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Let $X$ and $Y$ be jointly absolutely continuous random variables. Suppose $f_{X,Y}(x,y)=g(x)h(y)$. Prove that $X$ and $Y$ are independent.

So I want to show that $f_X(x)=g(x)$ and that $f_Y(y)=h(y)$

And I have that $$\int_{-\infty}^\infty f_{X,Y}(x,y)dy=f_X(x)=\int_{-\infty}^\infty g(x)h(y)dy=g(x)\int_{-\infty}^\infty h(y)dy$$

and similarly $$f_Y(y)=\int_{-\infty}^\infty f_{X,Y}(x,y)dx=h(y)\int_{-\infty}^\infty g(x)dx$$

But to say $$\int_{-\infty}^\infty g(x)dx=1$$ I would have to know that $g(x)$ is the density function already.

AColoredReptile
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    Technically, you can't prove $g,,h$ are the PDFs, as they could be e.g. double one & half the other. – J.G. Dec 01 '19 at 20:43

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As you've noted, the marginal PDFs are$$\begin{align}f_X(x)=\int_{\Bbb R}f_{X,\,Y}(x,\,y)dy&=g(x)\int_{\Bbb R}h(y^\prime)dy^\prime,\,\\f_Y(y)=\int_{\Bbb R}f_{X,\,Y}(x,\,y)dx&=h(y)\int_{\Bbb R}g(x^\prime)dx^\prime.\end{align}$$You need to prove $f_{X,\,Y}(x,\,y)=f_X(x)f_Y(y)$. Indeed, the right-hand side is$$g(x)\int_{\Bbb R}h(y^\prime)dy^\prime\cdot h(y)\int_{\Bbb R}g(x^\prime)dx^\prime=f_{X,\,Y}(x,\,y)\underbrace{\int_{\Bbb R^2}g(x^\prime)h(y^\prime)dx^\prime dy^\prime}_{\int_{\Bbb R^2}f_{X,\,Y}(x^\prime,\,y^\prime)dx^\prime dy^\prime=1}.$$

charlesreid1
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J.G.
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  • Can you explain the second part? I'm not sure what you mean by $f_{X,Y}(x)$ since isn't it a function of $2$ variables? And I'm not sure where $\int_{\mathbb{R}}h(y^\prime)dy^\prime$ is coming from. – AColoredReptile Dec 01 '19 at 20:44
  • @AColoredReptile Forgive my now corrected typo. The $^\prime$s are just a way of making sure the arguments $x,,y$ aren't used as dummy variables of integration. So $g(x)\int_{\Bbb R}h(y^\prime)dy^\prime$ is just the $f_X(x)$ factor. – J.G. Dec 01 '19 at 20:46
  • Ok I understand now thanks. – AColoredReptile Dec 01 '19 at 20:50