# Confidence intervals for expected value of almost independent variables

How may I calculate (or approximate with reasonable error) confidence intervals of expected value of non-normally distributed random variable $Y$ given its almost independent samples?

I have $n = 100$ samples of a random variables $$Y_j = \sum_{i=1}^{m} X_{i, j}^2 ,$$ where $X_{i, j}$ are almost independent random values and $m \gt 100$. By "almost independent" I mean that $$\sum_{j=1}^{n} \sum_{i=1}^{m} X_{i, j} = 0 .$$ For a given $i$ all $X_{i, j}$ have the same (possibly non-normal) distribution.

Were $Y_j$ independent, then from central limit theorem I would assume, that $$\bar{Y} \sim \mathcal{N}(\mu_Y,\,\frac{\sigma_Y^{2}}{n}) .$$

Unfortunately I know neither $\mu_Y$ nor $\sigma_Y^{2}$, so I have to approximate them by $\bar{Y}$ and $s_Y^2 = \frac{\sum_{j=1}^n (Y_j - \bar{Y})}{n - 1}$.

Was $Y$ normally distributed (and independent), confidence intervals would be given by $t$ distribution of $n - 1$ degrees of freedom.

If there is no formula for exact CIs, then how [un]safe would it be to ignore the dependence and/or non-normality of the distribution?

• The trouble is that the process that enforces the constraint may be very stringent, e.g., each "even-indexed" observation is the negative of the immediately preceding "odd-indexed" observation ($X_{i,2} = -X_{i,1}$), for example. This example is far more stringent than is needed to ensure the constraint is met, but, without knowing how that constraint is enforced, you're making a leap of faith if you ignore the correlation. – jbowman Feb 9 '18 at 23:17
• @jbowman the constraint originates from $X_{i, j}$ being kind of residuals in a model fitted to data from m distributions, each distribution sampled up to n times (in about 30% cases sampling failed which I am interpreting as zero-residual). – abukaj Feb 10 '18 at 1:30

The Berry Esseen theorem gives you an idea of how far off you are going to be from normality. This result is exact and does not rely on large $N$ asymptotics