Does absence of endogeneity problem means absence of heteroscedasticity?
For example, if an estimator is designed to correct for endogeneity does this mean that heteroscedaticity will be corrected for, too?
Solved – Does absence of endogeneity problem means absence of heteroscedasticity
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The article never assumed homoskadasticity in the definition. To put it in the context of the article, homoskedasticity would be saying $$ E\{(\hat x-x)(\hat x-x)^T\}=\sigma I $$ Where $I$ is the $n\times n$ identity matrix and $\sigma$ is a scalar positive number. Heteroscadasticity allows for
$$ E\{(\hat x-x)(\hat x-x)^T\}=D $$
Any $D$ diaganol positive definite. The article defines the covariance matrix in the most general way possible, as the centered second moment of some implicit multi-variate distribution. we must know the multivariate distribution of $e$ to obtain an asymptotically efficient and consistent estimate of $\hat x$. This will come from a likelihood function (which is a mandatory component of the posterior). For example, assume $e \sim N(0,\Sigma)$ (i.e $E\{(\hat x-x)(\hat x-x)^T\}=\Sigma$. Then the implied likelihood function is $$ \log[L]=\log[\phi(\hat x -x, \Sigma)] $$ Where $\phi$ is the multivariate normal pdf.
The fisher information matrix may be written as $$ I(x)=E\bigg[\bigg(\frac{\partial}{\partial x}\log[L]\bigg)^2 \,\bigg|\,x \bigg] $$ see en.wikipedia.org/wiki/Fisher_information for more. It is from here that we can derive $$ \sqrt{n}(\hat x -x) \rightarrow^d N(0, I^{-1}(x)) $$ The above is using a quadratic loss function but does not assuming homoscedasticity.
In the context of OLS, where we regress $y$ on $x$ we assume $$ E\{y|x\}=x'\beta $$ The likelihood implied is $$ \log[L]=\log[\phi(y-x'\beta, \sigma I)] $$ Which may be conveniently rewritten as $$ \log[L]=\sum_{i=1}^n\log[\varphi(y-x'\beta, \sigma)] $$ $\varphi$ the univariate normal pdf. The fisher information is then $$ I(\beta)=[\sigma (xx')^{-1}]^{-1} $$
If homoskedasticity is not meet, then the Fisher information as stated is miss specified (but the conditional expectation function is still correct) so the estimates of $\beta$ will be consistent but inefficient. We could rewrite the likelihood to account for heteroskacticity and the regression is efficient i.e. we can write $$ \log[L]=\log[\phi(y-x'\beta, D)] $$ This is equivalent to certain forms of Generalized Least Squares, such as Weighted least squares. However, this will change the Fisher information matrix. In practice we often don't know the form of heteroscedasticity so we sometimes prefer accept the inefficiency rather than chance biasing the regression by miss specifying weighting schemes. In such cases the asymptotic covariance of $\beta$ is not $\frac{1}{n}I^{-1}(\beta)$ as specified above.
Q1 "why [do] count data tend to be heteroscedastic"?
If we want to model counts as random, then the Poisson distribution, which is heteroscedastic, provides a natural characterisation of what 'random counts' might usefully mean. Hence one way to ask why count data is heteroscedastic is to ask why count data might be Poisson distributed. For this there are various derivations e.g. the 'Law of Rare Events' discussed in the link.
Poisson is not the only characterisation of 'random counts' that is possible, of which more below.
Q2 "is heteroscedasticity...something that [I] should be concerned about in [a] [P]oisson model if [I'm] using [dependent] variable that is consider to be count data?"
If you are running a regression that assumes that your dependent variable is Poisson distributed with a mean that depends on some covariates, e.g. a Generalised Linear Model, then you are already taking into account the heteroscedasticity due to being Poisson. However...
Overdispersion
This kind of model assumes that once the covariates have determined the expected mean then the remaining variation in your data is Poisson. But if you have missed out some important variables (which most of us do, most of the time) then the true mean might still be different for different values of those unseen variables, even if the variables that are in the model are the same. This is referred to as overdispersion and is a distinct variance-related issue you will want to think about. (Actually this is only one of several mechanisms that generates overdispersion, but it's enough for now).
The solution is to model the extra variation explicitly: Negative Binomial regression models are one class of models that do that.
Best Answer
No, not at all. Endogeneity is a first-moment problem, while heteroskedasticity is a second-moment problem.
For example, consider a linear model
$$ y_i=x_i\beta+u_i $$ An assumption like $$E(u_i|x_i)=0$$ implies no correlation of error term and regressor, so no endogeneity. No (conditional) heteroskedasticity in turn can be written as $$Var(u_i|x_i)=\sigma^2,$$ where $\sigma^2$ is a constant number.
There is no reason whatsoever that $$E(u_i|x_i)=0$$ would imply $$Var(u_i|x_i)=\sigma^2.$$