Pointwise probability limit Suppose we have a joint distribution of a random sample $(y_n,x_n)\in R^2$:
$$
y_n = \beta_0 x_n + \epsilon_n
$$
and the estimator
$$
\hat \beta = argmin_\beta \frac{1}{N} \sum_{n=1}^N(y_n - \beta x_n)^4
$$
with $x_n$ independent of $\epsilon_n$ and $\epsilon_n$ with symmetric distribution around 0.
What is the point-wise probability limit of the finite-sample objective function? When is $\beta_0$ the unique global minimizer of the limiting function and is $\hat \beta$ consistent for $\beta_0$?
 A: A) Probability Limit of Objective Function 
Using the functional relationship between the variables we get
$$Q_n = \frac{1}{N} \sum_{n=1}^N[\epsilon_n - (\beta-\beta_0) x_n]^4$$
which after a little algebra gives
$$Q_n = \frac{1}{N} \sum_{n=1}^N\left(\epsilon_n^4 -4(\beta-\beta_0)\epsilon_n^3 x_n-4(\beta-\beta_0)^3\epsilon_nx_n^3+6(\beta-\beta_0)^2\epsilon_n^2x_n^2 + (\beta-\beta_0)^4x_n^4\right)$$
For this to be well defined asymptotically, we need to assume the existence and finiteness of the fourth moments involved. Moreover, I guess "random sample" means i.i.d. Then, sample moments are consistent estimators of theoretical moments (the Weak law of large Numbers holds), which means that this large sum is broken down and each element converges to the associated expected values.  
Now, given independence between $\epsilon_n$ and $x_n$, expected values separate, and so the second and third component of the sum contain $E(\epsilon^3)$ and $E(\epsilon)$ respectively. Since the distribution of $\epsilon_n$ is symmetric around zero, odd-raw moments are zero, so these components will be eliminated. Therefore,
$$Q_n \xrightarrow{p} Q= E(\epsilon^4) +6(\beta-\beta_0)^2E(\epsilon^2)E(x^2) + (\beta-\beta_0)^4E(x^4)$$
B) Global minimizer
Taking the first derivative of $Q$ with respect to $\beta$ we have
$$\frac {\partial Q}{\partial \beta} = 12(\beta-\beta_0)E(\epsilon^2)E(x^2) + 4(\beta-\beta_0)^3E(x^4)$$
So for $\beta = \beta_0$ this becomes zero, providing a critical point of the limiting function.
Taking the second derivative (and then evaluating it at $\beta = \beta_0$) we have
$$\frac {\partial^2 Q}{\partial \beta^2} = 12E(\epsilon^2)E(x^2) + 12(\beta-\beta_0)^2E(x^4) \Rightarrow \frac {\partial^2 Q}{\partial \beta^2}|_{\beta = \beta_0} = 12E(\epsilon^2)E(x^2) > 0$$
so $\beta = \beta_0$ is indeed a global minimum. The only additional condition I see for this result, except of the assumptions of the question, is that fourth moments exist and are finite.
C) Consistency of $\hat \beta$
$\hat \beta$ is an Extremum estimator. The objective function is convex with respect to the parameter, it converges point-wise in probability to a limiting function, which in turn is uniquely minimized at $\beta_0$ (change the sign to talk about concavity and maximum). These three conditions, together with regularity conditions (measurability of the objective function with respect to the data, convex but not necessarily compact parameter space), guarantee that $\hat \beta$ is consistent for $\beta_0$.  
