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If R.V. A is independent of a random vector (B,C), is A necessarily independent of C?

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Write down the definition of independence, and you will successfully formalize a result quite intuitively obvious (if you are independent of what happens to $(B,C)$, you are independent of what happens to $C$). –  Davide Giraudo Oct 22 '12 at 13:03
    
By the way, welcome to math.stackexchange. –  Davide Giraudo Oct 22 '12 at 13:04
    
Thank you, I integrated both sides of the definition over variable B to get the result. –  Kevin Oct 22 '12 at 15:01
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You can answer your own question. –  Davide Giraudo Oct 22 '12 at 15:01

2 Answers 2

By independence of a and vector (b,c), we have $$f(a,b,c)=f(a)f(b,c)$$ So, integrating out b, $$\int_{-\infty}^{\infty} f(a,b,c)\,db = \int_{-\infty}^{\infty} f(a)f(b,c)\,db \implies f(a,c)=f(a)f(c) $$

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If you go back to the definition of independence, you can prove this statement without assuming the joint distribution of $(A,B,C)$ to be continuous.

$A$ and $(B,C)$ are independent if every $A$-measurable event is independent of every $(B,C)$-measurable event. Since every $C$-measurable event is $(B,C)$-measurable, conclude that every $A$-measurable event is independent of every $C$-measurable event. That is, $A$ and $C$ are independent.

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