Let $D=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq1\}$ be a unit disk and $(X,Y)$ be uniformly distributed on D. Find the probability density function of W for
- $W=X+Y$
- $W=Y/X$
I think I need to find a cdf and then differentiate it to find the pdf. However, I have never tried finding a cdf for a circle.
Should I split the circle into four parts and then find the pdf for each of them so that each part os either strictly increasing or stringly decreasing?
Best Answer
The joint probability density function is $f_{\small X,Y}(x,y)= \dfrac{1}{\pi}\mathbf 1_{x^2+y^2\leqslant 1}$.
Your solution requires integrating wrt $x$ and $y$ then deriving wrt the composite variable so... use the fundamental theorem of calculus to avoid some unnecessary calculations.
Thus $$\begin{align}f_{\small X+Y}(w) &= \frac{\mathrm d ~~}{\mathrm d w}\iint_{x^2+y^2\leq 1 , x+y\leqslant w}f_{\small X,Y}(x,y)\mathrm d x\mathrm d y\\&=\mathbf 1_{-\surd 2\leqslant w\lt \surd 2}\int_{x^2+(w-x)^2\leq 1}\left\lvert\dfrac{\partial (w-x)}{\partial w}\right\rvert f_{\small X,Y}(x,w-x)\,\mathrm d x\\&=\mathbf 1_{-\surd 2\leqslant w\lt\surd 2}\int_{(w-\sqrt{2-w^2})/2}^{(w+\sqrt{2-w^2})/2}\frac 1\pi\,\mathrm d x\\&~~\vdots\end{align}$$
Likewise $$\begin{align}f_{\small Y/X}(w) &= \frac{\mathrm d ~~}{\mathrm d w}\iint_{x^2+y^2\leq 1 , y/x\leqslant w}f_{\small X,Y}(x,y)\mathrm d x\mathrm d y\\&=\mathbf 1_{-1\leqslant w\lt 1}\int_{x^2+(xw)^2\leq 1}\left\lvert\dfrac{\partial xw}{\partial w}\right\rvert f_{\small X,Y}(x,xw)\,\mathrm d x\\&=\mathbf 1_{-1\leqslant w\lt 1}\int_{x^2\leq 1/(1+w^2)} \lvert x\rvert\cdot\frac 1\pi\,\mathrm d x\\&~~\vdots\end{align}$$