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Let $X$ and $Y$ be two random variable with a joint distribution $P_{(X,Y)}$ absolute continuous with respect to the Lebesgue measure on $\mathbb R^2$.
I want to show that $P_{(X,Y)}$ has a density, that it, there exists a measurable function $f_{(X;Y)}$:$\mathbb{R}^2 \rightarrow \mathbb{R}_+$ such that $E\left ( g(X,Y) \right )=\int_{\mathbb{R}^2}g(x,y)f_{(X,Y)}(x,y)$ for every positive measurable function $g:\mathbb{R}^2\rightarrow[0,\infty[$.
Question: What is the relation between this example and Radon-Nikodym theorem?

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closed as unclear what you're asking by Did, Math1000, user91500, Tom-Tom, user26857 Nov 26 '15 at 10:46

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    $\begingroup$ That is the Radon-Nikodym theorem. Do you want to do that yourself? It looks a bit like reïnventing the wheel. $\endgroup$ – drhab Nov 25 '15 at 9:27
  • $\begingroup$ @ drhab Thank you for your comment. But I couldn't derive probability density from Radon-Nikodyn theorem. Please write it for me. $\endgroup$ – linmu Nov 25 '15 at 9:47
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    $\begingroup$ I am not able to write it out more clearly than in the link in my former comment. A measurable function $f$ exists such that $P_{X,Y}(A)=\int_A fd\lambda$. This $f$ is the density you are looking for. Is it your question how to prove that $\mathbb Eg(X,Y)=\int gfd\lambda$ for each suited $g$? If so then you asked the wrong question. $\endgroup$ – drhab Nov 25 '15 at 10:11
  • $\begingroup$ @drhab thank you so much for your comment. $\endgroup$ – linmu Nov 25 '15 at 13:31
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To use the Radon-Nikodym theorem we need that:

(1) $P_{(X,Y)} \mbox{ is absolutely continuous w.r.t. the Lebesgue measure}$ $\lambda^2$ on $\mathbb{R}^2$

(2) $P_{(X,Y)}$ is $\sigma-$finite which is satisfied since it is even finite: $P_{(X,Y)}(\mathbb{R}^2) = 1 < \infty$.

(3) The Lebesgue measure $\lambda^2$ is $\sigma-$finite, for this we define for each $n \in \mathbb{N}$ the set $B_n := [-n,n] \times [-n,n]$ then we clearly have that $B_n \nearrow \mathbb{R}^2$ and for each $n: \lambda^2(B_n) = 4n^2 < \infty$. From this we conclude that $\lambda^2$ is indeed $\sigma-$finite.

Now the Radon-Nikodym theorem yields that there exists a measurable function $f_{(X,Y)}: X \rightarrow [0,\infty[$ which satisfies that what you need if you use Proposition 4.1 in the following: https://www.math.ksu.edu/~nagy/real-an/4-04-rn.pdf

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  • $\begingroup$ This is the complete answer, thank you. I ll add the complement for the last part. $\endgroup$ – linmu Nov 26 '15 at 6:17

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