Suppose there are two random sequences $\left\{X_n\right\}$ and $\left\{Y_t\right\}$. Both are i.i.d. and have zero first moment and finite second moment. Hence, \begin{equation} \begin{pmatrix}\sqrt{N}\bar{X}_N\\\sqrt{T}\bar{Y}_T\end{pmatrix}\overset{d}{\rightarrow}\begin{pmatrix}X_\infty\\Y_\infty\end{pmatrix}=\mathcal{N}\left(\mathbf{0},\begin{pmatrix}\mathrm{E}\left[X_1^2\right]&0\\0&\mathrm{E}\left[Y_1^2\right]\end{pmatrix}\right) \end{equation} as $\begin{pmatrix}N&T\end{pmatrix}\rightarrow\begin{pmatrix}\infty&\infty\end{pmatrix}$ by Central Limit Theorem and \begin{equation} \sqrt{N}\bar{X}_N+\sqrt{T}\bar{Y}_T\overset{d}{\rightarrow}X_\infty+Y_\infty \end{equation} by Continuous Mapping Theorem. However, the manipulation below may NOT be possible as both $N$ and $T$ go to infinity \begin{equation} \begin{pmatrix}\sqrt{NT}\bar{X}_N\\\sqrt{NT}\bar{Y}_T\end{pmatrix}\overset{d}{\rightarrow}\begin{pmatrix}X_\infty\\Y_\infty\end{pmatrix}=\mathcal{N}\left(\mathbf{0},\begin{pmatrix}T\mathrm{E}\left[X_1^2\right]&0\\0&N\mathrm{E}\left[Y_1^2\right]\end{pmatrix}\right) \end{equation} and hence this may be IMpossible as well. \begin{equation} \sqrt{NT}\bar{X}_N+\sqrt{NT}\bar{Y}_T\overset{d}{\rightarrow}\mathcal{N}\left(0,T\mathrm{E}\left[X_1^2\right]+N\mathrm{E}\left[Y_1^2\right]\right) \end{equation} However, for me it seems without any formal proof that at least I can tell \begin{equation} \frac{\sqrt{NT}\bar{X}_N+\sqrt{NT}\bar{Y}_T}{\sqrt{T\mathrm{E}\left[X_1^2\right]+N\mathrm{E}\left[Y_1^2\right]}}\overset{d}{\rightarrow}\mathcal{N}\left(0,1\right) \end{equation} and now I wonder if this is valid. Here I want to prove this, but I cannot apply Continuous Mapping Theorem anymore since both $N$ and $T$ are related to there and I do not know where to start. Can I ask any help regarding this strange matter?
2026-04-02 05:58:31.1775109511
Really Strange Manipulation of Convergence in Distribution
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You probably want to appeal to a uniform weak convergence result, such as is described in 'Uniform convergence of probability measure on classes of functions' (P. J. Bickel and P. W. Millar, Statistica Sinica Vol. 2, No. 1 (January 1992), pp. 1-15. Simplify your notation a bit, to assert that you know $V_{n,t}\xrightarrow{d}V$ where $V$ is a standard bivariate gaussian vector, as ($n,t)\to(\infty,\infty),$ and you want to know that $\alpha_{n,t}\cdot V_n\xrightarrow{d}W$ where $W$ is a standard univariate gaussian and $\alpha_{n,t}=(\cos(\theta_{n,t}),\sin(\theta_{n,t}))$ for certain $\theta_{n,t}$. It suffices to show that for each $f$ in some convergence determining class, you have $Ef(\alpha_{n,t}\cdot V_{n,t})\to Ef(W)$. To do this, consider the collection $F$ of functions of form $f_\theta:(x,y)\mapsto f(x\cos\theta +y\sin\theta )$; you want the convergence to be uniform over $F.$ By Proposition 2.2 in the cited paper, if you can show $F$ is compact for each of your $f$s, you are done. I would start by checking that if $f$ is in $C_0(\mathbb R)$, the corresponding $F$ is compact in $C_0(\mathbb R^2)$.
In other words, your vector quantity $V_{n,T} = \left(\sqrt n\bar{X}_n\big/\sqrt{EX_1^2},\sqrt T\bar{Y}_T\big/\sqrt{EY_1^2}\right) $ has a standard (mean zero, circularly symmetric) gaussian limit and you are interested in the limit distribution of a $(n,T)$-dependent linear function applied to $V_{n,T}$, whose coefficients are $\left(\sqrt T\sqrt {EX_1^2}, \sqrt n\sqrt{EY_1^2}\right)\big/\sqrt{T E(X_1^2)+nE(Y_1^2)}$. This looks complicated, but all we care about is that they are non-random and the sum of the squares of these coefficients is $1$, so we might as well think of the coefficients as being $(\cos\theta,\sin\theta)$ for some $(n,T)$-dependent $\theta$. For each fixed $\phi$, the dot product of $(\cos \phi,\sin\phi)$ with the vector $V_{n,T}$ has the desired limit. In your case, however, the coefficients vary with $(n,T)$, so as you correctly intuit, this poses a technical obstacle. But the set of all these linear functions is compact (it is a circle, after all), so for each $\epsilon$ there is a finite number of functions in the set that approximate any function in the set, to accuracy $\epsilon$. The rest is a routine compactness argument.
This is, perhaps daunting at first, and it is not helped by the intricacy of your particular problem statement. But if you want to master the theory and practice of convergence in distribution you must acquire a working knowledge of the basic tools of the subject, at say the level of Billingsley's Convergence of Probability Measures. Chief among them is the use of compactness.