If two pairs of random variables $(X,Y)$ and $(U,V)$ have the same joint pdf $f_{X,Y}(x,y)=f_{U,V}(x,y)$, can we conclude that $(X,Y)=(U,V)$?
Does joint pdf uniquely describes the random variables?
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probability-distributions
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1You can conclude that the are identically *distributed*, which is not quite the same as saying they are *identical*. – 2012-04-22
1 Answers
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You cannot conclude $(X,Y)=(U,V)$ holds pointwise, since they need not be defined on the same probability space. But you can conclude that they have the same distribution: $(X,Y)\sim (U,V)$, i.e. their distributions on $(\mathbb{R}^2,\mathcal{B}(\mathbb{R}^2))$ are equal.
Proof: Let $P_{(X,Y)}$ and $P_{(U,V)}$ denote the distribution on $(\mathbb{R}^2,\mathcal{B}(\mathbb{R}^2))$ of $(X,Y)$ and $(U,V)$ respectively. Then for any $A\in \mathcal{B}(\mathbb{R}^2)$ we have $ P_{(X,Y)}(A)=P((X,Y)\in A)=\int_A f_{X,Y}(x,y)dx dy=\int_A f_{U,V}(x,y)dx dy=P_{(U,V)}(A), $ which exactly shows that $(X,Y)\sim (U,V)$.
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0@broke: See edit. – 2012-04-22