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Consider $n$ iid samples $(X_i,Y_i)$ generated such that $X_i$ is a truncated normal of mean $\mu_X$ truncated to the left of the origin, and $Y_i$ is a truncated normal of mean $\mu_Y$ truncated to the right of the origin. All $X_i$, $Y_i$ independent of each other.

We define $\bar{W} = \frac{1}{2n} \sum_{i=1}^n (X_i + Y_i)$, and $\bar{Z} = \frac{1}{2n} \sum_{i=1}^n (X_i - Y_i)$. Clearly $\bar{W}$ is not independent of $\bar{Z}$.

However, applying CLT, we have that $$ \sqrt{2n}\left( \left(\begin{array}{c}\bar{W}\\\bar{Z} \end{array}\right) - \left(\begin{array}{c}\mu_W\\\mu_{Z} \end{array}\right) \right) \sim \mathcal{N}(0, I).$$

The fact that the covariance matrix is $I$ follows from the fact that $\sum_{i=1}^n X_i$ and $\sum_{i=1}^n Y_i$ are independent, and the linear transformation to obtain $\left(\begin{array}{c}\bar{W}\\\bar{Z} \end{array}\right) $ is a unitary transformation.

I don't quite see what's breaking here. And what would be a better way to approximate this while capturing dependence?

Edit: Question also posted here: https://stats.stackexchange.com/questions/354975/a-failure-of-convergence-of-conditional-distributions

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If by truncated you mean $X_i$ has a standard half-normal distribution and $Y_i$ has the same distribution but taking negative values, then

  • $X_i+Y_i$ and $X_i-Y_i$ are, as you say, not independent - in particular $X_i+Y_i$ is bounded between $-(X_i-Y_i)$ and $X_i-Y_i$ - but they have a covariance between them of $0$, since $\mathbb E[X_i+Y_i \mid X_i-Y_i =z]$ for all possible $z$. This then extends to $\bar{W}$ and $\bar{Z}$ and the zero covariance result you observed

  • $\sqrt{2n}(\bar{W} - \mu_W)$ and $\sqrt{2n}(\bar{Z} - \mu_Z)$ do not have variances of $1$ but variances of $\left(1-\frac{2}{\pi}\right) \approx 0.36338$. So too do each $X_i$ and $Y_i$

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  • $\begingroup$ Thanks. You're right about the variances. However, do you know any method I could use to obtain a tractable approximation of $\left(\begin{array}{c}\bar{W}\\\bar{Z} \end{array}\right)$ where the dependence between $\bar{W}$ and $\bar{Z}$ is not lost? $\endgroup$
    – Devil
    Commented Jul 7, 2018 at 20:08
  • $\begingroup$ @Devil Not really. I suspect the distribution is only interesting for small $n$ and I do not see an easy way to describe it even for $n=2$ $\endgroup$
    – Henry
    Commented Jul 8, 2018 at 0:18

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