Consider a sequence of random vectors $Z_n$ in $R^d$ converging in distribution to $N(0_d, S)$ for some psd matrix $S$.
Consider a sequence of square matrices $T_n$ lower triangular with diagonal elements all equal to 1 such that $T_nS T_n^\top \to^P C$ (convergence in probability to some psd matrix $C$ in $R^{d\times d}$). The sequence $T_n$ is further assumed bounded, $P(\|T_n\|_{op}\le b)=1$ for some non-random $b>0$.
A previous answer to a related question, https://math.stackexchange.com/a/4800263/484640, shows that $T_nZ_n\to^d N(0,C)$ in the special case $S=I_d$, thanks to the continuity of the Choesky factor.
Questions:
- is it always true that $T_n Z_n \to^d N(0, C)$, when $S$ is not invertible?
Edit: The case if $S$ invertible (positive definite) can be treated as follows: let $LL^T=S$ be the unique Cholesky decomposition of $S$. Then $\tilde T_n=T_nL$ and $\tilde Z_n=L^{-1} Z_n$ satisfy $\tilde Z_n\to^d N(0,I_d)$ and $\tilde T_n\tilde T_n^\top \to^P C$. Then https://math.stackexchange.com/a/4800263/484640 applied to $\tilde Z_n,\tilde T_n$ completes the proof that $T_n Z_n = \tilde T_n\tilde Z_n\to^d N(0, C)$.
Consider the LDLT decompositions $S=\triangle D \triangle ^T$ and $C=KEK^T$ with $D,E$ diagonal with non-negative entries and $\triangle ,K$ lower triangular matrices with only 1's on the diagonal. Then $T_nST_n^T\to^P C$ gives by the continuous mapping theorem $$L_n D L_n^T \to^P E $$ where $L_n = K^{-1}T_n\triangle$ is lower triangular with 1's on the diagonal. Also, $Z_n\to N(0_d, S)$ gives $\zeta_n\to^d N(0_d, D)$ where $\zeta_n=\triangle^{-1} Z_n$. In essence, we may work with diagonal matrices for the two covariance matrices.
Consider an index $i\in[d]$. Since $$ D_{ii} \le \sum_j (L_n)_{ij}^2 D_{jj} \to^P E_{ii} $$ we obtain $E_{ii}=0\Rightarrow D_{ii}=0$. Since we assume $P(\|T_n\|_{op}\le b)=1$, and $L_n,T_n$ both have 1's on the diagonal, $P(\|L_n\|_{op} + \|L_n^{-1}\|_{op} \le B)=1$ for some constant $B>0$ depending only on $b$ and the dimension $d$. Thus from $L_nDL_n^T= E + o_P(1)$ we obtain $D= L_n^{-1} E (L_n^{-1} )^T + o_P(1)$ and $$ E_{ii} \le \sum_j (L_n^{-1})_{ij}^2 E_{jj} \to^P D_{ii} $$ so $D_{ii}=0\Rightarrow E_{ii}=0$. This proves that $D_{ii}=0$ if and only if $E_{ii}=0$, that is, there exists a subset $I\subset[d]$ with $I=\{i\in[d]: D_{ii}\ne 0\}$ and $I=\{i\in[d]: E_{ii}\ne 0\}$.
It will be useful to restrict our matrices to rows or columns in $I$. To this end, consider the matrix $Q\in R^{[d]\times |I|}$ obtained from the identity $I_d$ and dropping the columns not indexed in $I$. Let also $\bar E = Q^T E Q$ and $\bar D = Q^T D Q$ be the smaller diagonal matrices, now with only nonzero positive entries. By the continuous mapping theorem, $$ (I_d- QQ^T)\zeta_n \to^d N(0_d, (I_d- QQ^T)D(I_d- QQ^T)) = N(0_d, 0_{d\times d}) $$ so $(I_d- QQ^T)\zeta_n\to^P 0_d$ in probability. Similarly, $$ Q^T \zeta_n \to^d N(0_{|I|}, Q^TDQ) = N(0_{|I|}, \bar D). $$ We have $Q^TL_nQ \bar D Q^TL_nQ = Q^TL_nDL_n^TQ \to^P Q^T E Q=\bar E$ so that $Q^TL_nQ\sqrt{\bar D} (Q^TL_nQ\sqrt{\bar D})^T \to^P \bar E$. By unicity and continuity of the Cholesky factor for positive definite matrices, the continuous mapping theorem gives $Q^TL_nQ\sqrt{\bar D} \to^P \sqrt{\bar E}$ or $$Q^TL_nQ \to^P \sqrt{\bar E \bar D^{-1}}$$ (here, $\bar E,\bar D$ are positive diagonal matrices that commute).
Now we come back to the object of interest, $L_n \zeta_n$, first restrict to indices in $I$: \begin{align*} Q^T L_n \zeta_n &= Q^T L_n (I_d-QQ^T)\zeta_n + (Q^T L_nQ) Q^T\zeta_n \\&= o_P(1) + (Q^T L_nQ) Q^T\zeta_n \end{align*} and $(Q^T L_nQ) Q^T\zeta_n \to^d N(0,\bar E)$, so that by Slutsky's theorem we obtain the desired limit for $e_i^T L_n\zeta_n$ on the components $i\in I$ (here, $e_i$ is the $i$th-canonical vector).
It remains to prove that the limit is 0 on the components $i\in[d]\setminus I$. Using again $L_n D L_n^T \to^P E$ restricted to a diagonal entry $i\notin I$, for any $j\in I$ we get $$ (L_n)_{ij}^2 D_{jj} \le \sum_{j} (L_n)_{ij}^2 D_{jj} = e_i^T L_n D L_n^T e_i \to^P E_{ii}=0 $$ so that $(L_n)_{ij}\to^P 0$ for any indices $j\in I$ and $i\notin I$ thanks to $D_{jj}>0$. Hence \begin{align*} e_i^T L_n \zeta_n &= e_i^T L_n (I_d-QQ^T)\zeta_n + e_i^T L_n Q Q^T\zeta_n \\&= o_P(1) + o_P(1) \end{align*} thanks to $e_i^T L_n QQ^T = \sum_{j\in I} (L_n)_{ij} e_j^T$. We have proved that $L_n\zeta_n\to^d N(0, E)$ and $T_nZ_n\to^d N(0,C)$ if we multiply again by $K^{-1}$ to come back to the initial non-diagonal psd matrices $S$ and $C$.