How to find the limits of integration for a convolution of pdfs?

Question

As the title suggests. I have been trying to figure this out for a long time now and I'm completely out of ideas.

Say we have two pdfs $f_X(x) = \begin{cases} 1/2 & 0 \le x \le 2 \\ 0 & elsewhere \\ \end{cases} $

and $f_Y(y) = \begin{cases} 1/3 & 0 \le x \le 3 \\ 0 & elsewhere \\ \end{cases}$

Obviously, the random variables $X$ and $Y$ have uniform distributions.

When I attempt to find the sum of these variables by the convolution, I cannot figure out how to work out the limits of the integral. Can anyone explain in detail how it works and possibly provide a general solution for two arbitary random variables?

EDIT: the convolution formula is given by $f_Z(z) = \int_{-\infty}^{\infty} f_X(x)f_Y(z-x) dx$

Answer 1

You are simply integrating over the joint support -- that is: where both functions are not zero.

$$\begin{align}f_Z(z)&=\int_{0\leq x\leq 2,0\leq z-x\leq 3}\tfrac 16\,\mathrm dx\\[1ex]&=\mathbf 1_{0\leq z\leq5}\int_{\max\{0,z-3\}\leq x\leq\min\{2,z\}}\tfrac 16\,\mathrm dx \\[3ex]&=\dfrac{1}{6}\cdot\begin{cases}\underline{\phantom{z\qquad}} &:& 0\leq z\lt 2\\\underline{\phantom{2\qquad}}&:& 2\leq z\lt 3\\\underline{\phantom{5-z~}}&:& 3\leq z\leq 5\\0&:&\text{elsewhere}\end{cases} \end{align}$$

Answer 2

Hints:

1. Figure out the range of values that $Z=X+Y$ can take on. Write down that $f_Z(z)$ for all $z$ outside the range you found.
2. Pick a number (say $1.12$) from the range found in Step 1.
3. Write the convolution integral $\displaystyle f_Z(1.12) = \int_{-\infty}^\infty f_X(x)f_Y(1.12-x) \,\mathrm dx.$
4. Notice that as $x$ sweeps from $-\infty$ to $\infty$, $f_X(x)$ has value $0$ everywhere except when $x\in [0,2]$ (where it has value $\frac 12$) and so the integral in Step 3 can be simplified to $$f_Z(1.12) = \int_{0}^2 \frac 12f_Y(1.12-x) \,\mathrm dx.$$
5. Notice that as $x$ from $-\infty$ to $\infty$, $f_Y(1.12-x)$ has value $0$ everywhere except when $1.12-x\in [0,3]$, that is, when $x\in [-2.88,1.12]$, and in this interval, $f_Y(1.12-x)$ has value $\frac 13$. Hence, the integral in Step 4 can be further simplified to $$f_Z(1.12) = \int_{0}^{1.12} \frac 12\cdot \frac 13 \,\mathrm dx.$$
6. Evaluate the integral in Step 5 to get the value of $f_Z(1.12)$.
7. Return to Step 2 to pick another number and repeat the whole process until you are sick and tired of the drudgery involved.

If you work carefully and systematically, writing down all results, after a while, you might observe a trend resulting in an "Hey Ma, I think I am beginning to see a pattern here!" moment, and be able to write down expressions (a.k.a. formulas) for $f_Z(z)$ that apply depending on the value of $z$. Be warned: there are three different formulas that you ought be coming up with and if you find fewer (or more!), you have done something wrong somewhere.

Answer 3

This should answer the question concerning the general case.

Let the pdfs $f$ respectively $g$ be zero outside the real intervals $[a,b]$ respectively $[c,d]$, where $a<b$ and $c<d$.

We want to restrict the integral

$$ (f * g) (z) = \int_{-\infty}^{\infty} f(x) g(z-x) dx$$

to appropriate intervals.

We know that

• $f(x)$ is zero if $x\notin[a,b]$ and that
• $g(z-x)$ is zero if $z-x \notin [c,d] \iff x\notin [z-d,z-c].$

Therefore, integration can be restricted to the interval $[a,b] \cap [z-d, z-c]$.

In other words, the integration variable needs only to run from $\max(a, z-d)$ to $\min(b, z-c)$.

Of course, if it turns out that $\max(a, z-d) > \min(b, z-c)$, then the convolution is zero.

In summary, $$ (f * g) (z) = \int_{\max(a, z-d)}^{\min(b, z-c)} f(x) g(z-x) dx$$ if $\max(a, z-d) < \min(b, z-c)$ and zero otherwise.