Let $a$ be a random vector with $\mathbb{E} a = b$ and $\mathbb{E} \|a\|_p^2 =\sigma^2$, where $1\leq p \leq \infty$. Is it true that $\mathbb{E} \|a-b\|_p^2 \leq 2\sigma^2$?
It is clear to me how to obtain the relation for $p=2$ (in fact, with $\sigma^2$ in place of $2\sigma^2$) by expressing the squared norm as a scalar product. However, I have trouble generalizing the property. I tried using the triangle inequality for $\|\cdot\|_p$, the Jensen's inequality for the expectation of a norm or a squared norm, but didn't arrive at the result. I also suspect that the Hölder's inequality might be useful. I would appreciate any help.
edit: actually, generalizing to $p\in[1,2]$ would already suffice for my purpose.
Best Answer
This is not correct, at least for $p=1$. This can be seen by letting $P(X=e_i)=1/n$, where $e_i \in \Bbb{R}^n$ is the $i$-th standard basis vector. For $n\to\infty$, this easily shows that one needs at least the constant 4 instead of 2.
The constant 4 in fact suffices, as can be seen since $(\alpha+\beta)^2\leq 2 (\alpha^2+\beta^2)$ and since $\|b\|_p^2 \leq (E \|a\|_p)^2 \leq E \|a\|_p^2 =\sigma^2$, and thus finally $$ E[\|a-b\|_p^2] \leq 2 E[\|a\|_p^2 + \|b\|_p^2] \leq 4 \sigma^2. $$