If $X_n\to X$ in $L^p$, then $E(X_n)^p \to E(X)^p$

Question

Here's what I did: I want to prove that $|E(X_n)^p -E(X)^p| \to 0$.

$|E(X_n)^p -E(X)^p| \leq E(|X_n^p-X^p|)$ But this isn't necessariliy $\leq E(|X_n-X|^p)$ (which converges to zero). Take as a counterexample $p=2$, $X_n = 3,99$ and $X=4$. So, what can I do now? Any hints? Thanks!

Answer 1

Think about a normed linear space. Any normed linear space. The norm is continuous since $$ \left|\|x_n\| - \|y_n\| \right| \le \|x_n - y_n\|.$$

Answer 2

$X_n \to X$ in $L^p$ for $p>1$ implies $X_n \to X$ in $L^1$ because $|| X||_p$ is an increasing function in $p$ (use Jensen), in particular $EX_n \to EX$. Since $x \mapsto x^p$ is continuous this also implie s$(EX_n)^p \to (EX)^p$.

Answer 3

For $0<p\le 1$, by the $c_r$ inequality,

$$ |\mathsf{E}|X_n|^p-\mathsf{E}|X|^p|\le \mathsf{E}|X_n-X|^p\to 0. $$

For $p>1$, the Minkowski inequality implies that

$$ \left|\|X_n\|_p-\|X\|_p\right|\le \|X_n-X\|_p\to 0. $$

Answer 4

Whenever $\Omega$ is a space of finite measure and $f$ is in $L^p(\Omega)$, it follows that $f$ is in $L^q(\Omega)$ for $1\leqslant q<p$, for $$\lVert f\rVert_q \leqslant \mu(\Omega)^{1/q-1/p} \lVert f\rVert_p$$

In particular for $\Omega$ a probability space we get $1\leqslant q\mapsto \lVert f\rVert_q$ is an increasing function. Thus $\lVert X-X_n\rVert_p\to 0 $ implies that $\lVert X-X_n\rVert_1\to 0$. Now $\lVert X\rVert_1=\mathbb E|X|$ so this entails that $\mathbb EX_n\to \mathbb EX$ for $|\lVert X\rVert_1-\lVert X_n\rVert_1|\leqslant \lVert X-X_n\rVert_1$. It follows that $(\mathbb EX_n)^p\to (\mathbb EX)^p$ by continuity.