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How can I calculate the value of the constant c, given the joint probability distribution below?

$$ f_{X,Y}(x,y) = \left\{\begin{array}{ll} ce^{-({\frac {x^2} {8}} + {4y})} & : -\infty \le x < \infty, y \ge 0\\ 0 & : otherwise\\ \end{array} \right. $$

I've attempted to plug the integral into tools such as `https://www.symbolab.com/, as well as my TI-89, but have not been successful thus far.

Also, are there any recommended materials for practicing these types of problems?

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You just need to remember the integration of the probability distribution is 1.

$$\int_{-\infty}^{\infty}\int_{0}^{\infty}f_{X,Y}(x,y)dydx=1$$

The followings are the calculations: $$\int_{-\infty}^{\infty}\int_{0}^{\infty}ce^{-(\frac{x^2}{8}+4y)}dydx\\=c\int_{-\infty}^{\infty}e^{-\frac{x^2}{8}}\int_{0}^{\infty}e^{-4y}dydx\\=\frac{c}{4}\int_{-\infty}^{\infty}e^{-\frac{x^2}{8}}dx=\frac{c}{4}*\sqrt{2\pi}2\frac{1}{\sqrt{2\pi}2}\int_{-\infty}^{\infty}e^{-\frac{1}{2}\frac{x^2}{4}}dx=\frac{c}{2}*\sqrt{2\pi}=1$$

$\therefore c=\frac{2}{\sqrt{2\pi}}=\sqrt{\frac{2}{\pi}}$

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  • $\begingroup$ Please don't give full answers to self-study questions, see stats.stackexchange.com/tags/self-study/info $\endgroup$ – Juho Kokkala Apr 19 '17 at 5:59
  • $\begingroup$ Would you mind explaining where the sqrt(2 * pi) * 2 came from? $\endgroup$ – statsaverse Apr 19 '17 at 6:14
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    $\begingroup$ that is from 1 i.e $\sqrt{2\pi}2*\frac{1}{\sqrt{2\pi}2}$which try to make a normal distribuiton $\endgroup$ – Deep North Apr 19 '17 at 6:18

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