How can I prove that this function is a bivariate density, meaning the double integral should equal $1$.
$$\iint_{\mathbb R^2} \:\frac{\left(1+\sin x\:\sin y\right)}{2\:\pi}e^{-\frac{\left(x^2+y^2\right)}{2}}dxdy$$
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What's the integration domain? Is ir $\mathbb R^2$? – Andrei Oct 14 '20 at 02:54
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@Andrei Yes, there is no restriction on domain. I have figured bivariate density is another way to call bivariate joint distribution, that's why I think over the whole domain, it should equal 1. – Agi Oct 14 '20 at 03:18
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$$\begin{align}\iint_{\mathbb R^2} \:\frac{\left(1+\sin x\:\sin y\right)}{2\:\pi}e^{-\frac{\left(x^2+y^2\right)}{2}}dxdy&=\frac{1}{2\pi}\iint_{\mathbb R^2}e^{-\frac{x^2+y^2}2}dxdy\\&=\frac1{2\pi}\int_0^{2\pi}d\phi\int_0^\infty e^{-r^2/2}r dr\\&=\int_0^\infty e^{-u}du\\&=-e^{-\infty}-(-e^0)\\&=1\end{align}$$
On the first line, I've noticed that the term containing $\sin$ is odd (in both $x$ and $y$), so integrating over a symmetric interval is $0$. On the second line I've changed to polar coordinates. Third line, I did a change of variable, $r^2/2=u$, so $du=rdr$.
Andrei
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Thanks, the explanation really helped with the understanding! How can I do $$\int_{\mathbb R} :\frac{\left(1+\sin x:\sin y\right)}{2:\pi}e^{-\frac{\left(x^2+y^2\right)}{2}}dy$$ and $$\int_{\mathbb R} :\frac{\left(1+\sin x:\sin y\right)}{2:\pi}e^{-\frac{\left(x^2+y^2\right)}{2}}dx$$ – Agi Oct 14 '20 at 03:44
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The integral of the sine part is zero, same as before. The first integral does not depend on $x$, so you can write it as $e^{-x^2/2}\int_{\mathbb R}e^{-y^2/2} dy$ – Andrei Oct 14 '20 at 03:49
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