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As the title says, I'm currently trying to find a distribution $p(\cdot)$ satisfying

  1. $p(x) = \int_0^\infty \exp(\lambda_1 y^2 + \lambda_2 x y) p(y) dy$
  2. $\forall x < 0 : p(x) = 0$

The only way I have so far thought of trying was to search for multivariate distributions with equal marginals where the conditional distribution $p(x \mid y)$ is of the form $\exp(\lambda_1 y^2 + \lambda_2 x y)$.

Any help on how I can tackle this problem is greatly appreciated!

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  • $\begingroup$ Have you tried differentiating both sides of equation? Maybe this function $p(x)$ satisfies some nice and solvable ODE? $\endgroup$ – Evgeny Aug 31 '16 at 16:55
  • $\begingroup$ I did that. With respect to $x$ this turns (by parts) into an integral involving the antiderivative of $p$. With respect to $y$ doesn't seem to yield a helpful equation either. $\endgroup$ – R.G. Aug 31 '16 at 17:20
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Hint: $$\int_0^\infty p(x)\exp(-y)\operatorname d y = p(x)$$

So a solution might be some $p(\;)$ such that: $p(x)\exp(-y) =p(y)\exp(y(\lambda_1 y+\lambda_2 x))$

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  • $\begingroup$ Great idea! But unless I'm missing something obvious wouldn't this require me to make $p(\cdot)$ a function of both $x$ and $y$? $\endgroup$ – R.G. Sep 1 '16 at 8:49
  • $\begingroup$ No, it is univariate. $\endgroup$ – Graham Kemp Sep 1 '16 at 8:58
  • $\begingroup$ Ah shoot. I shall resort to the excuse of not having had a coffee yet and I'll give your suggestion a try! $\endgroup$ – R.G. Sep 1 '16 at 9:03
  • $\begingroup$ The issue remains that your approach hints at having $p(y)$ of the form $\hat{p}(y) \exp(a_0 + a_1 y^2 + \ldots)$. But this choice leaves no option to remove the $\lambda_2 x y$ part in the exponent of $\exp(y(\lambda_1 y+\lambda_2 x))$. $\endgroup$ – R.G. Sep 1 '16 at 9:34

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