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Intuitively, the statement in the question body is true.

My definition of independence:

Let $X_1, \dots, X_k$ be random variables. Then they are called independent if for all Borel measurable sets $A_1, \dots, A_k$, we have

$$\mathbb{P}\{X_1 \in A_1, \dots ,X_k \in A_k\} = \prod_{i=1}^k \mathbb{P}\{X_i \in A_i\}$$

I tried to prove the result with this definition but I can't detach the two variables from the sum, which makes me unsure that the claim is even true.

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  • $\begingroup$ Try to express the probability of the sum as a sum of probabilities. $\endgroup$ – N74 Sep 29 '18 at 9:45
  • $\begingroup$ You mean try to write $\mathbb{P}\{X_1 + X_2 \in A \}$ as a sum of probabilities? $\endgroup$ – user370967 Sep 29 '18 at 9:55
  • $\begingroup$ Exactly. If you manage to do it, you can then use your definition of independence over the sum. $\endgroup$ – N74 Sep 29 '18 at 9:57
  • $\begingroup$ I tried writing the probability in another way but can't find anything meaningful. $\endgroup$ – user370967 Sep 29 '18 at 10:06
  • $\begingroup$ $X_1,X_2,X_3$ independent implies that any (measurable) function of $X_1,X_2$ is also independent of $X_3$. $\endgroup$ – StubbornAtom Sep 29 '18 at 10:08
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Let $A_{12}$ and $A_1$ be any two Borel measurable sets, let also $\mu_1$, $\mu_2$ and $\mu_3$ be the respective probability measures of $X_1$, $X_2$ and $X_3$. Since the latter are independent, then any joint probability measure of those would be the product and so we can do the Lebesgue integration

\begin{align*} \mathbb{P}(X_1 + X_2 \in A_{12},X_3\in A_3) &= \int \int\int \mathbb{1}(x_1 + x_2 \in A_{12},x_3\in A_3)d\mu_1d\mu_2d\mu_3\\ &=\int \int\int \mathbb{1}(x_1 + x_2 \in A_{12})\mathbb{1}(x_3\in A_3)d\mu_1 d\mu_2 d\mu_3\\ &=\int\int\mathbb{1}(x_1 + x_2 \in A_{12}) d\mu_1 d\mu_2 \int \mathbb{1}(x_3\in A_3)d\mu_3\\ &=\mathbb{P}(X_1+X_2\in A_{12}) \mathbb{P}(X_3\in A_3) \end{align*}

So yes $X_1+X_2$ and $X_3$ are independent.


Let's try without the Lebesgue integration, we can show that $Y=(X_1, X_2)$ is independent of $X_3$ since for any Borel mesurable sets $A_1$, $A_2$, $A_3$ \begin{align*} \mathbb{P}(Y\in A_1\times A_2,X_3\in A_3)&=\mathbb{P}(X_1\in A_1, X_2\in A_2, X_3 \in A_3)\\ &=\mathbb{P}(X_1\in A_1, X_2\in A_2) \mathbb{P}(X_3\in A_3)\\ &=\mathbb{P}(Y\in A_1\times A_2) \mathbb{P}(X_3\in A_3) \end{align*}

Now let $f$ be any deterministic function over the domain of $Y$ and $f^{-1}(Z)=\lbrace y | f(y)\in Z\rbrace$, then for Borel measurable sets $A_0$, $A_3$ \begin{align*} \mathbb{P}(f(Y)\in A_0,X_3\in A_3)&=\mathbb{P}(Y\in f^{-1}(A_0),X_3\in A_3)\\ &=\mathbb{P}(Y\in f^{-1}(A_0))\mathbb{P}(X_3\in A_3)\\ &=\mathbb{P}(f(Y)\in A_0)\mathbb{P}(X_3\in A_3)\\ \end{align*}

So $f(Y)$ is independent of $X_3$.

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  • $\begingroup$ Thanks. So the answer uses lebesgue integration? $\endgroup$ – user370967 Sep 29 '18 at 11:14
  • $\begingroup$ Well, it can. The Lebesgue integration permits to see the separation. Another way of justifying it intuitively would be to say that since $X_1$, $X_2$ and $X_3$ are independents, then $Y=(X_1, X_2)$ is independent of $X_3$, then any deterministic function of $Y$ preserves independence. $\endgroup$ – P. Quinton Sep 29 '18 at 11:16

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