Tail Probabilities of $L^p$ bounded martingale differences

Question

Assume that I have a probability space $(\Omega, \mathcal{A}, \mathbb{P})$ on which we define a sequence of martingale differences $X_1,X_2,\dots$ (w.r.t. to a certain filtration). Further let $p \in (1,2)$ and assume that the $X_i$ are bounded in $L^p$, i.e. there exists $M>0$ such that $$\Vert X_i \Vert_p \leq M$$ for all $i \in \mathbb{N}$.

Does this already imply that I find a random variable $X \in L^p(\Omega, \mathcal{A}, \mathbb{P})$ and a $C > 0$ such that $$\mathbb{P}(\vert X_i \vert > z) \le C \mathbb{P}(\vert X \vert > z)$$ for all $z > 0$, $i \in \mathbb{N}$?

Answer 1

The answer is negative. Suppose that for each $k>2$, the variable $X_k$ takes the values $\pm k^{1/p}$ with probability $k^{-1}$ each, and $P(X_k=0)=1-2k^{-1}$. (We can take $X_k=0$ for $k<2$.) Then $E[|X_k|^p]=2$ for all $k>2$. However, if $X$ satisfies $$\mathbb{P}(\vert X_k \vert > z) \le C \mathbb{P}(\vert X \vert > z)$$ for all $z > 0$ and $k\in \mathbb{N}$, then $$E[|X|^p]=\int_0^\infty\!\! P(|X|^p>x) \,dx \ge \sum_{k=3}^\infty \int_{k-1}^{k}\frac{P(|X_k|^p>x)}{C}=\frac2C\sum_{k=3}^\infty \frac1k=\infty \,.$$

Answer 2

Notice that $\mathbb{P}(\vert X_i \vert > z) \leqslant C \mathbb{P}(\vert X \vert > z)$ for some $X$ in $L^p$ would imply that $\{\lvert X_i\rvert^p,i\geqslant 1\}$ is uniformly integrable. Indeed, $$ \mathbb E\left[\lvert X_i\rvert^p\mathbf{1}_{\{\lvert X_i\rvert>R\}}\right] =\int_0^\infty pt^{p-1}\mathbb P\left(\{\lvert X_i\rvert>\max\{R,t\}\right)dt\leqslant C\int_0^\infty pt^{p-1}\mathbb P\left(\{\lvert X\rvert>\max\{R,t\}\right)dt=C\mathbb E\left[\lvert X\rvert^p\mathbf{1}_{\{\lvert X\rvert>R\}}\right].$$

Therefore, as pointed out by Yuval Peres, any martingale difference sequence which is bounded in $L^p$ but such the sequence of $p$ powers is not uniformly integrable would give an example.

Note that, even if $\{\lvert X_i\rvert^p,i\geqslant 1\}$ is uniformly integrable, an $X$ in $L^p$ dominating the tails up to a constant may fail to exist. Indeed, let $X_i$ taking the values $2^{i/p}$, $-2^{i/p}$ and $0$ with respective probabilities $1/(i2^i)$, $1/(i2^i)$, and $1-2/(i2^i)$, $i\geqslant 2$. Observe that $$ \mathbb E\left[\lvert X_i\rvert^p\mathbf{1}_{\{\lvert X_i\rvert>R\}}\right] =\frac 1i\mathbf{1}_{R<2^i} \leqslant \frac{1}{\log_2 R}$$ which gives uniform integrability. And if $Y\in L^p$ is such that $$\mathbb{P}(\vert X_i \vert > z) \leqslant C \mathbb{P}(\vert Y \vert > z),$$ then $$ \mathbb P\left(Y>2^{i/p}\right)\geqslant C^{-1}\mathbb P\left(\lvert X_{i+1}\rvert>2^{i/p}\right) =\frac 1{i2^i} $$ hence the series $\sum_i 2^i\mathbb P\left(Y>2^{i/p}\right)$ diverges.