Independence of iid samples when function is applied to to one member of sample [duplicate]

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

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

• 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. – dsaxton Jul 24 '15 at 21:53
• 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 $Y_i$?) – Dilip Sarwate Jul 24 '15 at 22:14
• @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. – Chester Jul 24 '15 at 22:55
• "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. – whuber Jul 25 '15 at 0:21