What are units for this metric? Suppose I want to prove that there is a correlation between number of bugs hitting the windshield per second at any given point in time $ b(t) = \frac{dB(t)}{dt} $ and the speed at which a car is going $ d(t) = \frac{d D(t)}{dt}$ where $B(t)$ is a cumulative function of the number of bugs and $D(t)$ is a cumulative function of the number of miles driven.
I am choosing to take an overlap of these two functions $\int{d(t)*b(t) dt}$ and then will normalize by $ \sqrt{\int{d(t)^2 dt} * \int{b^2(t) dt}}$ but don't know what the final units would be for this.
Note that I am also approaching this using PCC, covariance, etc. but am specifically interested in using this overlap methodology here. What would the units of $\int{d(t)*b(t) dt}$ be? After normalization, what would the units be?
 A: Letting,
$$\langle f, g \rangle := \int f(t) g(t) dt$$
The expression
$$\frac{\langle f, g \rangle}{\sqrt{\langle f, f \rangle \langle g, g \rangle}}$$
is unitless. It is the infinite dimensional extension of the cosine of the angle between two vectors, $f$ and $g$.
For your problem, just looking at the numerator, $b(t)d(t)$ has units of
$\frac{\text{bugs}}{\text{second}}\frac{\text{miles}}{\text{second}}$. The integration is over time, so we lose one unit of seconds (think about the definition of the integral, we are multiplying by $dt$ which has units of time and then just taking a limit), leaving us with, $$\frac{\text{bugs}\times\text{miles}}{\text{seconds}}$$
The denominator will have the same units after simplification.
By the way, if $f$ and $g$ are considered random variables and the integrals are taken with respect to their joint probability measure, then that expression is the PCC (Pearson correlation coefficient) and the numerator is their covariance.
Hope that clears things up!
