I guess in part what this all amounts to is what does the assumption {(x_i,y_i) : i=1,2,...,n} being i.i.d. imply about the i.i.d-ness of functions of it?

I am confused because for example I have seen online that the product of i.i.d. random variables are not necessarily i.i.d.

Thanks for all the help!


TLDR: $\{X_iY_i\}$ is i.i.d. as well, so the standard argument for consistency carries out here.

From the premises of the question, we know that $\{(X_i,Y_i)\}$ is iid.

  1. Independence of $X_iY_i$ from $X_jY_j$, $\forall i\ne j$. We can apply a standard result in probability theory that states that (measurable) functions of independent variables are independent as well (here's a concise proof of this result).

  2. Identical distribution. It follows from the definition of the distribution function of a function of a random vector ($(X_i,Y_i)$ in this case) and the fact that $(X_i,Y_i)\overset{d}{=}(X_j,Y_j)$ for all $i,j$.

The rest of your question is straightforward after this. Hope it helps.

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  • $\begingroup$ Hi there, thanks for your answer. For the identical distribution part - does the joint distribution determine the distribution function of the bivariate random variable determine the distribution of any functions of the two variables? $\endgroup$ – Dutchman Sep 20 '18 at 6:39
  • $\begingroup$ Not any two functions - they need to have some properties (measurability to begin with). There are theoretical results for certain classes of functions, for example one-to-one differentiable functions. I guess it really depends on the problem you have and the class you're working with. $\endgroup$ – MauOlivares Sep 20 '18 at 13:34

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