# Convolution of two distribution functions

The convolution of two distribution functions $F_1$ and $F_2$ is

$F(t) = \int_{0}^{t} F_2(t-x) dF_1(x)$.

This can also be expressed as

$F(t) = \int_{0}^{t} F_1(t-x) dF_2(x)$.

The similar expression holds for survival function also. Now, I am reading a book which mentions that

$\bar F(t-u) = \int_{-\infty}^{\infty} \bar F_1(t-s) f_2(s-u) ds$,

where $f_2$ is the density of $F_2$. I am unable to understand how are we getting the above step, esp. how is the upper limit changing to $\infty$. Can anybody please explain it?

• I don't think your limits of integration are right, to begin with, on the convolution of two distributions. It should be, I think, $\int_{-\infty}^\infty$. Some distributions won't actually require that whole range (depends on where/if they are zero for |x|>k), but normally the limits are infinite. – eSurfsnake Aug 21 '18 at 3:33
• Yes, you are right. I have changed the limits. Still, can you explain how are we getting that step? – User123 Aug 21 '18 at 3:36
• You can get displayed equations by enclosing them in double instead of single dollar signs. – joriki Aug 21 '18 at 3:59
• I didn't get your point @joriki – User123 Aug 21 '18 at 4:02
• You have three equations in your post that aren't inline. (These are called "displayed equations".) But you formatted them as inline equations by surrounding them with single dollar signs. That makes them come out cramped, harder to read and uncentred. Their formatting will be improved if you surround them with double dollar signs instead. – joriki Aug 21 '18 at 4:09