Azuma hoeffding inequality for supermartingale

Question

i have a simple question about Azuma-Hoffding inequality. If ${\displaystyle \left\{X_{0},X_{1},\dots X_n \right\}}$ is a martingale, since it is both a supermartingale and submartingale and $\lambda >0 $, we have the Azum-Hoeffding inequality : $$ P(\mid X_n - X_0 \mid \geq \lambda) \leq 2\exp(\frac{-\lambda²}{2\sum_{i=1}^{n} c_{k}^2}) $$ My question is, if ${\displaystyle \left\{X_{0},X_{1},\dots X_n \right\}}$ is just a supermartingale i can say that $$ P(\mid X_n - X_0 \mid \geq \lambda) \leq 2\exp(\frac{-\lambda²}{2\sum_{i=1}^{n} c_{k}^2}) $$ Thanks.

Answer 1

For a super-martingale, you have just $$ P( X_n - X_0 \geq \lambda) \leq \exp(\frac{-\lambda²}{2\sum_{i=1}^{n} c_{k}^2}) $$ So, i think you can't have the same bound for $P( X_n - X_0 \leq -\lambda)$. See https://en.wikipedia.org/wiki/Azuma%27s_inequality for more details.