Intuition behind convolution identity for Laplace transforms Convolutions, relatively speaking, are fairly straightforward for simple systems (from an applied perspective), but I cannot, at all, find the intuition behind the Laplace identity for convolutions. That is:
$$
\mathcal{L}\{f\star g\}=\mathcal{L}\{f\}\cdot \mathcal{L}\{g\}
$$
How is it possible to make sense of this? Perhaps there is some probability theory example that sheds light on it?
Thanks in advance.

Well convolutions are really DEFINED to make that identity true. Their use comes into play when you want to separate the elements of the equations that relate to the system itself,  those that relate to the initial conditions of the system, and those that relate to external forces. For example, consider
$$ ay''+by'+cy = g(x) $$
Transforming, 
$$a\big (s^2Y(s)-sy(0)-y'(0)\big ) + b\big (sY(s) - y(0) \big ) + cY(s) = G(s) $$
Now let 
$$H(s)=\frac 1{as^2+bs+c}$$
$$C(s) = sy(0)+y'(0)+y(0)$$
Then $Y(s) = G(s)H(s) + C(s)H(s) $ and using our definition of convolution, 
$$y(x) = g(x) ⋆ h(x) + c(x) ⋆ h(x) $$
If the convolution is easier to calculate than the inverse transform, then the convolution theorem can be used to simplify things and separate the aspects of the equations logically. 
I'll try to review this post when I'm not so tired in the morning. 
