Independence and homoskedasticity

Question

Does $E(u|x)=0$ imply homoscedasticity? If yes, why?

Also, does $E(u|x)=0$ mean that $u$ and $x$ are fully independent?

If answers to both these question are no:

Does full independence of $u$ and $x$ imply homoscedasticity? If yes, why?

EDIT

Motivation for my question is the following:

In Wooldridge's introductory econometrics it is suggested that you don't really need a homoskedasticity assumption to estimate the variance of OLS estimators if other simple linear regression assumptions hold.

According to the Author, the homoscedasticity assumption is added just to "simplify calculations".

However, this idea is not elaborated any further.

Answer 1

In order to see how the two assumptions does not imply eachother, we can simply run a small simmulation. Here I simmulate data such $E(u|X)=0$, but where $\sigma^2_u$ is not a constant.

n <- 100000
x <- rgamma(n, shape = 2)
u <- rnorm(n ,sd = abs(sin(x)))
y <- 5 + 7*x + u
# Regression:
lm(y ~ x)

Here we recover the true estimates, but the variance is wrong (so inference is wrong). We can however estimate the variance consistently using the a robust standard error matrix. There are many questions on this site that details how that works. In R we can use the package sandwich.

Answer 2

About this problem "Independence and homoskedasticity" some details more needed. However stating that we talk about a linear regression like:

$y = X’\beta + u$

Does $E(u|x)=0$ imply homoscedasticity? If yes, why?

In general no, but some forms of heteroskedasticity imply $E[u^2x] \neq 0$ and these go away.

Also, does $E(u|x)=0$ mean that $u$ and $x$ are fully independent?

No, stochastic independence is stronger condition.

If answers to both these question are no: does full independence of $u$ and $x$ imply homoscedasticity? If yes, why?

As before, in general no. However sometimes heteroskedasticity is defined as:

$V[\epsilon|X=x] = E[u^2|X=x] = \sigma^2(x)$ (a function of $X$)

and in this case stochastic independence between $u$ and $X$ imply homoscedasticity.

However sometimes the definition is like:

$V[u|Z=z] = E[u^2|Z=z] = \sigma^2(z)$

where $Z$ not necessarily include some elements of $X$. In this case: independence between $u$ and $X$, and heteroskedasticity; can hold simultaneously. Read here too Homoscedasticity and independence of errors

Finally

According to the Author, the homoscedasticity assumption is added just to "simplify calculations". However, this idea is not elaborated any further

OLS estimator can holds under heteroskedasticity also, because it can remain consistent. However in this situation robust standard error, as White version, are needed.