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The continuous random variables $X,Y$ have joint probability density function $f(x,y)=e^{-2x^2-2y^2}$. Then what is the marginal distribution of $X$

What i try: Marginal distribution of $X$ is given by $$\int^{\infty}_{-\infty}f(x,y)dy=\int^{\infty}_{-\infty}e^{-2x^2-2y^2}dy$$

$$=e^{-2x^2}\int^{\infty}_{-\infty}e^{-2y^2}dy$$

Please tell me is my process is right.

If it is right Then how do i solve that definite Integration, thanks

jacky
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  • Your argument is correct but your distribution is not normalized. – lcv Jun 05 '20 at 01:27
  • Can u please tell me How can i normalize it – jacky Jun 05 '20 at 01:48
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    $f$ is not normalized. To normalize it, divide by it's integral over the allowed values (here the plane $x,y$). The integral you encounter is called a Gaussian integral. It's one of the most famous result in the history of integration. I suggest you to Google it. – lcv Jun 05 '20 at 02:01
  • Is its answer is $$\displaystyle \frac{e^{-2x^2}}{\sqrt{2\pi}}$$, where $$\int^{\infty}_{-\infty}e^{-2y^2}dy=\sqrt{\frac{\pi}{2}}.$$ – jacky Jun 05 '20 at 02:10
  • @Ivc i did not understand that line. Divide it by integral over the allowed value. – jacky Jun 05 '20 at 02:15
  • You random variables $X,Y$ can presumably assume all values between $-\infty$ and $+\infty$. These are the allowed values. It should be an input of the problem, although i this case it seems clear enough from context. – lcv Jun 05 '20 at 04:25

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