Suppose $X$ is R.V with mean $\mu$ and variance $\sigma^2$, then $X - \mu$ is a Sub-Gaussian distribution having $\nu$ as the variance proxy. Now I want to find the concentration inequality for sample variance i.e.
$$\mathbb{P}\left(\frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2 - \sigma^2 \geq \epsilon\right)$$ where $\bar{X}$ is the sample mean
Work Done:
I begin like this
$$X_i - \bar{X} = X_i - \bar{X} + \mu - \mu = \underbrace{(X_i - \mu)}_A - \underbrace{(\bar{X} - \mu)}_B$$
where the variance proxy of part $A$ is $\nu$ and part $B$ is $\frac{\nu}{\sqrt{n}}$, thus the resultant proxy is $$\sqrt{\nu^2 + \frac{\nu^2}{n}} = \nu\sqrt{\frac{n+1}{n}}$$
Now if relook at sample variance we can say $$\frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2 - \sigma^2 = \frac{1}{n} \sum_{i=1}^n \left(Y_i^2 - \sigma^2\right)$$
where $Y_i = X_i - \bar{X}$ whose variance proxy is $\nu\sqrt{\frac{n+1}{n}}$. We already know that $Y_i^2 - \sigma^2$ is again a SG distribution. If we know the variance proxy of $Y_i$, how can we calculate the variance proxy of $Y_i^2 - \sigma^2$ and procced further?
