An interpretation: if you sample Y through a process of getting partial information step by step, the variance of each step adds up to the variance of sampling Y directly
The first two terms are V_{tot}(Y) and E[Var_{rem}(Y|X)] respectively, while the last part describes the “explained” variance.
To give an intuition for Var(E[Y|X]):
If X gives me some information about Y, then my new mean for Y should change depending on X. If X gives little information, then it should only wiggle my mean estimate of Y a little (low variance), but a very explanatory X will move my mean estimate of Y a lot (high variance)
If X gave no information, then E[Y|X] should have no variance (it’s always equal to the mean E[Y]).
If X completely explains Y, then E[Y|X] can equal any value in the domain of Y. Because every y has a corresponding x, that if sampled, means that P(Y=y|X=x) = 1. Indeed, E[Y|X] will have exactly the same distribution as Y, and so it will contain the full variance as Y
The relevant intuition I use comes from the [law of total variance](https://en.m.wikipedia.org/wiki/Law_of_total_variance) (or variance decomposition formula):
Var(Y)=E[Var(Y|X)]+Var(E[Y|X])
An interpretation: if you sample Y through a process of getting partial information step by step, the variance of each step adds up to the variance of sampling Y directly
The first two terms are V_{tot}(Y) and E[Var_{rem}(Y|X)] respectively, while the last part describes the “explained” variance.
To give an intuition for Var(E[Y|X]):
If X gives me some information about Y, then my new mean for Y should change depending on X. If X gives little information, then it should only wiggle my mean estimate of Y a little (low variance), but a very explanatory X will move my mean estimate of Y a lot (high variance)
If X gave no information, then E[Y|X] should have no variance (it’s always equal to the mean E[Y]).
If X completely explains Y, then E[Y|X] can equal any value in the domain of Y. Because every y has a corresponding x, that if sampled, means that P(Y=y|X=x) = 1. Indeed, E[Y|X] will have exactly the same distribution as Y, and so it will contain the full variance as Y