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

Convolution

• $f_{X+Y}(z)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{f}_X(y-z)f_Y(z)\delta z=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{f}_X(z)f_Y(y-z)\delta z$

Example

$f_X(x)=\{\begin{array}{cc}\frac{3x^2}{2}& x\in [-1,1]\\ 0& else\end{array}{f}_Y(y)=\{\begin{array}{cc}\frac{y}{9}& y\in [4,5]\\ 0& else\end{array}X\text{ and }Y\text{ are independent RVs, find }{f}_{X+Y}(z)$

• Start by converting the pdf's to indicator functions
  • $f_X(x)=\frac{3x^2}{2}\cdot 1_{[-1,1]}(x)f_Y(y)=\frac{y}{9}\cdot 1_{[4,5]}(y)$
    • 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
  • $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{f}_X(z)f_Y(y-z)\delta z=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{3z^2}{2}\cdot 1_{[-1,1]}(z)\frac{y-z}{9}\cdot 1_{[4,5]}(y-z)\delta z$
• Factor out any constants, in this case, a multiplier
  • $\frac{3}{18}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{z}^2\cdot 1_{[-1,1]}(z)(y-z)\cdot 1_{[4,5]}(y-z)\delta z=\frac{1}{6}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{z}^2\cdot 1_{[-1,1]}(z)(y-z)\cdot 1_{[4,5]}(y-z)\delta z$
• Factor out the indicator function for $(z)$ into the integral bounds
  • $\frac{1}{6}\underset{-1}{\overset{1}{\int }}{z}^2(y-z)\cdot 1_{[4,5]}(y-z)\delta z$
    • Note that $4\le y-z\le 5,(\text{ try to isolate }z)\to 4-y\le -z\le 5-y\to y-5\le z\le y-4$
• Now that have isolated the indicator for z, we can combine the entire integral for that indicator
  • $\frac{1}{6}\underset{-1}{\overset{1}{\int }}(z^2(y-z))\cdot 1_{[y-5,y-4]}(z)\delta z=\frac{1}{6}\underset{-1}{\overset{1}{\int }}(z^{2}y-z^3)\cdot 1_{[y-5,y-4]}(z)\delta z$
• Finally, split the integral into the separate cases based on the remaining indicator function
  • $f_{X+Y}(z)=\frac{1}{6}\{\begin{array}{cc}0& y<3\\ \underset{-1}{\overset{y-4}{\int }}(z^{2}y-z^3)\delta z& 3\le y<4\\ \underset{y-5}{\overset{y-4}{\int }}(z^{2}y-z^3)\delta z& 4\le y<5\\ \underset{y-5}{\overset{1}{\int }}(z^{2}y-z^3)\delta z& 5\le y<6\\ 0& y\ge 6\end{array}=\{\begin{array}{c}0\\ -\frac{255}{4}+43z+\frac{z^4}{12}-8z^2\\ \frac{369}{4}-\frac{122z}{3}+\frac{9z^2}{2}\\ 156-83z-\frac{z^4}{12}+\frac{25z^2}{2}\\ 0\end{array}$
    • 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 $3\le y<4$ 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 $\underset{a}{\overset{b}{\int }}(\cdot )1_{[c,d]}\delta z$ you will have $\{\begin{array}{c}y<(d-a=0)\\ (d-a=0)\le y<(c-a=0)\\ (c-a=0)\le y<(d-b=0)\\ (d-b=0)\le y<(c-b=0)\\ y\ge (c-b=0)\end{array}$