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I have a problem on independence of random variables: Indeed, if $(T_n)$ is a sequence of iid random variables, and $(S_n)$ is another sequence of iid random variables. Suppose the $(T_n)'s$ are independent of the $(S_n)'s$, then are the difference $(T_n - S_n)$ a sequence of iid random variables? My precise question there is about independence. Thanks.

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    $\begingroup$ the independence is within the sequence, i.e., $T_i$ is independent of $T_j$ for $i \neq j$, right? $\endgroup$ Nov 24, 2015 at 15:04
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    $\begingroup$ A general tool is that if $(X_i)_{i\in I}$ is a collection of independent random variables and if some sets $J(k)\subset I$ are disjoint, for $k$ in $K$, then the collection, $(Y_k)_{k\in K}$ is independent, where $Y_k=g_k(X_i;i\in J(k))$ for every $k$, for some measurable functions $g_k$. Your question is a special case of this. $\endgroup$
    – Did
    Nov 24, 2015 at 15:28
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    $\begingroup$ Dear MichaelChirico, yes for all i and j distinct, T_i and T_j are independent, and S_i and S_j are also independent. Next, for all i, T_i and S_i are independent. So, my question is: is T_i - S_i independent of T_j - S_j for distinct i and j? Thanks $\endgroup$
    – Martino
    Nov 24, 2015 at 16:03

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For any positive integer $n$, we have $T_n-S_n = g(T_n,S_n)$ where $g:\mathbb R^2\to \mathbb R$ is the map $(x,y)\mapsto x-y$. Given $t\in\mathbb R$, it is clear that $$g^{-1}((-\infty,t]) = \{(x,y):x-y\leqslant t\} $$ is a Lebesgue-measurable set in $\mathbb R^2$, and so $g$ is a measurable function. Since $\sigma(g(T_n,S_n))\subset \sigma(T_n,S_n)$, independence of $T_i$ and $S_j$ for $i\ne j$ implies independence of $g(T_i,S_j)$ for $i\ne j$. The "identical" part of iid for $g(T_n,S_n)$ follows from the "identical" part of iid for $(T_n)$ and $(S_n)$.

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