Implications of convergence in quadratic mean over convergence in the 4-th mean

Question

I am trying to understand whether the convergence in quadratic mean of a sequence $X_n$ to some $X$ has implications over its own convergence in the $4$-th mean to the same $X$. Are there any theoretical results on this subject?

Answer 1

If $\mathbb E\left[(X_n-X)^2\right]\to 0$, the 4-th moment may not exist. Say, let $X=0$, $X_n=Y/n$, where $Y$ has Pareto distribution with pdf $$ f_Y(t) = \frac{3}{t^4}I_{\{t\geq 1\}}. $$ Then $$\mathbb E\left[(X_n-X)^2\right] = \frac{1}{n^2}\mathbb E\left[Y^2\right] = \frac{3}{n^2}\to 0$$ and $$ \mathbb E\left[(X_n-X)^4\right] = \frac{1}{n^4}\mathbb E\left[Y^4\right] = \infty. $$ Vice versa, convergence in $4$-th mean implies convergence in quadratic mean since $$ \mathbb E[X^2] \leq \sqrt{\mathbb E[X^4]}. $$

ADDED

Even if a fourth moment exists, convergence in quadratic mean does not imply convergence in $4$-th order mean.

In the above example take $X=0$ and $$X_n=\frac{Y}{n}\cdot I_{\{Y<n^4\}}=\begin{cases}\frac{Y}{n}, & Y<n^4 \cr 0, & Y \geq n^4\end{cases} $$ Then $$\mathbb E[(X_n-X)^2]=\mathbb E[X_n^2] =\frac1{n^2} \int_1^{n^4}t^2 \frac{3}{t^4}\,dt \to 0$$ but $$\mathbb E[(X_n-X)^4]=\mathbb E[X_n^4] =\frac1{n^4} \int_1^{n^4}t^4 \frac{3}{t^4}\,dt = 3\frac{n^4-1}{n^4}\to 3.$$