# Bivariate Normal Distribution [duplicate]

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?

• 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? Mar 22, 2016 at 14:46
• For additional information and solutions, please see stats.stackexchange.com/search?q=normal+marginal+distribution .
– whuber
Mar 22, 2016 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. Mar 22, 2016 at 15:54
• 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, 2016 at 17:14
• I got the correct answer. Mar 27, 2016 at 14:20