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I was wondering if $$f_{X,Y}(x,y)=\frac{1}{2\pi}e^{\frac{-1}{2}(x^2+y^2)}$$ Is sufficient to show that X and Y have independent standard normal distributions? If not what else would I need to show?

Edit: I know that $\int f_{X,Y}(x,y)dx=f_Y(y)$ but I’m not sure if this assumes independence or not

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  • $\begingroup$ For a joint distribution with a density, showing $f_{X,Y}(x,y)=f_X(x)f_Y(y)$, i.e. the product of the marginal densities, is sufficient to say $X$ and $Y$ are independent. Showing what $f_X(x)$ and $f_Y(y)$ will tell you what distributions $X$ and $Y$ have. $\endgroup$
    – Henry
    Apr 16 at 13:31
  • $\begingroup$ What excatly is the need to downvote this question btw? Can someone please explain. These baseless downvotes are making this site less and less accessible to the people and is a major problem that has been creeping in slowly. There are less and less upvotes and more downvotes to questions which don't deserve one. $\endgroup$ Apr 16 at 13:55

2 Answers 2

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Edit: I know that $\int_\Bbb R f_{X,Y}(x,y)\,\mathrm dx=f_Y(y)$ but I’m not sure if this assumes independence or not

The Law of Total Probability uses the joint probability density function without making any assumption about its factorialisation.

However, in this particular case, the joint distribution is the product of two recognisable monovariate functions, so...

$\qquad\begin{align}f_X(x) &=\int_\Bbb R f(x,y)\,\mathrm d y \\[1ex]&=\int_\Bbb R \dfrac 1{2\pi} \mathrm e^{-x^2/2}\mathrm e^{-y^2/2}\,\mathrm d y\\&=\dfrac 1{\sqrt{2\pi}}\mathrm e^{-x^2/2}\int_\Bbb R \dfrac 1{\sqrt{2\pi}}\mathrm e^{-y^2/2}\,\mathrm d y\\&=\dfrac 1{\sqrt{2\pi}}\mathrm e^{-x^2/2}\\[1ex]f_Y(y)&=\dfrac 1{\sqrt{2\pi}}\mathrm e^{-y^2/2}\\[1ex]\therefore\ldots\end{align}$

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I would recommend you to notice that

\begin{align*} f_{X,Y}(x,y) = \frac{1}{2\pi}e^{-\frac{1}{2}(x^{2} + y^{2})} = \left(\frac{1}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}\right)\times\left(\frac{1}{\sqrt{2\pi}}e^{-\frac{y^{2}}{2}}\right) = f_{X}(x)f_{Y}(y) \end{align*} where $X\sim\mathcal{N}(0,1)$ and $Y\sim\mathcal{N}(0,1)$. Hence $X$ and $Y$ are independent according to this.

EDIT

Since $(X,Y)\sim\mathcal{N}_{2}(0_{2\times 2},I_{2})$, the covariance $\text{Cov}(X,Y) = 0$.

Given that we are dealing with normal distributions, this means that $X\perp Y$.

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