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Is it possible to find a function $F:\mathbb R^+\times \mathbb R^+\to\mathbb R$ such that \begin{align} \int_0^\infty\int_0^\infty F(x,y)e^{-xq-yq}dxdy=\frac{\alpha}{q^4}+\frac{\beta}{q^3}+\frac{\gamma}{q^2}, \end{align} with $\alpha, \beta, \gamma$ constants.

@Edit It is a problem that comes from probability. I have two independent random variables $X$ and $Y$ with exponential distribution of parameter $\frac{1}{\lambda}$ and I would like to understand with function $F$ of the two random variables has an expectation that is quadratic in $\lambda$. This means that I need a function $F$ such that \begin{align} \mathbb E(F(X, Y))=\int_0^\infty\int_0^\infty F(x, y)\frac{e^{-\frac{x}{\lambda}-\frac{y}{\lambda}}}{\lambda^2}dx dy=\alpha\lambda^2+\beta\lambda+\gamma. \end{align} If I call $q=\frac{1}{\lambda}$ the problem is reduced to the one written above, that means solving a double Laplace transform.

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  • $\begingroup$ Sure, a very simple one (constant) does the trick. $\endgroup$ Commented Nov 22, 2021 at 22:35
  • $\begingroup$ The constant gives $1/q^4$, not exactly that. $\endgroup$
    – user268193
    Commented Nov 23, 2021 at 6:39
  • $\begingroup$ Is $1$ the only constant in the world? Try again. $\endgroup$ Commented Nov 23, 2021 at 8:01
  • $\begingroup$ I edited the text because there was a mistake (it is just $q$ in the exponent and not $q^2$). If $F(x, y)=c$ the result of the double integral should be $c/q^2$. Am I correct? $\endgroup$
    – user268193
    Commented Nov 23, 2021 at 8:20
  • $\begingroup$ Yes that's right, can you think of a $c$ that would work? $\endgroup$ Commented Nov 23, 2021 at 8:27

1 Answer 1

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We already know that for $F(x,y)=\gamma$

$$\int_0^\infty\int_0^\infty \gamma e^{-xq-yq}\:dx\:dy = \frac{\gamma}{q^2}$$

Each of the terms $q^{-3}$ and $q^{-4}$ is (a scalar multiple of) a derivative of $q^2$. By Laplace transform properties in $1$D, we know that a derivative in the frequency domain is a multiplication by a monomial power in the time domain

$$\int_0^\infty tf(t)e^{-st}\:dt = -\frac{d}{ds}F(s)$$

In $2$D we now have options. Both $x\cdot1$ and $y\cdot 1$ accomplish a single derivative

$$\int_0^\infty\int_0^\infty \beta x e^{-xq-yq}\:dx\:dy = \int_0^\infty\int_0^\infty \beta y e^{-xq-yq}\:dx\:dy = \frac{\beta}{q^3}$$

And a second derivative can be accomplished one of three ways. This means that all functions with that specific Laplace transform are of the form

$$F(x,y) = \gamma+\beta(ax+by)+\frac{\alpha}{2}(ux^2+2vxy+wy^2)$$

where $a+b=u+v+w=1$

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  • $\begingroup$ The Laplace transform you have is linear but also not injective. Compute the Laplace transform of that $\beta$ term I have in my final answer and tell me what happens to the $a$ and $b$ $\endgroup$ Commented Nov 23, 2021 at 23:01
  • $\begingroup$ Sorry if I bother you but I have a question. How do you know that the function $F(x, y)=\gamma$ is the unique solution of $\int_0^\infty\int_0^\infty F(x, y)e^{-xq-yq}dxdy=\frac{\gamma}{q^2}$? and that it doesn't exists another function $F$ that satisfies that condition? $\endgroup$
    – user268193
    Commented Dec 6, 2021 at 17:01
  • $\begingroup$ @m91c That is an excellent question and truth be told, I don't know haha. I relied on $1$D Laplace transform properties to derive that but as you say, we have already proven that this Laplace transform is not injective. There may well be other solutions, but I can't think of any. If I do, I will let you know. However you only need a function right? There is no the unique answer so any answer will be as good as another. $\endgroup$ Commented Dec 6, 2021 at 17:27

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