I am looking for a definition of conditional convergence. Suppose that $X_1, X_2, \dots, X_n$ are $\mathbb R$-valued random variables with finite second moments, and $W_1, W_2, \dots, W_n$ are iid $\mathbb R$-valued random variables independent of the $X_i$s with finite second moments (w.l.o.g., assume $\mathbb EW_i = 0$ and $\mathbb EW_i^2 = 1$). Consider the sequence $(Z_n)_{n\in\mathbb N}$ of $\mathbb R$-valued random variables given by $$Z_n := \frac 1{\sqrt n}\sum^n_{i=1}W_iX_i.$$ Intuitively, this sequence should converge, conditional on the $X_i$s, to $\mathcal N(0,\sigma^2)$, where $\sigma^2 := \mathbb EX_i^2 - (\mathbb EX_i)^2$. However, I don't know how to show this rigorously. Part of the problem is that I don't even have a definition...

I have seen this definition: $(Z_n)_{n\in\mathbb N}$ converges weakly conditional on the $X_i$s to a $\mathbb R$-valued random variable $Z$ iff $(\Delta_n)_{n\in\mathbb N}$ converges to zero in probability for all bounded, continuous $\mathbb R$-valued functions on $\mathbb R$, where $$\Delta_n := \left\vert\mathbb E[f(Z_n)\mid\mathcal H] - \mathbb E[f(Z)]\right\vert$$ with $\mathcal H := \sigma(\{X_i : i\in\mathbb N\})$ denoting the $\sigma$-algebra generated by the $X_i$s. Are there other (equivalent but more straight forward) definitions of weak conditional convergence?

However, this definition does not really help me as computing the conditional expectation of $f(Z_n)$ given $\mathcal H$ for every bounded, continuous $\mathbb R$-valued functions is exhaustive.

  • $\begingroup$ Not necessarily $\endgroup$ Commented Dec 5, 2023 at 15:56
  • $\begingroup$ Are the $X_i$'s iid? If so, then, conditionally on the $X_i$'s, the $Z_n$'s are indeed almost surely asymptotically normal. Is this what you actually want? To get this conclusion, the iid condition on the $X_i$'s cannot be dropped. $\endgroup$ Commented Dec 5, 2023 at 18:00
  • $\begingroup$ well I could place some restrictions on the moments (e.g., $\mathbb EX_i\rightarrow 0$ and $\mathbb EX_i^2\rightarrow0$ to get the law of large of number to work. My confusion comes from how to even start (see other comments below) $\endgroup$ Commented Dec 5, 2023 at 22:23
  • $\begingroup$ You need, not only a LLN for the $X_i$'s (or, rather, for the $X_i^2$'s), but a bit more. Why don't you just say what your conditions on the $X_i$'s actually are? These conditions are the essence if you do want the $Z_n$'s to be almost surely asymptotically normal. As I said previously, it would be enough for the $X_i$'s to be iid. $\endgroup$ Commented Dec 5, 2023 at 22:33
  • $\begingroup$ I wonder whether you meant you want the $X\text{s}$ and the $W\text{s}$ each to be an infinite sequence rather than a sequence of length $n.$ $\endgroup$ Commented Dec 6, 2023 at 19:16

2 Answers 2


$\newcommand{\R}{\mathbb R}\newcommand{\N}{\mathbb N}\newcommand{\ep}{\varepsilon}\newcommand{\si}{\sigma}$This is to complement Nate River's answer by obtaining a simple and natural sufficient condition (which is close to a necessary one) for a central limit theorem for $Z_n$ holding almost surely (a.s.) conditionally on the $X_i$'s.

To have such a theorem, we should assume that the $X_i$'s and the $W_i$'s form infinite sequences of random variables (r.v.'s).

So, let $X:=(X_1,X_2,\dots)$ be a sequence of (real-valued) r.v.'s independent of iid zero-mean unit-variance r.v.'s $W_1,W_2,\dots$.

For $x=(x_1,x_2,\dots)\in\R^\N$, let \begin{equation*} Z_{x;n}:=\frac1{\sqrt n}\sum_{i=1}^n x_i W_i\quad\text{and}\quad s_n:=s_{x;n}:=\sqrt{\sum_{i=1}^n x_i^2}. \end{equation*} Then \begin{equation*} \frac{Z_{x;n}}{s_n/\sqrt n}\to Z\sim N(0,1) \tag{1}\label{1} \end{equation*} (in distribution as $n\to\infty$) if the following Lindeberg condition holds: \begin{equation*} \frac1{s_n^2}\sum_{i=1}^n x_i^2 EW_1^2\,1(x_i^2 W_1^2>\ep^2 s_n^2)\to0 \tag{2}\label{2} \end{equation*} for every real $\ep>0$. Moreover, \eqref{2} is necessary for \eqref{1} if the following natural "uniform smallness" condition holds: \begin{equation*} \frac1{s_n^2}\max_{i=1}^n x_i^2\to0. \tag{3}\label{3} \end{equation*} Furthermore, in our case, if \eqref{3} holds, then, by dominated convergence, $EW_1^2\,1(x_i^2 W_1^2>\ep^2 s_n^2)\to0$ uniformly in $i=1,\dots,n$, which implies \eqref{2}.

So, \eqref{3} implies \eqref{1}. Thus, letting $S_n:=\sqrt{\sum_{i=1}^n X_i^2}$, conditionally on $X:=(X_1,X_2,\dots)$ a.s. we have
\begin{equation*} \frac{Z_n}{S_n/\sqrt n}\to Z \tag{1a}\label{1a} \end{equation*} provided that a.s. \begin{equation*} \frac1{S_n^2}\max_{i=1}^n X_i^2\to0 \tag{3a}\label{3a} \end{equation*} If, moreover, the $X_i^2$'s satisfy the strong law of large numbers (SLLN) of the form \begin{equation} \frac{S_n^2}n\to\si^2 \tag{4}\label{4} \end{equation} a.s. for some $\si\in(0,\infty)$ (and \eqref{3a} still holds), then it follows that \begin{equation*} Z_n\to\si Z. \tag{1b}\label{1b} \end{equation*}

Clearly, the SLLN \eqref{4} will hold with $\si^2=EX_1^2$ if e.g. the $X_i$'s are iid with a finite second moment. However, the definition $\si^2 := EX_i^2 - (EX_i)^2$ in the OP is not what is needed unless $EX_i=0$ (consider e.g. the case when $X_i=1$ a.s. for all $i$).

  • $\begingroup$ Thank you! This is a great answer and exactly what I have been looking for. You give precise conditions of conditions I have to check for my $X_i$s (this is also the reason why I did not state any requirements of the $X_i$s in the first place). Just one thing is not yet entirely clear to me: why Lindeberg CLT? Since the $x_iW_i$s are independent but not identically distributed? $\endgroup$ Commented Dec 7, 2023 at 2:30
  • 1
    $\begingroup$ @SydAmerikaner : In the Lindeberg condition, the random variables are assumed to be independent but do not have to be identically distributed. $\endgroup$ Commented Dec 7, 2023 at 3:47

This is one of those things that is immediate if you use regular conditional probabilities. For $\mathbb x \in \mathbb R^{\mathbb N}$, and a random variable $Y$, let $\mu^{\mathbb x}_Y$ denote the regular conditional probability distribution of $Y$ given $(X_1, X_2, \dots) = \mathbb x$. We also use $\mu_X$ to denote the joint law of the $X_i$ on $\mathbb R^n$.

By independence of $W_i$ from $X_i$, we have that the conditional distribution of $Z_n$ given $X_i = \mathbb x$ is just $\frac{1}{\sqrt n} \sum_{i = 1}^n W_i x_i$, where I use $x_i$ to denote the $i$’th component of $\mathbb x$.

Now we may use the following definition of weak conditional convergence: $Z_n \to Z$ weakly conditional on $X_i$ if for $\mu_X$-almost every $\mathbb x$, we have that $\mu^{\mathbb x}_{Z_n} \to \mu^{\mathbb x}_{Z}$ weakly in the usual sense.

It can be checked that this definition is equivalent to yours (feel free to ask for details!). Using this equivalent definition, it seems that your $Z_n$ converge to a normal random variable with random parameter parametrized by the $X_i$. But due to the LLN on the $X_i$, the parameter should be deterministic in the limit.

  • $\begingroup$ "$Z_n\rightarrow Z$ weakly conditional on $X_i$ if for $\mu_X$-almost every $\mathbf x$, we have that $\mu^{\mathbf x}_{Z_n} \rightarrow\mu_Z^{\mathbf x}$ weakly in the usual sense". I think this is what confuses me: How can we have $\mu^{\mathbf x}_{Z_n} \rightarrow\mu_Z^{\mathbf x}$ weakly in the usual sense if a LLN at work at the same time. $\endgroup$ Commented Dec 5, 2023 at 16:04
  • $\begingroup$ More details: I would have to check that $\mu_X(\{\mu^{\mathbf x}_{Z_n} \rightarrow\mu_Z^{\mathbf x}\text{ for all $\mathbf x$}\}) = 1$ holds. So what I would do is to check $\mu^{\mathbf x}_{Z_n} \rightarrow\mu_Z^{\mathbf x}$ for all $\mathbf x$ and then compute the probability of those $\mathbf x$ for which no convergence could be shown. However, this feels like it's very difficult (exhaustive) to do, too. In the last paragraph you gave a nice heuristic: $Z_n$ converge conditional to normal and LLN applies to $X_i$. I think this argument spelled out rigorously is what I am looking for. $\endgroup$ Commented Dec 5, 2023 at 16:11
  • $\begingroup$ @SydAmerikaner Yes the full proof is rather difficult to do. I’ll see if I have time to try to work it out. $\endgroup$
    – Nate River
    Commented Dec 6, 2023 at 5:23

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