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Suppose that $(X,Y)$ has the probability density function given below:

$f(x,y)=\frac{1}{\sqrt{3}{\pi}} e^{-\frac{2}3(x^2-xy + y^2)}, (x,y)\in \mathbb{R}^2$

a) I want to find the density function of $X$.

b) I want to find the density function of $Y$.

Answer: I know $f_X(x)=\int_{-\infty}^{\infty}f(x,y)dy$.Similarly the PDF of $Y$ also can be given $f_Y(y)=\int_{-\infty}^{\infty}f(x,y)dx$. But how to proceed further?

Dhamnekar Winod
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    So you know the integrals. Why don't you take them? That's how you proceed: you take the integrals. Do you want us to do this for you? – Aksakal Mar 22 '16 at 14:46
  • For additional information and solutions, please see http://stats.stackexchange.com/search?q=normal+marginal+distribution . – whuber Mar 22 '16 at 15:08
  • @whuber I put this question in online integral calculator, but in the answer ERROR FUNCTION is also included. and answers provided to me do not include ERROR FUNCTION.I know ERROR FUNCTION is required to find normal distribution density. – Dhamnekar Winod Mar 22 '16 at 15:54
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    The density is not an error function (it is the *derivative* of ERF). Your online calculator likely was not sufficiently sophisticated to simplify its result. As the duplicate (and other answers) show, the calculation is simply a matter of completing the square; it requires no knowledge of any particular techniques of integration. – whuber Mar 22 '16 at 17:14
  • I got the correct answer. – Dhamnekar Winod Mar 27 '16 at 14:20

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