The random variable $X_{e_n}$ and $Y_{e_n}$ are defined as the sum of independent random variables $x_{e_i}$ and $y_{e_i}$ as follows: $X_{e_n} = \sum_{i=1}^n x_{e_i}$, $Y_{e_n} = \sum_{i=1}^n y_{e_i}$.
The random variable $x_{e_i}^{\mathrm {'s}}$ are generated using the following process. At each time step $i$, I choose an index $e$ from a finite set of indices $E$ with probability $p_{e_i}$. These probabilities $p_{e_i} \; \forall e \in E$ is known to me. $x_{e_i}$ is the reward value obtained by choosing the index $e$ at time step $I$ and $y_{e_i}$ is defined as follows, for $0 \leq a < 1$. $$ y_{e_j} = \begin{cases} \frac{x_{e_j} + a}{p_{ej}},& \text{ if } e \text{ is the index chosen at } j\\ \frac{a}{p_{ej}},& \text{ otherwise}. \end{cases} $$ The reward values are not deterministic. The rewards are determined by the previous index chosen till time step $j$ i.e. $x_{e_i}$ is a r.v. which is defined in terms of the past chosen indexes $e_1 \dots e_{i-1}$ (Here I use $e_1$ to denote the index chosen at time step 1) (The reward and index selection can be seen as part of the adversarial repeated game where reward is set by the adversary).
How can I show that the sum of the random variables does not deviate much in expectation i.e. $$\mathbf{E}\left[e^{a\left(X_{e_n} - Y_{e_n}\right)}\right] \leq 1.$$
I think this is a false statement,
I can prove that $ \textbf{E}\left[a(X_e - Y_e)\right] \leq 0 $.
But since $\text{exp}(x)$ is a concave function $\textbf{E}\left[e^{a(X_e - Y_e)}\right] \leq 1 $ is not necessarily true.
-------------- Proof of $ \textbf{E}\left[a(X_e - Y_e)\right] \leq 0 $ --------------
For each $e,j$,
$$ \begin{aligned} \textbf{E}\left[x_{ej} - y_{ej}\right] = & x_{ej} - \left[p_{ej} \cdot \frac{x_{ej}+a}{p_{ej}} + (1-p_{ej})\cdot\frac{a}{p_{ej}}\right] \\ = & x_{ej} - \left[x_{ej} + \frac{a}{p_{ej}}\right] = - \frac{a}{p_{ej}} \end{aligned} $$
Thus, $$ \textbf{E}\left[a(X_e - Y_e)\right] = -a^2\sum_{j=1}^n \frac{1}{p_{ej}} \leq 0 $$
----------------- Counter Example -----------------------
Since each step and each event is independent, it is sufficient to show for one event and one timestamp.
For example, a certain event at a certain step (we will use $x$, $p$, $y$ to denote the value since for simplicity).
$$ \begin{aligned} \textbf{E}\left[e^{a(x- y)}\right] = & p \cdot \exp\left[a(x - \frac{x+a}{p})\right] + (1-p) \cdot \exp\left[a(x - \frac{a}{p})\right] \\ = & \exp\left[-\frac{a^2}{p}\right]\left[p\cdot \exp\left( - \frac{1-p}{p}ax\right) + (1-p)\exp(ax)\right] \end{aligned} $$
If $x=10, a=1, p=0.5$ $$ \textbf{E}\left[e^{a(x- y)}\right] = \exp(-2) \cdot [0.5*\exp(-10) + 0.5*\exp(10)] \approx 1490 > 1 $$