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This question already has an answer here:

I have tuples, $\{(X_i,Y_i)\}_{i=1}^n$, which are iid.

Suppose I introduce a new variable, $T_i = g(X_i)$, where the function $g$ is arbitrary. Is it true that $\{(T_i, Y_i)\}_{i=1}^n$ are iid as well? Is the following argument correct?


My reasoning:

for $i \neq j$:

\begin{align*} P(T_i \in A, Y_i \in B, T_j \in C, Y_j \in D) &= P(g(X_i) \in A, Y_i \in B, g(X_j) \in C, Y_j \in D) \\ &= P(X_i \in g^{-1}(A), Y_i \in B, X_j \in g^{-1}(C), Y_j \in D) \\ &= P(X_i \in g^{-1}(A), Y_i \in B) P(X_j \in g^{-1}(C), Y_j \in D) \\ \end{align*}

where $g^{-1}(A) = \{ x: g(x) \in A \}$.

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marked as duplicate by whuber Jul 25 '15 at 0:22

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  • $\begingroup$ Yes, if $Y$ and $X$ are independent, then so are $Y$ and $g(X)$. This makes sense intuitively since the function $g$ itself doesn't carry any information about $Y$, so merely applying that function to $X$ can't somehow induce dependence. $\endgroup$ – dsaxton Jul 24 '15 at 21:53
  • $\begingroup$ Makes sense. I guess I just needed some reassurance. Thanks. $\endgroup$ – Chester Jul 24 '15 at 21:56
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    $\begingroup$ What exactly is meant by iid when you say that $(X_i,Y_i)$ are iid? Are the $2n$ random variables a collection of mutually independent identically distributed random variabes? Are the $n$ pairs $(X_i,Y_i)$ identically distributed and independent of each other? (nothing being implied about the independence of $X_i$ and $Y_i$?) $\endgroup$ – Dilip Sarwate Jul 24 '15 at 22:14
  • $\begingroup$ @DilipSarwate, The last one: $i\neq j \Rightarrow P(X_i,Y_i,X_j,Y_j) = P(X_i, Y_i) P(X_j,Y_j)$. I should've been more precise in my statement. $\endgroup$ – Chester Jul 24 '15 at 22:55
  • $\begingroup$ "iid" means identical and independently distributed. Obviously the first "i" no longer holds upon applying $g$ to just one component. The remaining "i" is addressed even more generally in the duplicate, which is applied here by transforming one variable with $g$ and the other with the identity transformation. $\endgroup$ – whuber Jul 25 '15 at 0:21

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