pepe Silvia Jun 15 '20 at 0:18
  • $\begingroup$ @PepeSilvia I corrected some typos, were you referring to those as incorrect ? It's hard to explain fully where this problem came from, it's related to Brownian motion, large deviation and entropy minimization. $\endgroup$ – W. Volante Jun 15 '20 at 0:42
  • $\begingroup$ Sorry it's possible you didn't make a mistake, I didn't realise you substituted the integral for $f$, in any case it looks fine to me now. I'm familiar with large deviation theory and Brownian motion but I haven't seen equations of this type before. $\endgroup$ – Pepe Silvia Jun 15 '20 at 0:49
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    $\begingroup$ This can be simplified to $$g(y)\int_0^1 \frac{e^{xy}}{\int_0^1 e^{xz}g(z)\:dz}\:dx = 1$$ by expanding the powers and letting $g(y) = f(y)e^{-\frac{y^2}{2}}$ $\endgroup$ – Ninad Munshi Jun 15 '20 at 5:22
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    $\begingroup$ @NinadMunshi: well, "tautological" may be the wrong term, but you may find it interesting that a function satisfying $$g(x)\int_0^1 e^{xz}g(z)\:dz=1$$ satisfies $$g(y)\int_0^1 \frac{e^{xy}}{\int_0^1 e^{xz}g(z)\:dz}\:dx = 1$$ as well. And the former equation is far more amenable to analysis or numerical experiments. $\endgroup$ – Professor Vector Jun 15 '20 at 17:45
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Things begin to clear up, a bit: you have a system of equations for $f$ and $\hat f,$ and while $f=\hat f$ would give a solution, that's probably nonsense from the point of view of your model. I investigated your problem numerically, starting, however, from the form $$g(y)\int_0^1 \frac{e^{xy}}{\int_0^1 e^{xz}\,g(z)\:dz}\:dx = 1$$ proposed by Ninad Munshi. Here, $g(y) = f(y)e^{-\frac{y^2}{2}},$ so $g(1) = e^{-\frac{1}{2}}g(0)$ corresponds to $f(0)=f(1),$ and that's a symmetry present in the original problem, obviously (your equations are invariant under $x\to1-x, y\to1-y,$ so you must have $f(x)=f(1-x)$ and $\hat f(1-x)=\hat f(x)$ due to uniqueness). So we have a system of equations $$h(x)=\frac1{\int_0^1 e^{xz}\,g(z)\:dz},$$ $$g(x)=\frac1{\int_0^1 e^{xz}\,h(z)\:dz}.$$ The most obvious try to solve that numerically would be a simple iteration, $$h_{n+1}(x)=\frac1{\int_0^1 e^{xz}\,g_n(z)\:dz},$$ $$g_{n+1}(x)=\frac1{\int_0^1 e^{xz}\,h_{n+1}(z)\:dz},$$ approximating the integrals by some sort of quadrature formula. Interestingly, this process (with the initial guess $g_0(x)=e^{-x^2/2}$) converges with impressive speed, the difference to the previous value is $\le10^{-15}$ after about five iterations. And even though that numerical process knows nothing about $f(0)=f(1),$ the relation $g(1) = e^{-\frac{1}{2}}g(0)$ is satisfied with said accuracy.

So we can calculate those functions rather precisely, but... I doubt there is a closed solution in elementary functions. And while there are algorithms identifying constants known with sufficient accuracy (quite beyond 15 or 16 digits, though), I don't know of anything similar for functions.

EDIT: as it turns out, the functions $g$ and $h$ differ only by a multiplicative constant. So there is a solution with $f=\hat f$ for your equations! Here is a graph:

enter image description here

We have $f(0)\approx 1.1234005998770296,$ but I can't identify the constant.

EDIT 2: Factorizations aren't always obvious, if $q(x,y)=e^{-(x-y)^2/2},$ we have $$\frac{q(x,y)}{e^{xy}}=e^{-\frac{x^2}2}\cdot e^{-\frac{y^2}2}.$$ If the integration interval were $(-\infty, \infty)$ instead of $[0,1],$ the solution would be just constant, and I'm pretty sure that fact has a probabilistic interpretation. And since that constant is sufficiently close to the solution you seek, that may explain the rapid convergence. However, that's all mere speculation without knowing more details from the probabilistic background.

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  • $\begingroup$ Thanks for your post. I ran the simulation following what you wrote and indeed I get a very fast convergence. How do you explain that theoretically, it is impossible to separate the variables, but the simulations show that indeed $f=\hat f$ ? $\endgroup$ – W. Volante Jun 17 '20 at 19:55
  • $\begingroup$ @W. Volante I don't have a theoretical explanation for the rapid convergence, maybe you should tell us more about the derivation of those equations. Don't be afraid, my PhD was about stochastic processes, and you may have heard of my supervisor, Skorokhod. $\endgroup$ – Professor Vector Jun 17 '20 at 20:10
  • $\begingroup$ Wow a student of Anatoliy! I meant we know that we cannot split $\frac{q(x,y)}{e^{xy}}$ into the product of two functions $f(x)f(y)$ that separate the variables because of $e^{xy}$ (see our discussion above), so this means we cannot have $f=\hat f$. Yet in the simulations, we see that $f=\hat f$ for the right constant, how is that possible ? So either there is a magical way of splitting $\frac{q(x,y)}{e^{xy}}$ or $f$ and $\hat f$ are so close to each other (but different) that the simulations don't see the difference ? This discussion is independent of how I get the equations $\endgroup$ – W. Volante Jun 17 '20 at 21:51

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