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.


1 Answer 1


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).

  • $\begingroup$ Thanks for your response. I agree with the explanation. $\endgroup$
    – Mike Kehoe
    Jun 7, 2018 at 4:54

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