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?

PS: question also posted here: https://math.stackexchange.com/questions/2843340/a-failure-of-convergence-of-conditional-distributions

  • 1
    $\begingroup$ Cross-posting between stackexchange sites so quickly is generally not seen as a good idea. I have made more substantial points at math.stackexchange $\endgroup$ – Henry Jul 7 '18 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.