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This is question 3.8.4 of An Introduction to Mathematical Statistics and Its Applications, 5th Edition, by Larsen and Marx. This is not homework for a class I am taking now, but might someday be for a class I'll take in the future.

The goal could be expressed in terms of probabilities, to try and show that for every interval A and B, P(V $\in$ A and X+Y $\in$ B) = P(V $\in$ A) $\centerdot$ P(X+Y \in B).

But, I've been trying to answer the question in terms of pdf's -- to show that f$_{V,X+Y}$(v,w) = f$_V$(v) $\centerdot$ f$_{X+Y}$(w), for W = X + Y, without success.

The other thing I was considering was the equivalence of f$_{V+(X+Y)}$(w), f$_{(V+X)+Y)}$(w), and f$_{(V+Y)+X}$(w), for W = X + Y + V.

Is this the wrong tack? Is there some easy way of proving this, say, using graph theory, with random variables as nodes and dependency relationships as edges?

Thanks in advance for any hints / help.

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A measure-theoretic proof seems the easiest:

$V \perp X,Y \Longleftrightarrow \sigma(V) \perp \sigma(X,Y)$, where $\sigma(Z)$ is the sigma-field generated by the random variable $Z$.

As $\sigma(X+Y) \subset \sigma(X,Y)$, you have $\sigma(V) \perp \sigma(X+Y)$. As whuber points out, independence of random variables is equivalent to independence of their generated sigma-fields, so you are done.

I'm also including the measure theoretic definition of independence of random variables. Let $(\Omega,\mathcal{F},\mathcal{P})$ denote a probability space.

Definition: Random variables $X_1,X_2,\ldots$ are independent iff the corresponding $\sigma$-fields $\sigma(X_1),\sigma(X_2),\ldots$ are independent.

Note that using a few other results, it can be shown that the following more common understanding of independence of random variables can be derived:

Corollary: Two random variables $X$ and $Y$ are independent iff for any $x,y \in \mathbb{R}$ $P(X \le x,Y \le y)=P(X \le x)P(Y \le y)$

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  • $\begingroup$ +1 for an excellent idea. I would encourage you, though, to revisit the notation on the last line, which although suggestive is clearly false (and irrelevant): since $V$ is a random variable but $\sigma(V)$ is a sigma algebra, it cannot possibly be the case that $V\in\sigma(V)$. Likewise the second part of the statement is false. I believe you intended something like $\sigma(X+Y)\subset\sigma(X)\cap\sigma(Y)\subset\sigma(X)\cup\sigma(Y)$. Since independence is defined in terms of elements of sigma fields, a fortiori $V\perp X+Y.$ $\endgroup$
    – whuber ♦
    Jun 28, 2014 at 14:20
  • $\begingroup$ @whuber: Yes you are absolutely right I realized later on that it was an abuse of notation. $\endgroup$
    – ved
    Jun 28, 2014 at 14:25
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    $\begingroup$ Were you aware that you can edit your answer? In fact, we encourage that: one thing that distinguishes this site from others is our drive to improve the answers (and the questions). By custom, we tend to refrain from making direct edits to other people's answers (although that is permissible), preferring instead only to suggest how the answers could be improved. After all, many answers are technical--like this one--and it's easy for readers (like myself) to misunderstand or misinterpret something that turns out to be perfectly correct. $\endgroup$
    – whuber ♦
    Jun 28, 2014 at 14:31
  • $\begingroup$ @whuber: Is the above edit fine? $\endgroup$
    – ved
    Jun 28, 2014 at 14:37
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    $\begingroup$ It sure is. I do have another suggestion, though, which you can take or leave as you wish. Many people seem to think that this measure-theoretic formulation and understanding of basic probability concepts is "sophisticated" or "abstract." As such, there is potentially a mismatch between the knowledge exhibited in the question and the knowledge required to understand the answer. You can help bridge that gap by quoting the definition of independence formulated in terms of sigma fields and showing how simple and understandable it is. That would make your answer more complete and relevant. $\endgroup$
    – whuber ♦
    Jun 28, 2014 at 14:42

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