Sum of (non-centered) Subgaussian Random Variables

Question

If we have a finite sum of independent subgaussian (or subexponential) random variables which are not mean zero, i.e. random variables $(X_i)_{i=1, \dots, N}$ such that $\forall i=1, \dots, N$
$$ \mathbb{P}(X_i \geq t) \leq 2 \exp{\left(\frac{-t^2}{C^2}\right)} $$
for some $C > 0$, do we know that the sum is again subgaussian (or subexponential)? I know this is the case when they are centered, since we can then consider the moment-generating functions, but does this extend to the more general case?

Answer 1

Let me first show this for $n = 2.$

We have, by expanding the square and Cauchy-Schwarz, that$$ \mathbb{E}[\exp( (X_1 + X_2)^2/K^2)] = \mathbb{E}[\exp(X_1^2+X_2^2/K^2) \cdot \exp(2 X_1X_2/K^2)] \\ \le\mathbb{E}[\exp(2X_1^2/K^2) \cdot \exp(2X_2^2/K^2)]^{1/2} \cdot \mathbb{E}[\exp(4X_1X_2/K^2)]^{1/2}.$$

But $2X_1X_2\le X_1^2 + X_2^2,$ so using independence, $$ \mathbb{E}[\exp( (X_1 +X_2)^2/K^2] \le \mathbb{E}[\exp(2X_1^2/K^2)] \mathbb{E}[\exp(2X_2^2/K^2)].$$ Now, if $\mathbb{E}[\exp(X^2/K^2)] \le 2,$ then $\mathbb{E}[\exp(2X^2/4K^2)] \le \sqrt{2}$ by using Jensen's inequality for the concave maps $u \mapsto \sqrt{u}$. This shows that $\|X_1 + X_2\|_{\psi_2} \le 2\max(\|X_1\|_{\psi_2}, \|X_2\|_{\psi_2}).$ You can use this in pairs, and iterate to show that $\|\sum X_i\| \le C n \max (\|X_i\|_{\psi_2}).$ It should probably be possible to improve this to something like $C\sum\|X_i\|_{\psi_2},$ not sure right now (see below).

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Alternatively: using the fact that $\|\mathbb{E}[X]\|_{\psi_2} \le C \|X\|_{\psi_2},$ and the norm property, we have $$ \left\| \sum X_i \right\|_{\psi_2} \le \left\| \sum (X_i - \mathbb{E}[X_i])\right\|_{\psi_2} + \left\| \sum \mathbb{E}[X_i]\right\|_{\psi_2}\\ \le \left\| \sum (X_i - \mathbb{E}[X_i])\right\|_{\psi_2} + \sum_i \|\mathbb{E}[X_i]\|_{\psi_2} \\ \le \left\| \sum (X_i - \mathbb{E}[X_i])\right\|_{\psi_2} + C \sum \|X_i\|_{\psi_2}.$$ Using the fact that for centred independent random variables, the square of the subGaussian norm of their sum is bounded up to constants by the sum of the squares of their subGaussian norms, we get $$ \left\| \sum X_i \right\|_{\psi_2} \le \sqrt{C \sum \|X_i\|_{\psi_2}^2} + C' \sum \|X_i\|_{\psi_2}.$$