correlation operation written using the convolution operator: h(-w) * R(w)

Question

Below is a convolution integral.
$$\int \limits_{-\infty}^{\infty} h(\alpha)~R_{XY}(\tau+\alpha)~d\alpha = h(-\tau) * R_{XY}(\tau)$$

But, How did they get the minus sign in $h(-\tau)$, as in $h(-\tau) * R_{XY}(\tau)$?

I thought for the convolution operator, it was a requirement that the parameter to the function on the left side and right side are the same? which is why they write $(f*g)(t) = f(t)*g(t)$... but here we have $f(-t) * g(t)$ ... is this right?

does:

$f(t) \star g(t) = f(-t) * g(t)$

Answer 1

Yes, they are equivalent. Assuming $\star$ for correlation and * for convolution.

$f(t) \star g(t) = \int_{-\inf}^{\inf} f(\tau) g(\tau + t) d\tau$

Let $\alpha = -\tau$ so $d\tau = -d\alpha$:

$= \int_{\inf}^{-\inf} f(-\alpha) g(- \alpha + t) (- d\alpha )$

$= - \int_{\inf}^{-\inf} f(-\alpha) g(- \alpha + t) d\alpha$

$= \int_{-\inf}^{\inf} f(-\alpha) g(t - \alpha) d\alpha$

$= f(-t) * g(t)$