Relation between Correlation and Convolution We have two functions of time $f(t)$ and $g(t)$, for which convolution and correlation are defined as following:
Convolution: $(f(t)\ast g(t))(\tau) = \int_{-\infty}^\infty{f(t)g(\tau-t)dt}$
Correlation: $(f(t)\star g(t))(\tau) = \int_{-\infty}^\infty{f^\ast(t)g(\tau+t)dt}$
In the english wikipedia and in other sources I found that the following relationship should hold:
$(f(t)\star g(t))(\tau) = (f^\ast(-t)\ast g(t))(\tau)$
Is this correct? If so, how can i prove this? Usually, i would try substitution, but how to change the $g(\tau+t)$ to $g(\tau-t)$?
 A: I figured out the answer while writing down the question. Here it is:
$(f^\ast(-t)\ast g(t))(\tau) = \int_{-\infty}^\infty{f^\ast(-t)g(\tau-t)dt} = -\int_{\infty}^{-\infty}{f^\ast(t)g(\tau+t)dt} = \int_{-\infty}^{\infty}{f^\ast(t)g(\tau+t)dt} = (f(t)\star g(t))(\tau)$
In the second step, the substitution $t\to -t$ took place.
A: We have -
Convolution - $ f(t) * g(t) = \int_{-\infty}^{\infty}f(\tau) . g(t-\tau) d\tau$
Correlation - $ R_{fg}(\tau) = \int_{-\infty}^{\infty}f(t) . g(t-\tau) dt$
Now, 
$$ f(t) * g(t) = \int_{-\infty}^{\infty}f(\tau) . g(t-\tau) d\tau$$
Replace $\tau$ by $u$-
$$ f(t) * g(t) = \int_{-\infty}^{\infty}f(u) . g(t-u) du$$
Replace $t$ by $k$-
$$f(t) * g(t) = \int_{-\infty}^{\infty}f(u) . g(k-u) du
              = \int_{-\infty}^{\infty}f(t) . g(k-t) dt
              = \int_{-\infty}^{\infty}f(t) . g(-(t-k)) dt = f(t) \star g(-t)$$
which is the correlation between $f(t)$ and $g(-t)$.
This can be seen as -
$$f(t) \star g(-t) = \int_{-\infty}^{\infty}f(t) . g(-(t-\tau)) dt$$

So, $(f(t)⋆g(-t))(\tau)=(f(t)∗g(t))(\tau)$.

Similarly, we can prove that $(f(t)⋆g(t))(\tau)=(f(t)∗g(-t))(\tau)$.
