# Can independent non-identically distributed random variables be converted into i.i.d. random variables?

This paper gives the i.ni.d. example of "measurements accurate to the nearest foot may be combined with measurements accurate to the nearest inch". Can we simply pool that data to satisfy i.i.d.? (The disadvantage being that pooling throws away information.) More generally, can we combine i.ni.d. random variables into i.i.d. random variables by using the weighted convolution of the non-identical distributions as our new identical distribution?

• The title is much more general than the example. Do you want the full generality of the title or something much more like the specifics of the example? – Glen_b -Reinstate Monica Nov 15 '14 at 4:07
• I'd prefer a general answer. However, if the answer is no in general, I'd be interested to hear under which conditions the answer is yes. – Antlers Nov 15 '14 at 4:15
• This question needs clarification. For instance the sum of a collection of variables is iid (because there is only one of them!). – whuber Nov 15 '14 at 14:25

Sure, at least for continuous variables. Let $X,Y$ be two independent random variables with distribution functions $G_1$ and $G_2$ respectively. Then $U=G_1(X)$ and $V=G_2(Y)$ are i.i.d. standard uniform.
If $F$ is any continuous invertible cdf then $W=F^{-1}(G_1(X))$ and $Z=F^{-1}(G_2(Y))$ are i.i.d. with distribution $F$.