# If X, Y, Z are independent random variables, then X + Y, Z are independent random variables. [duplicate]

This question already has an answer here:

I found the same question (X,Y,Z are mutually independent random variables. Is X and Y+Z independent? here), but the answer uses characteristic functions and fourier inversion theorem, but this is exercise in chapter long before characteristic functions.

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Note that $X, Y, Z$ are mutually independent if and only if $$\Pr\{X \leq x, Y \leq y, Z \leq z\} = \Pr\{X \leq x\}\Pr\{Y \leq y\}\Pr\{Z \leq z\}$$ for any $x, y, z \in \mathbb{R}$.
Now, for any $w, z \in \mathbb{R}$
\begin{align}\Pr\{X + Y \leq w, Z \leq z\} & = \int_{-\infty}^{+\infty} \Pr\{X + Y \leq w, Z \leq z|X = x\}dF_X(x) \\ & = \int_{-\infty}^{+\infty} \Pr\{x + Y \leq w, Z \leq z\}dF_X(x) \\ & = \int_{-\infty}^{+\infty} \Pr\{x + Y \leq w\}\Pr\{Z \leq z\}dF_X(x) \\ & = \Pr\{Z \leq z\}\int_{-\infty}^{+\infty} \Pr\{x + Y \leq w\}dF_X(x) \\ & = \Pr\{Z \leq z\}\Pr\{X + Y \leq w\} \\ \end{align} where the first and the fifth equalities are using the law of total probability, the second and the third equalities are using the given mutual independence, and the fourth equality is pulling out the term that is independent of the integrating variable $x$. Here $F_X(x) = \Pr\{X \leq x\}$ is the CDF of $X$.