Convergence in Probability $\leadsto$ Convergence Almost Sure | In the context of conditional convergence in distribution

Question

Let $Z_1, Z_2, ...$ be a sequence of independent but not necessarily identically distributed real-valued random variables with finite second moments, and let $Y$ be a real-valued random variable that is independent of all of the $Z_i$'s. Define new random variables $X_n := Z_nY$ for all $n\in\mathbb N$, and suppose that these newly defined random variables satisfy the Lindeberg condition in probability, that is, for all $\epsilon>0$ and all $\eta>0$ it holds that $$P\left(\frac{1}{s_n^2}\sum_{k=1}^n\operatorname{E}\left[(X_k - \mu_k)^2\cdot 1_{\left(\frac{X_k - \mu_k}{s_n}\right)^2>\epsilon}\middle|\, Y\right]\geq\eta\right)\rightarrow0\tag{$\star$}$$ as $n\rightarrow\infty$, where $\mu_k := \operatorname E[X_k\mid Y]$ and $$s_n := \sqrt{\sum_{k=1}^n\operatorname{E}\left[(X_k - \mu_k)^2\mid Y\right]}.$$

I want to show that ($\star$) implies $$P\Big(\rho(Q_n,Q)\geq\delta\Big)\rightarrow 0$$ as $n\rightarrow\infty$, where

• $\rho$ denotes the Lévy-Prokhorov metric on the set $\mathcal M$ of all probability measures on the Borel $\sigma$-algebra over $\mathbb R$,
• $Q_n$ is the distribution of $\frac{\sum_{k=1}^n(X_k - \mu_k)}{s_n}$ conditional on $Y$ for all $n\in\mathbb N$, and
• $Q$ is the probability measure associated with the standard normal distribution on $\mathbb R$.

Since $\mathbb R$ is separable, $\rho$ metrizes the topology of convergence in distribution on $\mathcal M$. That is, $\rho(Q_n, Q\big)\rightarrow 0$ as $n\rightarrow\infty$ is equivalent to $\frac{\sum_{k=1}^n(X_k - \mu_k)}{s_n}$ converging to a standard normal random variable in distribution as $n\rightarrow\infty$. Since $\rho(Q_n,Q)$ is a random variable (as it still depends on $Y$), I want to show that the probability of $\rho(Q_n,Q)$ being larger than any $\delta>0$ converges to $0$ in probability.

My approach: Let $\epsilon>0$. Put $$W_n := \frac{1}{s_n^2}\sum_{k=1}^n\operatorname{E}\left[(X_k - \mu_k)^2\cdot 1_{\left(\frac{X_k - \mu_k}{s_n}\right)^2>\epsilon}\mid\,Y\right]$$ for each $n\in\mathbb N$. By Sub sub sequences and a relation between convergence in probability and a.s convergence (Final Result), the condition ($\star$) guarantees that for every subsequence $W_{n_1}, W_{n_2}, ...$ there is a sub-subsequence $W_{n_{m_1}}, W_{n_{m_2}}, ...$ such that $$P\left(\text{$W_{n_{m_\ell}} \rightarrow 0$ as $\ell\rightarrow\infty$}\right) = 1.$$ Since the Lindeberg condition holds, the Lindeberg CLT implies that $$P\left(\text{$\rho(Q_{n_{m_\ell}},Q) \rightarrow 0$ as $\ell\rightarrow\infty$}\right) = 1,$$ i.e., the sub-subsequence $Q_{n_{m_1}}, Q_{n_{m_2}}, ...$ converges to $Q$ almost surely. Since almost sure convergence implies convergence in probability, it follows immediately that for all $\delta>0$: $$P\left(\rho(Q_{n_{m_\ell}},Q) \geq\delta\right)\rightarrow 0$$ as $\ell\rightarrow\infty$. That is, for every subsequence there is a sub-subsequence such that the distance in the Lévy-Prokhorov metric tends to $0$ in probability. Consequently, by Sub sub sequences and a relation between convergence in probability and a.s convergence (Claim 1), the Lévy-Prokhorov metric tends to $0$ in probability for the full sequence, i.e., $$P\Big(\rho(Q_n,Q)\geq\delta\Big)\rightarrow 0$$ as $n\rightarrow\infty$.$\qquad\blacksquare$

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Is my reasoning correct? I am unsure because it feels a bit like I cheated: I have ($\star$) for a subsquence of the sequence $W_1, W_2, ...$, while in the last step I draw my conclusion based on on subsequences of $Q_1, Q_2, ...$. Certainly, both sequences are related, but the former 'dictates' the subsequence for the latter, and there may be the possibility that the 'dictated' subsequences of the former really contain all subsequences of the latter. On the other hand, there is a clear relationship (also used in the third last step) between $W_n$ and $Q_n$ for all $n\in\mathbb N$, and this relationship bijective. So extracting any possible subsequence of $W_1, W_2, ...$ will eventually cover any possible subsequence of $Q_1, Q_2,...$.

Anyway, I would be grateful if any doubts would be resolved and I have a clear answer (also, if there are any mistakes in my proof).

Answer 1

The statement is true if and only if $Y\neq0$ almost surely. Indeed if $Y=0$ then $s_n=0$ and a number of parts of the question are not well-defined.

So assume that $Y\neq0$ almost surely. We have $\mu_k=\mathbb E[X_k|Y] = \mathbb E[Z_k]Y$, and similarly $$s_n=|Y|\sqrt{\sum_{k=1}^n (Z_k-\mathbb EZ_k)^2}$$ So $$\frac{\sum_{k=1}^n (X_k-\mu_k)}{s_n}=\mathrm{sgn}(Y)\frac{\sum_{k=1}^n (Z_k-\mathbb EZ_k)}{\sqrt{\sum_{k=1}^n (Z_k-\mathbb EZ_k)^2}}$$ And similarly the Lindeberg condition can be rewritten purely in terms of the $Z_k$. All in all it reduces to applying the unconditional form of the Lindeberg condition to the $Z_k$. And the $\mathrm{sgn}(Y)$ ends up not mattering since the standard normal distribution is symmetric around zero.