I faced a limiting distribution with zero covariance between two variables but their correlation is $1$. Is there such a distribution? How it can be explained?

You are right may I need give more detail. OK, X and Y are bivariate normal distribution with different variances and means (free of n) but corr=1-(1/n), now investigate the limiting distribution of Yn|Xn=x.

**Answer**

Following a clarification by the OP, it appears that a) we assume that the two variables follow jointly a bivariate normal and b) our interest is in the conditional distribution, which is then

$$Y_n\mid X_n=x \ \sim\ \mathcal{N}\left(\mu_y+\frac{\sigma_y}{\sigma_x}\rho_n( x – \mu_x),\, (1-\rho_n^2)\sigma_y^2\right)$$

Then we see that as $n \to \infty$, we have $\rho_n \to 1$, and the variance of the conditional distribution goes to zero. Intuitively, if correlation goes to unity, “knowing $x$” is enough to “know $y$” also.

But nowhere in the above do we get that $\text{Cov}(Y_n, X_n)$ is zero. Even at the limit covariance will remain equal to $\text{Cov}(Y_n, X_n) \to \sigma_y \sigma_x$.

Note that the *conditional* covariance (and then also the conditional correlation) is *always* zero, because,

$$\text{Cov}(Y_n, X_n \mid X_n =x) = E(Y_nX_n\mid X_n =x) – E(Y\mid X_n =x) E(X\mid X_n =x)$$

$$=xE(Y_n\mid X_n =x) – xE(Y\mid X_n =x) =0$$

This happens because by examining $X_n = x$ we have turned one of the random variables into a constant, and constants do not co-vary with anything.

**Attribution***Source : Link , Question Author : Behgol , Answer Author : Alecos Papadopoulos*