University of Alberta Guide/STAT/222/Combining Continuous Random Variables

Convolution

• $f_{X+Y}(z)=\int _{-\infty }^{\infty }f_{X}(y-z)f_{Y}(z)\delta z=\int _{-\infty }^{\infty }f_{X}(z)f_{Y}(y-z)\delta z$

Example

$f_{X}(x)={\begin{cases}{\frac {3x^{2}}{2}}&x\in [-1,1]\\0&else\end{cases}}\qquad f_{Y}(y)={\begin{cases}{\frac {y}{9}}&y\in [4,5]\\0&else\end{cases}}\qquad X{\mbox{ and }}Y{\mbox{ are independent RVs, find }}f_{X+Y}(z)$

• Start by converting the pdf's to indicator functions
  • $f_{X}(x)={\frac {3x^{2}}{2}}\cdot 1_{[-1,1]}(x)\qquad f_{Y}(y)={\frac {y}{9}}\cdot 1_{[4,5]}(y)$
    • Now $f_{X}(x)\,$ is defined only when $x\in [-1,1]\,$ and $f_{Y}(y)\,$ is defined only when $y\in [4,5]\,$
• Use the convolution formula above to write out the integral
  • $\int _{-\infty }^{\infty }f_{X}(z)f_{Y}(y-z)\delta z=\int _{-\infty }^{\infty }{\frac {3z^{2}}{2}}\cdot 1_{[-1,1]}(z){\frac {y-z}{9}}\cdot 1_{[4,5]}(y-z)\delta z$
• Factor out any constants, in this case, a multiplier
  • ${\frac {3}{18}}\int _{-\infty }^{\infty }z^{2}\cdot 1_{[-1,1]}(z)(y-z)\cdot 1_{[4,5]}(y-z)\delta z={\frac {1}{6}}\int _{-\infty }^{\infty }z^{2}\cdot 1_{[-1,1]}(z)(y-z)\cdot 1_{[4,5]}(y-z)\delta z$
• Factor out the indicator function for $(z)\,$ into the integral bounds
  • ${\frac {1}{6}}\int _{-1}^{1}z^{2}(y-z)\cdot 1_{[4,5]}(y-z)\delta z$
    • Note that $4\leq y-z\leq 5,({\mbox{ try to isolate }}z)\rightarrow 4-y\leq -z\leq 5-y\rightarrow y-5\leq z\leq y-4$
• Now that have isolated the indicator for z, we can combine the entire integral for that indicator
  • ${\frac {1}{6}}\int _{-1}^{1}\left(z^{2}(y-z)\right)\cdot 1_{[y-5,y-4]}(z)\delta z={\frac {1}{6}}\int _{-1}^{1}\left(z^{2}y-z^{3}\right)\cdot 1_{[y-5,y-4]}(z)\delta z$
• Finally, split the integral into the separate cases based on the remaining indicator function
  • $f_{X+Y}(z)={\frac {1}{6}}{\begin{cases}0&y<3\\\int _{-1}^{y-4}(z^{2}y-z^{3})\delta z&3\leq y<4\\\int _{y-5}^{y-4}(z^{2}y-z^{3})\delta z&4\leq y<5\\\int _{y-5}^{1}(z^{2}y-z^{3})\delta z&5\leq y<6\\0&y\geq 6\\\end{cases}}\qquad ={\begin{cases}0\\-{\frac {255}{4}}+43z+{\frac {z^{4}}{12}}-8z^{2}\\{\frac {369}{4}}-{\frac {122z}{3}}+{\frac {9z^{2}}{2}}\\156-83z-{\frac {z^{4}}{12}}+{\frac {25z^{2}}{2}}\\0\\\end{cases}}$
    • When $y<3\,$ the integral has no bounds since $(y<3)-4<-1\,$ so the upper bound would be less than $-1\,$ which would be $0\,$.
    • When $3\leq y<4\,$ the integral is bound between $-1\,$ and $y-4\,$since $y-4\,$ will be at least $-1\,$ but less than $0\,$
    • As you can see there is a pattern here, it goes as follows:
      • Given $\int _{a}^{b}(\cdot )1_{[c,d]}\delta z\,$ you will have ${\begin{cases}y<(d-a=0)\\(d-a=0)\leq y<(c-a=0)\\(c-a=0)\leq y<(d-b=0)\\(d-b=0)\leq y<(c-b=0)\\y\geq (c-b=0)\\\end{cases}}$