Convolution of a function with itself I was going through this question. In the answer, the limits of the integral was transformed from (-infinity)-(+infinity) to 0 to x. Can anyone explain how this happened?
Asked
Active
Viewed 673 times
0
-
They used the fact that the function is zero outside of $[0,1]$. – Cameron Williams Jul 24 '16 at 00:25
-
@CameronWilliams In the first case, when $x$ is between 0 and 1, how did they come up with the limits of $0$ and $x$? – meta_finance Jul 24 '16 at 00:29
1 Answers
2
The function is non-zero only in $[0,1]$. So, for the convolution to be non-zero you need $$ 0 < x - y < 1 \implies x-1<y<x \ $$ $$ \ \text{and} \ 0<y<1.$$ If $0<x<1$ then clearly $$0<y<x$$ has to hold by the two inequalities above.
Andrew Whelan
- 2,138