Consider independent random variables $X_i$ each with density given by
$f_{X_i}(x) = \frac{\phi}{\sqrt{2\pi \sigma_i^2}} e^{-\frac{1}{2} \frac{(x - \mu_i)^2}{\sigma_i^2}} + \frac{1 -\phi}{\sqrt{2\pi \sigma_i^2}} e^{-\frac{1}{2} \frac{(x + \mu_i)^2}{\sigma_i^2}}$
where $\phi \in (0, 1)$, $\sum_{i = 1}^n \sigma_i^2 = 1$, $\sum_{i=1}^n \mu_i = \mu \neq 0$, $\mu_i > 0$. Notice that $\mathbb{E}[X_i] = (2\phi -1) \mu_i$.
Question: How tightly does $\sum_{i=1}^n X_i$ concentrate around its mean? Is $\sum_{i=1}^n X_i - \mathbb{E}[X_i]$ subgaussian? Or at least sub-gamma or sub-exponential? (See pages 24-30 of Gabor Lugosi's Concentration Inequalities textbook).
My attempt: Computer simulations indicate that it might be at least sub-gamma. In an attempt to use the Chernoff bound, I calculate:
$\mathbb{E}\left[e^{\lambda \sum_{i=1}^n X_i - \mathbb{E}[X_i]} \right] = e^{-\lambda (2\phi - 1) \mu} \prod_{i=1}^n \left(\phi e^{\lambda^2 \sigma_i^2/2 + \lambda \mu_i } + (1-\phi) e^{\lambda^2 \sigma_i^2 / 2 - \lambda \mu_i} \right) = e^{\lambda^2/2-\lambda (2\phi - 1) \mu} \prod_{i=1}^n \left(\phi e^{\lambda \mu_i } + (1-\phi) e^{- \lambda \mu_i} \right)$
but I can't upper bound this to look like a subgaussian mgf. I have also tried showing the sufficient condition that $\mathbb{E}[X^{2q}] \leq q!C^q$ (see Theorems 2.1 and 2.3 of Lugosi's Concentration Inequalities book) but have not succeeded in obtaining a tight enough bound. I have obtained a $\mathbb{E}[X^{2q}] \leq 2q!(2(\sigma_i + 2\phi \mu_i)^2)^q$ bound, but this is loose and implies no concentration around the mean as the subgaussian proxy grows just as fast.
For reference:
Subgaussianity: Suppose random variable follows the inequality:
$\mathbb{E}[e^{\lambda X}] \leq \exp(\lambda^2 \sigma^2 / 2)$,
then we say that X is $\sigma$-subgaussian.