Let $(X,Y)$ be bivariate normal distributed and
$$
S_X^2 = \sum_i (X_i – \bar{X})^2 \\
S_Y^2 = \sum_i (Y_i – \bar{Y})^2.
$$
Now I want to derive the limit of
$$
T_n:= \left ( \sqrt{n} \left ( \frac{\bar{X}}{S_X} – \frac{\mu_X}{\sigma_X} \right), \sqrt{n} \left ( \frac{\bar{Y}}{S_Y} – \frac{\mu_X}{\sigma_Y} \right) \right).
$$
Then
$$
\lim_{n \to \infty} T_n = N(0,\Sigma).
$$
The multivariate delta method would tell me that
$$
\Sigma =
\begin{pmatrix}
1 +\frac{\mu_X ^2}{2 \sigma_X^2} & \rho +\frac{\rho^2 \mu_X \mu_Y}{2 \sigma_Y \sigma_X} \\
\rho +\frac{\rho^2 \mu_X \mu_Y}{2 \sigma_Y \sigma_X} & 1 +\frac{\mu_Y ^2}{2 \sigma_Y^2}
\end{pmatrix}
$$
while I would expect from Slutsky's theorem
$$
\Sigma =
\begin{pmatrix}
1 & \rho\\
\rho & 1
\end{pmatrix}.
$$
Where is the error in my thinking?
Best Answer
So, I think I figured it out. While the application of the Slutsky theorem can give us the limits of $$ \lim_{n \to \infty} \sqrt{n} \left ( \frac{\bar{X}-\mu_X}{S_X} \right ) \\ = \lim_{n \to \infty} \sqrt{n} \left ( \frac{\bar{X}-\mu_X}{\sigma_X} \right ), $$ it cannot be used to obtain the limit of $$ \sqrt{n} \left ( \frac{\bar{X}}{S_X} - \frac{\mu_X}{\sigma_X} \right). $$ Therefore we need the Delta method and also obtain a different covariance matrix.