The Least Squares Assumptions Assume the following linear relationship:
$Y_i = \beta_0 + \beta_1 X_i + u_i$, where $Y_i$ is the dependent variable, $X_i$ a single independent variable and $u_i$ the error term.
According to Stock & Watson (Introduction to Econometrics; Chapter 4), the third least squares assumption is that the fourth moments of $X_i$ and $u_i$ are non-zero and finite $(0<E(X_i^4)<\infty \text{ and } 0<E(u_i^4)<\infty)$.
I have three question:


*

*I do not fully understand the role of this assumption. Is OLS biased and inconsistent if this assumption does not hold or do we need this assumption for inference?  

*Stock and Watson write "this assumption limits the probability of drawing an observation with extremely large values of $X_i$ or $u_i$." However, my intuition is that this assumption is extreme. Are we in trouble if we have large outliers (such that the fourth moments are large) but if these values are still finite? By the way: What is the underlying definition an outlier?

*Can we reformulate this as follows: "The kurtosis of $X_i$ and $u_i$ are nonzero and finite?"
 A: *

*This is a sufficient assumption, but not a minimal one [1]. OLS is not biased under these conditions, it is just inconsistent. The asymptotic properties of OLS break down when $X$ can have extremely large influence and/or if you can obtain extremely large residuals. You may not have encountered a formal presentation of the Lindeberg Feller central limit theorem, but that is what they are addressing here with the fourth moment conditions, and the Lindeberg condition tells us basically the same thing: no overlarge influence points, no overlarge high leverage points [2].

*These theoretical underpinnings of statistics cause a lot of confusion when boiled down for practical applications. There is no definition of an outlier, it is an intuitive concept. To understand it roughly, the observation would have to be a high leverage point or high influence point, e.g. one for which the deletion diagnostic (DF beta) is very large, or for which the Mahalanobis distance in the predictors is large (in univariate stats that's just a Z score). But let's return to practical matters: if I conduct a random survey of people and their household income, and out of 100 people, 1 of the persons I sample is a millionaire, my best guess is that millionaires are representative of 1% of the population. In a biostatistcs lecture, these principals are discussed and emphasized that any diagnostic tool is essentially exploratory[3]. A nice point is made here: when exploratory statistics uncover an outlier, the "result" of such an analysis is not "the analysis which excludes the outlier is the one I believe", it is, "removing one point completely changed my analysis."

*Kurtosis is a scaled quantity which depends upon the second moment of a distribution, but the assumption of finite, non-zero variance for these values is tacit since it is impossible for this property to hold in the fourth moment but not in the second. So basically yes, but overall I have never inspected either kurtosis or fourth moments. I don't find them to be a practical or intuitive measure. In this day when a histogram or scatter plot is produced by the snap of one's fingers, it behooves us to use qualitative graphical diagnostic statistics, by inspecting these plots.
[1] https://math.stackexchange.com/questions/79773/how-does-one-prove-that-lindeberg-condition-is-satisfied
[2] http://projecteuclid.org/download/pdf_1/euclid.ss/1177013818
[3] http://faculty.washington.edu/semerson/b517_2012/b517L03-2012-10-03/b517L03-2012-10-03.html
A: You do not need assumptions on the 4th moments for consistency of the OLS estimator, but you do need assumptions on higher moments of $x$ and $\epsilon$ for asymptotic normality and to consistently estimate what the asymptotic covariance matrix is.
In some sense though, that is a mathematical, technical point, not a practical point. For OLS to work well in finite samples in some sense requires more than the minimal assumptions necessary to achieve asymptotic consistency or normality as $n \rightarrow \infty$.
Sufficient conditions for consistency:
If you have regression equation:
$$ y_i = \mathbf{x}_i' \boldsymbol{\beta} + \epsilon_i $$
The OLS estimator $\hat{\mathbf{b}}$ can be written as:
$$ \hat{\mathbf{b}} = \boldsymbol{\beta} + \left( \frac{X'X}{n}\right)^{-1}\left(\frac{X'\boldsymbol{\epsilon}}{n} \right)$$
For consistency, you need to be able to apply Kolmogorov's Law of Large Numbers or, in the case of time-series with serial dependence, something like the Ergodic Theorem of Karlin and Taylor so that:
$$ \frac{1}{n} X'X \xrightarrow{p} \mathrm{E}[\mathbf{x}_i\mathbf{x}_i'] \quad \quad \quad \frac{1}{n} X'\boldsymbol{\epsilon} \xrightarrow{p} \mathrm{E}\left[\mathbf{x}_i' \epsilon_i\right] $$
Other assumptions needed are:


*

*$\mathrm{E}[\mathbf{x}_i\mathbf{x}_i']$ is full rank and hence the matrix is invertible. 

*Regressors are predetermined or strictly exogenous so that $\mathrm{E}\left[\mathbf{x}_i \epsilon_i\right] = \mathbf{0}$.


Then $\left( \frac{X'X}{n}\right)^{-1}\left(\frac{X'\boldsymbol{\epsilon}}{n} \right) \xrightarrow{p} \mathbf{0}$ and you get $\hat{\mathbf{b}} \xrightarrow{p} \boldsymbol{\beta}$
If you want the central limit theorem to apply then you need assumptions on higher moments, for example, $\mathrm{E}[\mathbf{g}_i\mathbf{g}_i']$ where $\mathbf{g_i} = \mathbf{x}_i \epsilon_i$. The central limit theorem is what gives you asymptotic normality of $\hat{\mathbf{b}}$ and allows you to talk about standard errors. For the second moment $\mathrm{E}[\mathbf{g}_i\mathbf{g}_i']$ to exist, you need the 4th moments of $x$ and $\epsilon$ to exist. You want to argue that $\sqrt{n}\left(\frac{1}{n} \sum_i \mathbf{x}_i' \epsilon_i \right) \xrightarrow{d} \mathcal{N}\left( 0, \Sigma \right)$ where $\Sigma = \mathrm{E}\left[\mathbf{x}_i\mathbf{x}_i'\epsilon_i^2 \right]$. For this to work, $\Sigma$ has to be finite.
A nice discussion (which motivated this post) is given in Hayashi's Econometrics. (See also p. 149 for 4th moments and estimating the covariance matrix.)
Discussion:
These requirements on 4th moments is probably a technical point rather than a practical point. You're probably not going to encounter pathological distributions where this is a problem in everyday data? It's for more commonf or other assumptions of OLS to go awry.
A different question, undoubtedly answered elsewhere on Stackexchange, is how large of a sample you need for finite samples to get close to the asymptotic results. There's some sense in which fantastic outliers lead to slow convergence. For example, try estimating the mean of a lognormal distribution with really high variance. The sample mean is a consistent, unbiased estimator of the population mean, but in that log-normal case with crazy excess kurtosis etc... (follow link), finite sample results are really quite off.
Finite vs. infinite is a hugely important distinction in mathematics. That's not the problem you encounter in everyday statistics. Practical problems are more in the small vs. big category. Is the variance, kurtosis etc... small enough so that I can achieve reasonable estimates given my sample size?
Pathological example where OLS estimator is consistent but not asymptotically normal
Consider:
$$ y_i = b x_i + \epsilon_i$$
Where $x_i \sim \mathcal{N}(0,1)$ but $\epsilon_i$ is drawn from a t-distribution with 2 degrees of freedom thus $\mathrm{Var}(\epsilon_i) = \infty$. The OLS estimate converges in probability to $b$ but the sample distribution for the OLS estimate $\hat{b}$ is not normally distributed. Below is the empirical distribution for $\hat{b}$ based upon 10000 simulations of a regression with 10000 observations. 
The distribution of $\hat{b}$ isn't normal, the tails are too heavy. But if you increase the degrees of freedom to 3 so that the second moment of $\epsilon_i$ exists then the central limit applies and you get:

Code to generate it:
beta = [-4; 3.7];
n = 1e5;    
n_sim = 10000;    
for s=1:n_sim
    X = [ones(n, 1), randn(n, 1)];  
    u  = trnd(2,n,1) / 100;
    y = X * beta + u;

    b(:,s) = X \ y;
end
b = b';
qqplot(b(:,2));

