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We know that $Y$ is independent of a vector $X$ means it's independent of all linear combinations of components of $X$. Then I'm curious about the correctness of the following statement:

If $Y$ is independent of $X_{1}$ and $X_{2}$, $Y$ is also independent of $X_{1}+X_{2}$.

If this is true, does it mean if we know $Y$ is independent marginally of all components of $X$, then we'll come to $Y$ is independent of $X$?

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    $\begingroup$ It depends on whether you read "$Y$ is independent of $X_{1}$ and $X_{2}$" as (a) "$Y$ is independent of $X_{1}$ and $Y$ is independent of $X_{2}$" so pairwise independent rather than as (b) "$Y$ is independent of $X=(X_{1},X_{2})$" which would be jointly independent. With joint independence, $Y$ would also be independent of $X_1+X_2$ $\endgroup$
    – Henry
    Commented Jun 18, 2021 at 15:02
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    $\begingroup$ Your initial characterization of independence is misleading: instead of "means" you should state "implies." $\endgroup$
    – whuber ♦
    Commented Jun 23, 2021 at 17:33
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    $\begingroup$ You should state X1 is independent from X2 else trivial solutions like X1 = -X2; Y = X1 + X2 show up. $\endgroup$
    – Vorac
    Commented Aug 1, 2021 at 16:51

1 Answer 1

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Somewhat surprisingly, this is not necessarily true. For example, consider the joint probability distribution described by the following table.

$Y$ $X_1$ $X_2$ prob
0 1 0 0.25
1 1 1 0.25
1 0 0 0.25
0 0 1 0.25

Each of the three random variables follows marginally a Bernoulli($0.5$) distribution, and it is very easy to confirm that $Y \perp X_1$ and $Y \perp X_2$. However, consider that the probability $$P(Y=1, \ X_1 + X_2 = 1) = 0$$ while on the other hand $$P(Y=1)P(X_1+X_2 = 1) = \frac{1}{2}\times\frac{1}{2} = 0.25.$$ Thus, $Y$ is not independent of $X_1 + X_2$.

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    $\begingroup$ This answer implicitly supposes "independent of $X_1$ and $X_2$" means separately independent of each variable. Many readers will understand this phrase instead to be equivalent to "independent of $(X_1,X_2)$"--but in this latter sense, the conclusion is true, as shown at stats.stackexchange.com/questions/94872. Your example is one in which $Y$ is a function of $(X_1,X_2)$ (with no random error). $\endgroup$
    – whuber ♦
    Commented Jun 23, 2021 at 17:33
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    $\begingroup$ @whuber, I understand your point (which is also captured in a highly upvoted comment on the question). In my opinion, the final sentence of the OPs question makes it clear that this is the intention ("marginally independent"). $\endgroup$
    – knrumsey
    Commented Jun 23, 2021 at 18:18
  • $\begingroup$ Thank you for pointing out the "marginally" qualifier. $\endgroup$
    – whuber ♦
    Commented Jun 23, 2021 at 18:49

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