Consider a sequence of i.i.d. random variables $\{\xi_j\}_{j=1}^J$.

Consider a random variable $X$ with support $\mathbb{R}$.

Consider the map $f: \mathbb{\mathbb{R}}\rightarrow \mathbb{R}^J$. Let $f_j(X)$ be the $j$-th element of the $J\times 1$ random vector $f(X)$.

Assume that $E(\xi_j| X)=0$ $\forall j=1,...,J$.

I want to show that $$ \frac{1}{J} \sum_{j=1}^J f_j(X) \xi_j\rightarrow_p0 \text{ as $J\rightarrow \infty$} $$

The book I'm reading claims that this holds because:

  • The random variables in the sequence $\{f_j(X) \xi_j\}_{j=1}^J$ are i.i.d. conditional on $(f_j(X) \text{ }\forall j=1,...,J)$.

  • Hence, under the law of large numbers for triangular arrays, $ \frac{1}{J} \sum_{j=1}^J f_j(X) \xi_j\rightarrow_p0$ as $J\rightarrow \infty$

I'm struggling to understand this proof.

The law of large number for triangular arrays states:
Consider the triangular sequence $\{(Y_{J,j})_{j=1}^J\}_{J \in \mathbb{N}}$. Assume $Y_{J,1},\ldots,Y_{J,J}$ are i.i.d random variables with mean $\mu_J$. Then, under some conditions [?], it holds that $$ {1 \over J}\sum_{j=1}^J Y_{J,j}-\mu_J \to_p 0 \text{ as $J\rightarrow \infty$} $$ (see here for example).

But what is $ Y_{J,j}$ in my example? Could you help me to clarify?

  • 1
    $\begingroup$ Does the book give any conditions that $f$ has to satisfy? After all, if $f_j(x)=2^jx$ then $\frac{1}J\sum_{j=1}^Jf_j(X)\eta_j$ might not converge to zero. $\endgroup$ Jun 9, 2019 at 12:06

1 Answer 1


My interpretation:

First of all, $f$ should really be $f^J$ since the domain of $f$ is $\mathbb{R}^J$ so different $J$'s $\implies$ different $f$'s.

Then the thing you want to prove becomes (edit in red):

$$\frac{1}{J} \sum_{j=1}^J f^\color{red}{J}_j(X) \xi_j\rightarrow_p0 \text{ as $J\rightarrow \infty$}$$

Then comparing to your theorem for triangular array, we identify $Y_{J,j} \equiv f^J_j(X) \xi_j$, and note that:

$$\mu_J = E[Y_{J,j}] = E[f^J_j(X) \xi_j] = E_X [ E_\xi [\xi_j | f^J_j(X)]] = E_X [ 0] = 0$$

However, as pointed out by the comment of @AngelaRichardson, you might be missing some condition on $f$ (and/or on $Y$). E.g. consider:

$$f^J_j(X) = \begin{cases} JX, & j = J\\ 0, & j < J \end{cases} $$


$$\frac{1}{J} \sum_{j=1}^J f^J_j(X) \xi_j = \frac{1}{J} (JX) \xi_j = X \xi_j \not\to_p 0$$

Even if you additionally impose the condition that $f_j^J = f_j^K$ for all $j,J,K$ in range, this would rule out the above example but the example by Angela that $f_j(X) = 2^j X$ is still likely to lead to $\not\to_p 0$.

  • $\begingroup$ Thanks. Can you provide some sufficient conditions for the triangular law of large numbers to hold (e.g., finiteness and existence of some moments, etc....)? Maybe the book considers them implicit. Thanks. $\endgroup$
    – Star
    Jun 10, 2019 at 8:19
  • $\begingroup$ Also, it is still unclear to me how we get i.i.d.ness of $\{f_j^J(X)\xi_j\}_{j=1}^J$ from the assumptions. $\endgroup$
    – Star
    Jun 10, 2019 at 11:00
  • $\begingroup$ I personally have not heard of the triangular law until this post. I was simply trying to fit the first part of what you said into the second part of what you said. :) However, you have a good point... I'd think $\{f^J_j(X) \xi_j\}_{j=1}^J$ are NOT i.i.d. because the components $f_j$ can certainly correlate. Even your book only claimed they are i.i.d. when conditioned on ${f^J_j(X)}_{j=1}^J$, but even that seems false: they might be independent when conditioned, but how can they be identically distributed? The $\xi$'s are i.d. but when scaled by $f_j$ they shouldn't be i.d. $\endgroup$
    – antkam
    Jun 10, 2019 at 15:39

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