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Say I have i.i.d. samples $X_1, \ldots, X_n \sim p \mathcal{N}(\mu_1, \sigma^2) + (1 - p) \mathcal{N}(\mu_2, \sigma^2)$. Then suppose I estimate the mean with the sample mean

$$ \widehat{\mu} = \frac{1}{n} \sum_{j = 1}^n X_j $$

Clearly $\widehat{\mu}$ is unbiased since $\mathbb{E}[\widehat{\mu}] = p\mu_1 + (1 - p)\mu_2$. How can I get a concentration inequality similar to a gaussian tail bound? Recall that if we had i.i.d. samples $Y_1, \ldots, Y_n \sim \mathcal{N}(\mu, \sigma^2)$ then we know

$$ \mathbb{P}\left(\left\lvert\frac{1}{n}\sum_{j=1}^n Y_j - \mu\right\lvert > \varepsilon\right) \leq \exp\left(-\frac{n\varepsilon^2}{2\sigma^2}\right) $$

Can we get a similar style concentration inequality for $\widehat{\mu}$? That is, I'm looking for something like

$$ \mathbb{P}\left(\left\lvert\widehat{\mu} - (p\mu_1 + (1 - p)\mu_2)\right\lvert > \varepsilon\right) \leq \: ??? $$

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