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Suppose I have a Lebesgue integrable function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ with the following two properties

$$ \int_{\mathbb{R}} f(x,y)\,\textrm{d}x = g(y) $$


$$ \int_{\mathbb{R}} f(x,y)\,\textrm{d}y = h(x) $$

Is $f$ uniquely defined by the two equations? I.e., if I have another function $k$ with the same two one dimensional integrals, is it true that $f=k$ in $L^1(\mathbb{R}^2)$?

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Did you intend one integral to be with respect to $x$ and another with respect to $y$? – Michael Hardy Oct 3 '12 at 17:51
Yes, thank you for reading my mind. I've corrected it now. – Carl Morris Oct 3 '12 at 20:46
up vote 2 down vote accepted

This question has a nice probabilistic interpretation. If we require that $$\int \int f(x,y)dxdy=1$$ and that $f(x,y) \geq 0$ then your question is "is it the case that the joint distribution of two random variables is determined by their marginal distributions?"

An easy way to see that this is not the case is to consider a standard normal random variable in $\mathbb{R}^2$ (which means that the covariance matrix of $(X,Y)$ is the identity matrix $I$) and another mean zero Gaussian vector with a covariance matrix that has $1$ on the diagonal and anything but $0$ on the off diagonal. These two distributions are different (they have different densities) but because the variances of $X$ and $Y$ are one in both cases, the marginals are the same

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I think your answer is particularly interesting, since I encountered this idea while working on some probability problems. – Carl Morris Oct 3 '12 at 20:47

Suppose \begin{align} \int_R f(x,y)\,dx & = g(y) \\ \int_R f(x,y)\,dy & = h(x) \end{align}

Then $f_2(x,y) = \dfrac{g(y)h(x)}{\int_{\mathbb R^2}f(u,v)\,du\,dv}$ is a function that has these same two "marginals", i.e. if you put $f_2$ where $f$ appeared above, the integrals come out the same. So must $f=f_2$? Certainly not: for example, let $$ f = f_2 + 1_{[0,1]^2} - 1_{[-1,0]^2} $$ where the subscripted $1$s are indicator functions of the sets identified in the subscripts.

PS: Maybe you intended both integrals to be with respect to $x$, as you wrote, but in that case your question is rather oddly phrased. You said "Suppose I have [ . . . ] a function $f$". Meaning $f$ is given. The value of $\int_\mathbb R f(x,y)\,dx$ is then uniquely determined as a function of $y$; it doesn't make sense to speak of it's being equal to two different functions. If the question is whether $f$ is determined by that function, the answer is no: suppose $f=0$ except within $[0,1]\times\mathbb R$, and let $k(x,y) = f(x+5,y)$. Then $k$ differs from $f$ but their integrals with respect to $x$ are equal.

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