Bernstein inequality Let us assume $X_1, ..., X_n$  are independent random variables bounded by the interval $[a_i, b_i]$ and $S_n = X_1 + ... + X_n$. When $|X_i - E[X_i]|\leq M$,  the Bernstein's inequality suggests the following.  It can be asumed that $M = \max_{i} \big\{b_i - E[X_i]\big\}$.
$$P(S_n - E[S_n] > t) \leq \exp{\left(\frac{-t^2}{2\sum_{i=1}^{n}\operatorname{Var} (X_i) + \frac{2}{3} Mt}\right)}.$$
Now, I have a case where $Y_1, ..., Y_{n_1}$  are independent random variables bounded by the interval $[c_i, d_i]$ and $S_{n_1} = Y_1 + ... + Y_{n_1}$.  In addition, $Z_1, ..., Z_{n_2}$  are independent random variables bounded by the interval $[e_i, f_i]$ and $S_{n_2} = Z_1 + ... + Z_{n_2}$.  $Y_i$'s and $Z_i$'s are also independent.  Let, $M_1 = \max_{i} \big\{d_i - E[Y_i]\big\}$ and $M_2 = \max_{i} \big\{f_i - E[Z_i]\big\}$.  I would like to have a Bernstein's bound for $P(S_{n_1}+S_{n_2} -E[S_{n_1}+S_{n_2}] > t)$.
My try:
$$\begin{align}
&P(S_{n_1}+S_{n_2} -E[S_{n_1}+S_{n_2}] > t) \\
&\leq \exp{\left(\frac{-t^2}{2\big[\sum_{i=1}^{n_1}\operatorname{Var} (Y_i)+ \sum_{i=1}^{n_2}\operatorname{Var} (Z_i)\big] + \frac{2}{3} [M_1 + M_2] t}\right)}. \end{align}$$
I am wondering whether the above equation is correct.
 A: I can think of at least two "direct applications" of Bernstein inequality, and they are different from yours. I wouldn't say yours is incorrect, but to me it is not a "direct application".
First Direct Application
Consider combining all $Y_i$ and all $Z_i$ as just one set. In short, this gives the term $\frac23 \max\{ M_1, M_2\}\cdot t\,$ instead of $\frac23 [ M_1 + M_2 ]\cdot t\,$ in your expression, with other terms all the same.
Since this is in the denominator of negative exponent, your $M_1 + M_2 > \max\{ M_1, M_2\}$ is more conservative, with the whole $\exp(-\text{blah})$ being larger.
Formal justification of the above if needed:
Since $Y_i$ and $Z_i$ are independent within each set and to each other, along with the upper bounds $d_i$ and $f_i$ being distinct to begin with, we can combine $Y_i$ and $Z_i$ as just one set.
That is, we have a set for $i = 1,2,\ldots, (n_2+n_1)$ that shall be denoted $W_i$, which bounding intervals are $[c_i, d_i]$ for the first $n_1$ terms and $[e_{i-n_1}, f_{i-n_1}]$ for the remaining $i = 1+n_1,2+n_1,\ldots,n_2+n_1$. (the $c_i, d_i, e_i, f_i$ are given as in your question statement)
Thus, applying the definition (quoting your statement in the question post) $M = \max_{i} \big\{b_i - E[X_i]\big\}$, here we have the "relevant $M$" as
$$\max\left\{ \max_{i=1\sim n_1} \big\{d_i - E[Y_i]\big\} ~, ~ \max_{i=1\sim n_2} \big\{f_i - E[Z_i]\big\} \right\} = \max\{ M_1, M_2\}$$
Second Direct Application
Consider the equivalent statement of the inequality in terms of the complement (CDF instead of the tail):
$$P\left( S_{n_1} - E[S_{n_1}] \leq x \right) > \mathcal{P}_1(x) \equiv  1 - \exp\left[ -x^2 \left( 2\sum_{i=1}^{n_1}\operatorname{Var} (Y_i) + \frac{2}{3} M_1 x \right)^{-1} \right] \\
P\left( S_{n_2} - E[S_{n_2}] \leq x \right) > \mathcal{P}_2(x) \equiv 1 - \exp\left[ -x^2 \left( 2\sum_{i=1}^{n_2}\operatorname{Var} (Z_i) + \frac{2}{3} M_2 x \right)^{-1} \right] $$
again, all the $S_{n_1}$ etc are as defined by you.
The desired probability is a convolution-like integral, due to the direct product of probabilities from independence:
\begin{align*}
&\phantom{{}={}} P\left( S_{n_1}+S_{n_2} -E[S_{n_1}+S_{n_2}] > t \right) \\
&= 1 - P\left( S_{n_1}+S_{n_2} -E[S_{n_1}+S_{n_2}] \leq t \right)\\
&= 1 - \int_{u = -\infty}^{ \infty} P\left( S_{n_1} -E[S_{n_1}] \leq t \right)\cdot P\left( S_{n_2} -E[S_{n_2}] \leq t-u \right)\,\mathrm{d} u \\
&\leq 1 - \int_{u = -\infty}^{ \infty} \mathcal{P}_1(u) \mathcal{P}_2(t-u) \,\mathrm{d} u
\end{align*}
Once you figure out the proper range for $t$ to replace the integration lower limit $-\infty$ and upper $\infty$, this integral is not difficult.
Anyway, this is what I consider a "direct application" of Bernstein inequality, and it's not the same as the one presented (unless there's some more steps pushing the inequality in a way I cannot imagine).
