Let $X_1, X_2, \ldots , X_n$ be a random sample from $U(0, θ)$, where $θ > 0$ is the unknown parameter. Let $X_{(n)} = max\{X_1,X_2, \ldots , X_n \}$. Then which of the following is (are) consistent estimator(s) of $θ^3$?
(A) $8X_n^3$
(B) $X_{(n)}^3$
(C) $(\frac{2}{n}∑_{i=5}^n X_i)^3$
(D) $\frac{nX_{(n)}^3 +1}{n+1}$
How do I test option (A) and (C)? I have been stuck for a while and some help would be appreciated.
My approach so far:
$$f(X_{(n)})=\frac{nx^n}{\theta^n}$$ Using the above density function I proved that
$E(X_{(n)})\rightarrow\theta \ as \ n\rightarrow\infty$
$Var(X_{(n)})\rightarrow0 \ as \ n\rightarrow\infty$
Thus, sufficient conditions for consistency have been met and $X_{(n)}$ is a consistent estimator for $\theta$. Since $\theta^3$ is a continuous function of $\theta$, and $X_{(n)}$ is a consistent estimator for $\theta$, it can be said that $X_{(n)}^3$ must be a consistent estimator for $\theta^3$. Validity of (B) and (D) now follows from this fact.
For (C), we can write $$\dfrac{2}{n}\sum_{i = 5}^{n}X_{i} = 2\left(\dfrac{1}{n}\sum_{i = 1}^{n}X_{i}\right) - 2\dfrac{X_{1} + \dots + X_{4}}{n}$$ which using the law of large numbers, converges in probability to $2\dfrac{\theta}{2} = \theta$. Then by applying the continuous mapping theorem, we can show that the estimator is consistent for $\theta^{3}$.
For (A), it is not consistent because it doesn't converge in probability to anything (it will have some scaled beta distribution).