Might be a straightforward question, but I've been getting some mixed answers and I'd like to know for sure if my understanding is correct. Suppose I've some random variable $X$ and the expected value of this random variable is to be used in an equation. For example, let $X$ be the amount of apples per person, and I'm interested in calculating the amount of apples eaten by a group of people over time. A differential equation for the amount of apples in the population, $A$, may look something like this:
$$\dot{A} = - \alpha A \sum_x xp(x),$$ where each individual in the population has probability $p(x)$ of possessing $x$ apples, and $\alpha$ is the apple-eating rate. The sum is just the expected value of the number of apples in the population, and we let this be denoted by $\mu$, so our differential equation becomes $$\dot{A} = -\alpha \mu A$$ but now the units don't quite work out, as $[\alpha] = [1/time]$ and $[A] = [apples]$. Is $\mu$ unitless or am I formulating the equation incorrectly? I've even seen instances where the mean is dependent on a state variable, e.g. $\mu = N/A$ for some other state variable $N$, which would make the above differential equation $$\dot{A} = -\alpha A \left(\frac{N}{A}\right) = -\alpha N,$$ which can be a bit confusing dimension-wise, especially when $N$ has nothing to do with apples! This question applies to higher moments as well, e.g. problems of the form $$\dot{A} = -\alpha A \sum_x x^2 p(x) = -\alpha A E(X^2).$$
The mean has the same units as the random variable, period. Then, for a dimensionally consistent equation, you should have $[\alpha]=([X][{\rm time}])^{-1}$
The same goes for the units of the moments, i.e. $[ E(X^k)] = [X]^k$