Let $X_1,\dots,X_n$ be $n$ (real-valued) positive random variables independent and identically distributed. The distribution depends on $n$. For example suppose $\chi^2(n)$ for each variable.
I want to show that
$$ plim ~ E\left(\dfrac{n^2X_iX_j}{\left(\sum_{k=1}^n X_k\right)^2}\right)= 1 ~ \forall i,j$$
Simulations show that it is true but I tried to prove it in vain. I can prove if the distribution were not depend on $n$. Can someone help me?
Can you confirm that this is true?
I noticed that what I wrote is true only for $i\ne j$.
Assume $E(X_i) = \mu$ and $V(X_i)=\sigma^2$.
Case 1: $i=j$ $$plim ~ E\left(\dfrac{n^2X_i^2}{\left(\sum_{k=1}^n X_k\right)^2}\right) = plim ~ E\left(\left(\dfrac{X_i}{\frac{1}{n}\sum_{k=1}^n X_k}\right)^2\right)$$ I do not know under which conditions one can permute the $plim$ and the expectancy. But I believe when the function under the expectancy is uniformly bounded in probability (by analogy to the deterministic case). Is it correct?
If so, I can permute the expectancy and the $plim$ because the variable under the expectancy is uniformly bounded in probability. To prove that, I will just show that its $plim$ is bounded.
$$ plim\left(\dfrac{X_i}{\frac{1}{n}\sum_{k=1}^n X_k}\right)^2 = \left(plim\dfrac{X_i}{\frac{1}{n}\sum_{k=1}^n X_k}\right)^2 \sim \dfrac{X_i^2}{\mu^2}$$ By that, I mean the $plim$ does not exist. It is still a random variable. $\dfrac{X_i}{\mu}$ is bounded in probability: It is not possible to increase the high values probabilities faster than the mean (the variable is positive). Thus, the $plim$ is bounded and is interchangeable with the expectancy.
$$E\left(\dfrac{X_i^2}{\mu^2}\right)=\dfrac{\sigma^2 + \mu^2}{\mu^2}= \dfrac{\sigma^2}{\mu^2} + 1 > 1$$
$$ plim ~ E\left(\dfrac{n^2X_i^2}{\left(\sum_{k=1}^n X_k\right)^2}\right)=\dfrac{\sigma^2}{\mu^2} + 1 > 1$$ Case 2: $i\ne j$ Using the same proof as previously, the expectancy and the $plim$ are interchangeable.
$$plim ~ E\left(\dfrac{n^2X_i X_j}{\left(\sum_{k=1}^n X_k\right)^2}\right) = E~ plim~ \left(\dfrac{X_i X_j}{\left(\frac{1}{n}\sum_{k=1}^n X_k\right)^2}\right)$$
$$plim~ \left(\dfrac{X_i X_j}{\left(\frac{1}{n}\sum_{k=1}^n X_k\right)^2}\right) \sim \dfrac{X_iX_j}{\mu^2}$$
This $plim$ is also a random variable and bounded in probability as in the case 1. But in this case, we have $$ E\left(\dfrac{X_iX_j}{\mu^2}\right)=1 \implies plim ~ E\left(\dfrac{n^2X_i X_j}{\left(\sum_{k=1}^n X_k\right)^2}\right) = 1$$
Perhaps this proof is wrong. Can someone confirm it?