Think of a linear regression model: $$ Y=X\beta+\epsilon $$ where $\epsilon|X\sim N\left(0,\sigma^{2}\right)$. The vector of parameters $\beta$ can be consistenly estimated by OLS. I have one conceptual question here: first, $\epsilon$ has units of $Y.$ However, when we minimize the (unweighted) squared sum of residuals, we treat each observation the same. Conceptually, think of two different values of $X.$ If the true conditional mean function is linear, that is in the population: $$ \mathbb{E}\left(Y|X\right)=X\beta $$ then the conditional mean of $Y$ changes with values of $X.$ If $\beta>0$ , then the value of the conditional mean grows with the values of $X.$ However, under homoskedasticity, the variance of the error term is assumed to be constant. However, for large values of the conditional mean, the percentage deviation of the disturbance decreases and tends to 0. In other words, the error dispersion becomes more and more miniscule. If $X$ is non-stochastic, then the conditional variance of $Y$ is $\sigma^{2}$ as well. As such, the conditional coefficient of variaion of $Y$ is: $$ \frac{\sigma^{2}}{X\beta} $$ which goes to $0,$ simply as $X$ increases.. I guess my question is isn't the heteroskedastic assumption restrictive in its basic principle. In other words, even if heteroscedasticity is not related to the value of the regressors, isn't this conceptually incorrect as we are imposing that the average percentage deviation of the error term from the conditional mean decreases without bound? Should the error term not also be scaled by the value of $Y?$ Another way of stating is: an error of an equal magnitude means different things for different values of the regressand/regressors. An error of 1 when Y=10 is much more than an error of 1 when Y=100000; but OLS treats these as symmetric.
5 Answers
It appears you are describing the well-studied case of the error term being conditionally heteroskedastic, $\text{Var}(\epsilon_i) = h(\mathbf x_i)$. Usual functional forms assumed for $h(\mathbf x_i)$ are
$$ h(\mathbf x_i) = (\mathbf a' \mathbf x)^2, \;\;\;h(\mathbf x_i) = \exp\{\mathbf a' \mathbf x\}$$
These usually imply that the variance of $\epsilon$ increases with the level of the regressors, and so maintain some degree of proportionality.
After H. White developed the theory of heteroskedasticity-robust and consistent estimation without the need to specify a functional form for the heteroskedastic variance, such specifications have been largely abandoned, at least in the field of econometrics, partly because of misspecification worries due to the arbitrary assumption about the functional form of $h(\mathbf x_i$).
It is standard practice to assume an heteroskedastic error and apply White's estimator or some variant of it to estimate the covariance matrix.
If you have an application where you believe that the conditional coefficient of variation of $Y$ should be constant in $X$, then you probably want a log-log regression model like:
\begin{align} \ln{Y} = \ln{X} \beta + \epsilon \end{align}
$Y$ then has conditional mean and variance:
\begin{align} E\{Y|X\} &=\exp{(\ln{X}\beta)} \cdot E\{\exp(\epsilon)\}\\ V\{Y|X\} &=(\exp{(\ln{X}\beta)})^2 \cdot V\{\exp(\epsilon)\} \end{align}
The conditional coefficient of variation of $Y$ is then:
\begin{align} \frac{\sqrt{(\exp{(\ln{X}\beta)})^2 \cdot V\{\exp(\epsilon)\}}}{\exp{(\ln{X}\beta)} \cdot E\{\exp(\epsilon)\}\\} = \frac{\sqrt{V\{\exp(\epsilon)\}}}{E\{\exp(\epsilon)\}\\} \end{align}
This is invariant to $X$.
You have a minor typo in your question. The coefficient of variation has the standard deviation of $Y$ in the numerator, not the variance of $Y$.
I think you may be confused with the term "error". Once conditioning $X$, the conditional variance does not depend on $X$ in the constant variance assumption. It only tells you how accurate your "measurement on the each $X$" not about how accurate your "final value" itself.
Here is a simple example. Let say I measure the height of my house. I have a poor tool so whenever I measure the height, it looks like 5m +- 1m. Here my "error" looks large (it is about 20% of the height of my house.) However, let say I know my house is on a hill with exactly 1000m altitude. Then, my measure for altitude of my house is 1005m +- 1m. Hence I have a very precise number for the altitude of my house (error is less than 0.1%). However, still variances of my measurements for the height of my house and altitude of my house are same to each other.
An error of 1 when Y=10 is not much more than an error of 1 when Y=100000.
They are both the same error of 1.
The model is just stating what is. You have variables Y that are a sum of the mean term $X\beta$ plus some noise $\epsilon$ distributed according to $N(0,\sigma^2)$.
You may feel that the noise level is 'worse'. But that is a subjective term, unlike 'more' which can be objectively quantified.
The big worry seems to be that $\sigma$ is not concerned with the scale of your Y, you are specifically worried about the proportion $\sigma/Y$. As already mentioned, classical statistics typically focuses on the magnitude $\sigma$, not $\sigma/Y$. This, in part, is because we can simply calculate that proportion if so desired. For example, when working with zero inflated or overdispersed distributions it is not unusual to discuss the $\bar{x}/\sigma^2$, which is certainly tied to what you are interested in.
One reason (I suspect) we do not use that measure is because practitioners often scale your Y variables in linear regression. This renders the interpretation of the proposed proportion $\sigma/Y$ as universal standard very tricky. Furthermore many regressions (and ML processes) do request standardized inputs prior, which renders the estimates very similar to the measure you're requesting.
@Alecos Papadopoulos has also correctly pointed out that your concern about proportionality of $\sigma/Y$ is in the neighborhood of other concerns about heteroskedacity and heteroskedacity-robust estimators. If you find that there is strong proportionality between $\sigma/Y$, you should investigate heteroskedacity.
