# 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 $g(\tau-t)$?

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.

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