In the paper http://www.mat.univie.ac.at/~schachermayer/pubs/preprnts/prpr0154.pdf

you can find a trajectorial version of Doob's inequality. It is given by:

$$\bar{s}^2_T+4\sum_{k=0}^{T-1}\bar{s_k}(s_{k+1}-s_k)\le 4s^2_T$$

 

$$ \bar{s}_k=\max (s_1,...,s_k) $$

The proof should be straightforward but I am not able to prove it, please help me with that more or less simple inequality.

In the paper http://www.mat.univie.ac.at/~schachermayer/pubs/preprnts/prpr0154.pdf

you can find a trajectorial version of Doob's inequality. It is given by:

$$\bar{s}^2_T+4\sum_{k=0}^{T-1}\bar{s_k}(s_{k+1}-s_k)\le 4s^2_T$$

$$ \bar{s}_k=\max (s_1,...,s_k) $$

The proof should be straightforward but I am not able to prove it, please help me with that more or less simple inequality.

In the paper http://www.mat.univie.ac.at/~schachermayer/pubs/preprnts/prpr0154.pdf

you can find a trajectorial version of Doob's inequality. It is given by:

$$\bar{s}^2_T+4\sum_{k=0}^{T-1}\bar{s_k}(s_{k+1}-s_k)\le 4s^2_T$$

$$ \bar{s}_k=\max (s_k) $$

The proof should be straightforward but I am not able to prove it, please help me with that more or less simple inequality.

In the paper http://www.mat.univie.ac.at/~schachermayer/pubs/preprnts/prpr0154.pdf

you can find a trajectorial version of Doob's inequality. It is given by:

$$\bar{s}^2_T+4\sum_{k=0}^{T-1}\bar{s_k}(s_{k+1}-s_k)\le 4s^2_T$$

$$ \bar{s}_k=\max (s_1,...,s_k) $$

The proof should be straightforward but I am not able to prove it, please help me with that more or less simple inequality.