Let $X$ be a random variable with $E[X] = \mu < \infty$.


For $n=1,2,\dots$, construct a triangular array of random variables as


\begin{equation}
Y\_{n,i} = X\_i \frac{\sqrt{\mu}}{\sqrt{\sum\_{j=1}^n X\_j/n}}.
\end{equation}


Then, does the following hold?
\begin{equation}
\frac{1}{n} \sum\_{i=1}^n Y\_{n,i} \overset{p}{\to} \mu,
\end{equation}
where $\overset{p}{\to}$ denotes the convergence in probability.


Or more generally, for any bounded measurable function $f$, does the following hold?
\begin{equation}
\frac{1}{n} \sum\_{i=1}^n f(Y\_{n,i}) \overset{p}{\to} E[f(X)].
\end{equation}


If not, is there any necessary condition?


Thanks in advance.



Weak law of large numbers for triangular arrays
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\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}} 
\newcommand{\EE}{\mathcal E}
\newcommand{\F}{\mathcal F}
\newcommand{\I}{\mathcal I}
\newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}
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\newcommand{\pow}{\text{power}}
\renewcommand{\bar}{\overline}$


It appears that in the OP's question the condition that the $X\_i$ be independent copies of $X$ is missing; let us assume this condition. To make (1) meaningful and (2) nontrivial, we also need to assume that $\mu>0$. In addition, we need to define $Y\_{n,i}$ on the event $\{\sum\_{j=1}^n X\_j\le0\}$; let then define $Y\_{n,i}$ as $0$ on this event (of vanishing probability as $n\to\infty$). 


Note first that (3) will fail to hold in general. E.g., suppose that $\PP(X=1)=\mu=1-\PP(X=0)$, where $\mu$ is any transcendental (non-algebraic) number in the interval $(0,1)$; for instance, one may take $\mu=e-2$. Then $\E X=\mu$ and all the nonzero values taken by $Y\_{n,i}$ with nonzero probability are transcendental. Let now $f(y):=\ii{y=1}$ for all real $y$, where $\ii\cdot$ denotes the indicator. Then $\frac1n\sum\_{i=1}^n f(Y\_{n,i})$ is $0$ almost surely, whereas $\E f(X)=\PP(X=1)=\mu\ne0$, so that (3) fails to hold. 




---


The positive news is that, as will be shown here, (3) will hold -- even with the almost sure (a.s.) convergence (rather than just in probability) -- if $f$ is continuous, which will be henceforth assumed. Take any positive real numbers $a$ and $\eta$. Since $f$ is continuous, it is uniformly continuous on the interval $[-2a,2a]$. So, for some $\ep\in(0,1]$ and all $x,y$ in $[-2a,2a]$ such that $|x-y|\le a\ep$, we have $|f(y)-f(x)|<\eta$. Take now any $\de>0$ such that 
\begin{equation\*}
 |x-\mu|<\de\implies\big|\sqrt{\tfrac\mu x}-1\big|<\ep. \tag{4}
\end{equation\*}


Now, introduce events 
\begin{equation\*}
 A\_n:=A\_{n,\de}:=\{|\bar X\_n-\mu|<\de\},\quad B\_i:=\{|X\_i|\le a\},
\end{equation\*}
where $\bar X\_n:=\frac1n\,\sum\_{j=1}^n X\_i$. By the strong law of large numbers, $\ii{A\_n^c}\to0$; everywhere here ${}^c$ denotes the complement and the convergence is for $n\to\infty$; the convergence of sequences of random variables (r.v.'s) is a.s. 


On the event $A\_n\cap B\_i$, we have $|X\_i|\le a\le 2a$ and, by (4),  

\begin{equation\*}
 |Y\_{n,i}-X\_i|=|X\_i|\Big|\frac{\sqrt{\mu}}{\sqrt{\bar X\_n}}-1\Big|\le a\ep 
\end{equation\*}
and hence also 
$|Y\_{n,i}|\le a(1+\ep)\le2a$. 
So, by the mentioned uniform continuity of $f$, 
\begin{equation\*}
 |f(Y\_{n,i})-f(X\_i)|<\eta\quad\text{on the event}\quad A\_n\cap B\_i. 
\end{equation\*}


Introduce now the r.v.'s 
\begin{equation\*}
 Z\_n:=\frac1n\sum\_1^n f(Y\_{n,i}),\quad 
 T\_n:=\frac1n\sum\_1^n f(Y\_{n,i})\ii{B\_i}, 
\end{equation\*}
\begin{equation\*}
 V\_n:=\frac1n\sum\_1^n f(X\_i),\quad 
 W\_n:=\frac1n\sum\_1^n f(X\_i)\ii{B\_i}. 
\end{equation\*}
Since the function $f$ was assumed to be bounded, we have $|f|\le M$ for some real $M>0$. Therefore, 
\begin{align\*}
 &|Z\_n-T\_n|\le\frac Mn\sum\_1^n \ii{B\_i^c}\to M\PP(B\_1^c)=M\PP(|X|>a), \\ 
 &|T\_n-T\_n\ii{A\_n}|=|T\_n|\ii{A\_n^c}\le M\ii{A\_n^c}\to0, \\ 
 &|T\_n\ii{A\_n}-W\_n\ii{A\_n}|\le \frac1n\sum\_1^n |f(Y\_{n,i})-f(X\_i)|\ii{A\_n\cap B\_i}\le\eta, \\ &|W\_n\ii{A\_n}-W\_n|=|W\_n|\ii{A\_n^c}\le M\ii{A\_n^c}\to0, \\ 
 &|W\_n-V\_n|\le\frac Mn\sum\_1^n \ii{B\_i^c}\to M\PP(B\_1^c)=M\PP(|X|>a), \\ 
 &V\_n\to\E f(X). 
\end{align\*}
Thus, 
\begin{equation\*}
 \limsup\_n\Big|\frac1n\sum\_1^n f(Y\_{n,i})-\E f(X)\Big|
 =\limsup\_n\Big|Z\_n-\E f(X)\Big|
 \le2M\PP(|X|>a)+\eta. 
\end{equation\*}
Letting now $a\to\infty$ and $\eta\downarrow0$, we see that indeed a.s. 
\begin{equation\*}
 \frac1n\sum\_1^n f(Y\_{n,i})\to\E f(X). 
\end{equation\*}


