0
$\begingroup$

I have two functions $f(x)$ and $g(x)$ that are nonzero everywhere except in the interval $[a,b]$ (i.e. there are a $h(x)$ and $j(x)$ such that $f(x) = \mathbb{1}_{a\leq x\leq b} \cdot h(x)$ and $g(x) = \mathbb{1}_{a\leq x\leq b} \cdot j(x)$).

The general convolution definition is $$\left(f*g\right)(x)=\int_{-\infty}^{+\infty}f(y) g(x-y)\text dy$$

When $a=0$ and $b=+\infty$ $$\left(f*g\right)(x)=\int_{0}^{x}f(y) g(x-y)\text dy=\int_{0}^{x}f(x-y) g(y)\text dy$$

What are the integration limits for arbitrary $a$ and $b$ (with $a < b$)?

$\endgroup$
0
$\begingroup$

You just have to check where the functions could be both nonzero. In particular you need $$y\in [a,b]\quad\mathrm{and}\quad x-y\in[a,b]$$ that is $$a\leq y\leq b\quad \mathrm{and}\quad x-b\leq y\leq x-a$$ that is equivalent to $$y\in[a,b]\cap[x-b,x-a].$$ Depending on $x$ this could be even the emptyset.

$\endgroup$
  • $\begingroup$ So the limits would be $\max(a,x-b)$ and $\min(x-a,b)$ for $x\in[2a,2b]$, seems like. $\endgroup$ – Pedro Carvalho Sep 23 '16 at 18:38
  • $\begingroup$ @Pedrocarvalho it should be that for $x\in [2b,2a]$ $\endgroup$ – Del Sep 24 '16 at 7:03
  • $\begingroup$ I don't think so? First because, since $a<b$, then $[2b,2a]=\varnothing$. Second, if $f$ is the pdf of a random variable $X$ and $g$ that of a random variable $Y$ then the convolution is the pdf of the random variable $X + Y$ which can take values between $2a$ and $2b$ in that order, so that's also the support of the convolution. $\endgroup$ – Pedro Carvalho Sep 24 '16 at 13:53
  • $\begingroup$ @PedroCarvalho Yes you are right, I had confused the two extremes for a moment $\endgroup$ – Del Sep 24 '16 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.