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?


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


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

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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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