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The joint probability density function of two continuous random variables X and Y is given that $f_{X,Y}(x,y)=\alpha$ when $x^2+y^2 \le r^2$ and $f_{X,Y}(x,y)=0$ otherwise. How to compute the constant $\alpha$?

Could anybody give me a hint? As the two of continuous random variables are not specified to uniformly distributed, I hardly can picture a X,Y plane to illustrate the limits of integration. Thanks!

user8185923
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  • The limits of $x$ go from 0 to $r$ while the limits of $y$ also go from $0$ to $r$. Pardon my ignorance, but isn't this a fairly simple double integral? – Gokul Sep 09 '17 at 03:51
  • this is what exactly I doubt, that if i can define the $0 \le x \le r$, $0 \le y \le r$. – user8185923 Sep 09 '17 at 03:54
  • The integral of a constant is just the constant times the area of the domain. The domain is a disk of radius $r$. Choose the constant so that the integral evaluates to $1$. It's that simple. – quasi Sep 09 '17 at 03:54
  • @quasi, but will the disk definitely centered at original point and why? – user8185923 Sep 09 '17 at 03:57
  • $x^{2} + y^{2} = r^{2}$ is a circle centered at $(0,0)$. So I am guessing you wouldn't be incorrect in assuming the limits of $x$ go from $0$ to $r$ and similarly for $y$ too. – Gokul Sep 09 '17 at 04:00
  • The solution set in $\mathbb{R}^2$ of the inequality $x^2+y^2 \le r^2$ is the set of points inside or on the circle with equation $x^2+y^2=r^2$, which is a circle of radius $r$, centered at the origin. – quasi Sep 09 '17 at 04:02
  • When I said "the domain is a disk of radius $r$", what I meant is that the support of $f$ (i.e., the set of points where $f \ne 0$) is a disk of radius $r$. – quasi Sep 09 '17 at 04:06
  • @quasi: I realised I was incorrect in taking the limits - if you could post the correct answer, it'd be helpful for me to learn. – Gokul Sep 09 '17 at 04:10

1 Answers1

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Let $D\;$be the disk of radius $r$, centered at the origin.

Since $f$ is a probability mass function, \begin{align*} &\iint_{\mathbb{R}^2} f(x,y)\,dA=1\\[4pt] \implies\;&\iint_D f(x,y)\,dA=1&&\text{[since $f$ is zero outside of $D$]}\\[4pt] \implies\;&\iint_D \alpha\,dA=1\\[4pt] \implies\;&\alpha \iint_D 1\,dA=1\\[4pt] \implies\;&\alpha(\pi r^2)=1\\[4pt] \implies\;&\alpha=\frac{1}{\pi r^2}\\[4pt] \end{align*}

quasi
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