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 $\begin{cases}y < (d - a = 0)\\ (d - a = 0) \leq y < (c - a = 0)\\ (c - a = 0) \leq y < (d - b = 0)\\ (d - b = 0) \leq y < (c - b = 0)\\ y \geq (c - b = 0)\\ \end{cases}$